How can water evaporate at room temperature?

Question: Boiling point of water is 100 degree Celsius. The temperature at which water in liquid form is converted into gaseous form. Then how it possible for water to evaporate at room temperature?
Comment: The water molecules at the surface of the water don't need as much energy to evaporate.
Comment: Boiling is different than simple evaporation.. Water vapours can exist at zero degrees Celsius.
Comment: Possible duplicate of: http://physics.stackexchange.com/q/10470/
Comment: pressure can be low enough for water to evap at room temperature.
Answer: Think of temperature as average kinetic energy of the water molecules. While the average molecule doesn't have enough energy to break the inter-molecular bonds, a non-average molecule does.
Water is a liquid because the dipole attraction between polar water molecules makes them stick together. At standard atmospheric pressure (acting somewhat like a vice), you need a comparatively large temperature of 100°C (translating to a high average energy distributed among the micsroscopic degrees of freedom, most relevantly the kinetic ones) for water molecules to break free in bulk, creating bubbles of water vapour within the liquid.
However, at the surface of the liquid, lone molecules may end up getting enough kinetic energy to break free due to the random nature of molecular motion at basically any temperature. On the flip side, water molecules in the atmosphere may enter the liquid at the surface as well, which is measured by equilibrium vapour pressure.
Comment: @JohnRennie: I expanded the answer and made it community wiki - feel free to improve upon it as yous ee fit...
Comment: Got a spark but not clear enough. like @JohnRennie said.
Comment: This is basically the answer I would give. Christoph, would you consider fleshing this out a bit? At the moment it's so brief only people who already know the answer will be able to see what you're getting at.
Comment: Is the "flip side" the reason why evaporation can't be considered a Maxwell's demon of sorts?
Answer: Imagine spinning a roulette wheel, but instead of dropping in one ball, you drop in 100. They all rattle around at different speeds, like the molecules in water. You can cool them down by spinning the wheel slower, so they bounce about less; heat them up by spinning faster so they bounce more; you can freeze them by stopping the wheel and waiting till they're all stationary; and you can boil them by spinning the wheel so fast that they all fly out of the top.
Now pick up all the balls and throw them back in with the wheel spinning at a moderate speed. If you watch for a while you'll see that although the average speed of the balls is below the "boiling point" where they all fly out the top, every now and again one ball will ricochet off another with enough force to send it flying out of the wheel. If you watch for long enough eventually all the balls will be gone. Your balls just evaporated.
Comment: good thinking. This analogy really helps.
Comment: This analogy should be summed up and added to the most visible one. It basically explains that's because sometimes molecules ricochet off other molecules. Thanks for the example.
Answer: Temperature is a measure for how much kinetic energy the molecules in a substance have. If the temperature is high, they are moving pretty fast, if the temperature is low, they are moving a lot slower. If molecules are moving slow, they bundle up and you get a solid. Once you heat it up a bit, the substance starts to become liquid. When you heat it up even more, the molecules will start to move so fast they will spread out into the entire space (gas).
However, this is all averages. In a liquid all molecules are moving, some faster than others. If a molecule happens to break through the 'surface' of the water, it'll have escaped the inter-molecular forces holding the water together and it'll be evaporated. This can also happen with solids, there it is called sublimation.
If you're heating up water, you're adding energy so this process will start to go faster. Then at boiling point, you'll reach the point where molecules will want to start moving so fast they start to form gas bubbles inside the liquid.
disclaimer: this is just what I remember from high school.
Comment: Good memory there.
Answer: The boiling temperature of a liquid is not the temperature at which it can enter the gaseous state.  Rather, it is the temperature at which the saturation vapor pressure $e_s$ is equal to ambient atmospheric pressure.  This is why, for example, water boils at lower temperatures at higher altitudes.
Furthermore, water is always evaporating.  It is also always condensing.  You can picture a cup of liquid sitting in a room.  The evaporation rate will be driven by the saturation vapor pressure $e_s$ calculated with the temperature of the liquid water.  The condensation rate will be determined by the vapor pressure $e$ of the water vapor in the air.  Typically when someone says a liquid is evaporating (or condensing) they are referring to net evaporation.  Net evaporation is when evaporation exceeds condensation (liquid is decreasing) and net condensation is when condensation exceeds evaporation (liquid is increasing).
You will typically hear humidity expressed as "relative humidity", which is the ratio of vapor pressure $e$ to saturation vapor pressure $e_s$ and thus when a the relative humidity $\dfrac{e}{e_s} < 1$, net evaporation is occurring and when $\dfrac{e}{e_s} > 1$ net condensation is occurring.  In both cases however, evaporation is constantly occurring.
Answer: At the boiling point the gas is produced inside the liquid, but at the surface you constantly have molecules going in and out. If the environment is kept quite dry, then few molecules will come back in with respect to the ones that leave. Off course the higher the temperature, the easier will be for a molecule to get enough energy to break free, but this can happen at any temperature - at the surface.
Answer: It can also be understud by the idea of partial pressure. Water will evaporate in an atmosphere until its partial pressure has reached the vapour pressure given for the ambient temperarture (relative humidity of 100%).
Answer: These answers account for the kinetics of the process.  Thermodynamics provides an alternate picture, well suited for any question involving phase transitions. For the system of liquid plus empty volume, the free energy can be lowered by trading some enthalpy (to take molecules from the liquid into the  gas phase) for the increase in entropy (all the states available in the gas phase).  For a given temperature, this balance determines the equilibrium vapor pressure. There will always be some vapor in equilibrium (at all non-zero temperatures).
Answer: All water moleucles contain energy, in accordance with the temperature.  Hot water has enough energy to escape the liquid as vapour.
Even though a body of water is below boiling point, the molecules with more energy (relative to the body of water),rise to the top and can escape, as vapour
Tis easy!
Answer: In Science class, my teacher put a flask of water on his worktable, covered it with a glass dome with a rubber rim on it so as to seal the dome to the desk. He then hooked a hose to the dome, the other end to a vacuum device and proceeded to remove the air from the dome creating a very low pressure zone within the dome... as the pressure dropped, the water began to boil - at room temperature.  This is why mountaineers have difficulty getting hot water at higher altitudes.
Comment: Welcome to phys SE! Your information is interesting and relevant, but doesn't quite address the question.


What was the reason for some plant and animals to become giant in course of evolution?

Question: The dinosaurs, mammoths, giant plants etc are known to be bigger than modern animals. I wonder why they had been lived and why they are not living now? I really don't know much but is it something about oxygen balance or something similar?
Comment: But note that the largest animal ever to have lived (blue whale) is alive right now!
Comment: The wording of your question is somewhat obscure. Do you mean "what was the reason for some plant and animals to become giant in course of evolution?" or probably "What are the advances and drawbacks of being a giant plant or animal organism?"
Comment: Sorry for my bad English. I mean "what was the reason for some plant and animals to become giant in course of evolution?"
Comment: Your English is not at all bad, I just wasn't sure what you mean and wanted to ask before I correct the title of your question. My native language is Russian, by the way, so don't take my English for "granted"  :-)
Comment: You might find the following [question](http://biology.stackexchange.com/questions/111/do-large-animals-often-evolve-into-smaller-animals) helpful, but (replying to the duplicate flag) I don't think that these questions are exact duplicates...
Comment: consider the current era is also a time of giants, the largest whales, the largest cephalopods, and the largest trees all all alive right now. Basically you can find something giant in every era.
Answer: Thanks for asking an interesting question which made me think.
The short answer is that something evolves if there is an advantage to the genes involved, and, by 'advantage' I mean it produces more copies of the genes in the next generation so more individuals with that characteristic will be present in the population.
As to what particular advantage increased body mass had to any particular species that would depend on the particular species and its environment at the time.  There is almost certainly not a 'one-fits-all' answer.
Most very big land mammals are herbivores and there is an advantage in having a large digestive tract when you eat a lot of plant material with a fairly low nutrient level.  However, when there are also predators around, getting big and slow might be more of a disadvantage than staying small and fast. With evolution there is often a balance of competing forces which are traded of against one another.
Where large mammals have become isolated on islands with no predators, such as a species of now-extinct elephant on the Flores Island, they tend to become smaller so this suggests avoiding predators by simply being too big for them to kill might be an important factor.
And of course with many species we have the effect of sex-selection. If females select for the biggest males, there will be a selection pressure towards greater size.
With plants it could be something else entirely. Trees, for example, have long trunks basically for one of two reasons.
Firstly to put their leaves and seeds out of the reach of herbivores, so we have a kind of arms race like that between giraffes and acacia trees in Africa.
Secondly, when they grow in close clumps like woods and forests, there will be competition for sunlight and soil resources like water. A species which can reach higher and put it's leaves above those of it's neighbours will win the competition and there will also then be pressure on it's neighbours to evolve longer trunks.  Long trunks mean that water must be pumped a long way to the leaves so the tree with the longest trunks will also need bigger root systems.  Again a trade off between making a long trunk and dominating the water supply underground.
Many trees are also believed to have formed symbiotic associations with fungi in their root systems which help them take in the nutrients and water required to 'service' a long trunk and high-up leaf crown so they may be the only types of plants capable of doing this anyway.
One thing you might like to think about is why some whales are very big and some are relatively small (dolphins, porpoises, killer whales). What is the difference in their life-style, food, habitat, and so on, which could drive these evolutionary differences.
Comment: Getting big is not necessarily disavantageous when there are carnivours arounds. If you are big enough there needs to be a big number of predators to bother you, especially if you travel in big groups (think about elephants, for instance).
Comment: another big advantage is resource and environmental fluctuation , a big animal can survive sudden changes in the environment or resource availability better than small ones, https://pubmed.ncbi.nlm.nih.gov/16701331/
Answer: to expand on Rose's answer, certain environmental factors are what people cite with unusual size changes in an animal over time as it diversifies into new species.
Gigantism (and dwarfism) on islands are a particular example.  In this case an animal may arrive on an island and some of the predators or competitors that it is used to will not be there.  Then a single species will spin off many species, each one will fill one of the niches that are unfilled on the island - eating bugs from bark, breaking open nuts.  The lack of large carnivores will allow a nice juicy herbivore to get larger without fear of being eaten.  Similarly the Moa was a bird that became an aggressive carnivore in its island habitat, where it's very rare to see an aggressive land bird when a big cat or other carnivore is already present.  Such large changes in environment can also result in dwarfism - smaller species as smaller animals like squirrels or mice might not be present to consume all the food that are available.  The Hobbits recently discovered in Indonesia are an example of island dwarfism.
That being said, selective pressure can probably make many animals larger or smaller - lots of the dog breeds are smaller than wolves that they descended from.  I think steers are as large as we can make them and are larger than native cattle stock.
As far as the environment that caused dinosaurs to be so large, the higher levels of oxygen on the planet at the time is a common theory for why they were so large.  I also think that at least early on, the selection conditions that favored larger animals were probably put in place as animals evolved without any competitors whatsoever (first animals on land or sea).  If this were to happen again, its not clear whether dinosaurs would have been so large - perhaps we could have had gigantic mammals or descendants of trilobites.
Answer: It needs to be recognized that this is an outstanding question in biology, i.e., there is no certain answer.
This review considers the various theoretical and empirical evidence for phenotypic selection on body size:

Blanckenhorn WU (2000) The Evolution of Body Size: What Keeps
Organisms Small? Quarterly Review of Biology 75: 385-407.

Some have hypothesized a connection between body size and the evolution of endothermy:

Clarke A, Portner HO (2010) Temperature, metabolic power and the
evolution of endothermy. Biological Reviews 85: 703-727.

And others have suggested a link to oxygen in the atmosphere:

Harrison JF, Kaiser A, VandenBrooks JM (2010) Atmospheric oxygen
level and the evolution of insect body size. Proceedings of the Royal
Society B: Biological Sciences 277: 1937.

And I'll throw this hypothesis into the ring.  The large dinosaurs only went extinct 65 million years ago.  That's basically yesterday in terms of the geological time scales across which body size may evolve - they (dinosaurs) originally evolved 230 million years ago.  That means we've got at least 165 million years to go before you can see if our current diversity of life on earth can evolve similarly large organisms!
Answer: One significant point not considered in previous answers is the reproductive rate and environmental variability.  Larger animals have longer times between generations and generally reproduce less rapidly.  In a less stable environment, more rapid reproduction allows more rapidly adapting population to the availability of resources, providing a larger population so that random events are less likely to eliminate the entire species (genetic variation can be a factor here), and minimizes the absolute amount of resources needed for a stable population.
There are significant benefits to being larger.  In addition to interactions with other animals, a larger animal has lower surface area to mass ratio and so will lose less heat per unit mass (which might have been significant for ice-age mammals?).  A larger animal can also support larger organs (not just digestive tract as mentioned by Rosa Rubicondior, large brains would not be possible in small animals) or substantial-sized specialized features (where utility is not a linear function of size).


How to clean grain bags?

Question: I did a partial mash tonight, and I want to reuse my grain bag for future uses. I tried to rinse it out, but pieces of grain are still in there. What do you guys do to clean it out fully? Throw it in the wash?
Comment: What type of bag are you using, nylon?
Comment: Sorry, yes nylon mesh.
Answer: My process usually looks like this:

Empty as much of the bag out into the trash as I can
Rinse the bag under running water, getting off whatever bits I can from the outside
Turn the bag inside-out
Continue to rinse under running water, while mashing and rubbing the bag together in my hands

I find that the last step does a pretty good job, and the rubbing makes it go so much faster than just trying to "push" the bits of grain off with the sprayer.
Answer: I neeeeever quite get all the little bits unstuck.  Fortunately brewing is rather forgiving.  Any little bits that still make it will just end up going along for the ride in the next batch.  The wort is still being boiled, so any baddies hanging out in the few specks of grain aren't going to make it through.  Dunk it in star-san just beforehand if you want to feel extra warm and fuzzy.  The little bits of husk left aren't going to leech any noticable amount of tannin or anything.  Just make sure you clean everything you can readily get, and RDWHAHB about the rest.
Answer: After my first partial mash I just let my nylon bag sit in a bowl of hot water for about 15 minutes and then just rinsed the bag really well with the sprayer attachment on my faucet.  All the color and bits of grain came right out.  I think it also helped that I turned the bag inside out to clean it as most of what was stuck to the bag was on the inside.

How to clean grain bags?


What measures are commonly used for the complexity of an organism?

Question: I'm aware of measures like number of distinct cell types being used as a measurement of complexity in biology, for example in the G-value paradox. But this doesn't really help for unicellular organisms. Is it possible to define a unit of complexity to make comparisons between different organisms?
Answer: This is a very interesting question. Not sure if there is any way to answer without specifying a definition for complexity.
It is clear that everything is made of only the same few elements, so complexity must involve more than that. Do the number and position of identical molecules matter? If so, size matters and we would label a large jelly-fish as more complex than a cat and the air in a large hot-air ballon as more complex than any single living organism (assuming it is bigger).
I think it is clear that this is not what is meant by biological complexity. Molecules acting identically should not count multiple times; the number of different molecules matters more than the mere number of molecules. But are immune cells really more complex simply because of a relatively minor difference that allows them to generate a wide variety of genetic variants?
What about the variable coat color of a calico cat. Does the complex and non-repeatable (even with cloning) pattern mean it is complex, or does the fact that a simple genetic process underlies it mean it is simple.
The difference in complexity of different representation of the same thing is a classic problem in information theory (see Kolmogorov complexity). The ratio of the circumference of a circle to its diameter requires an infinite number of decimal digits to describe, but it is just a single geometric concept and it can be represented with one or two characters, as pi or π. This applies to the problem of using phenotypes to measure complexity -- the language and definitions we use to describe or to differentiate phenotypes determines what is complex and what is simple.
Moving beyond a simple static snapshot case, biological complexity involves time-dependent evolution. The "game of life" demonstrates how simple rules + initial conditions yield apparent complex behaviors, and the beauty of fractals arises the same way. Many apparently complex organisms grow from small seeds plus time-evolved exposure to simple things like water and light and CO2. Are trees only as complex as the seed they grow from?
Finally, what about higher order complexity? One human brain is very similar in general makeup to another, but the concepts and memories stored seem complex and very different. Comparing the complexity of stored information between two people is the whole complexity problem over again.
It seems to me that complexity can only be cleanly measured within a specific context, and then only relatively. One can measure the relative genomic complexity of two organism as the edit distance between the alphabetic sequences representing their genomes. The relative regulatory complexity can be measured as mentioned above by @user1682. Each of these says something about an organisms complexity. But what if organism A has higher genomic complexity and organism B has higher regulatory complexity? It might be possible to derive something like a principle component "basis set" from several of these kinds of measurements using a training set and a list of organisms with predefined complexity values. Applying this "standard" scoring metric to another organism would allow ranking it on a "standard" scale, but it would be difficult to generate and obtain consensus on the initial "expert" ranked organism scale. Such a standard might have some value, but is not really theoretically satisfying in that it describe an organisms complexity in the same way that the average of a small sample describes the distribution of a large population.
Comment: I think your answer raises a few deficencies in the question, so I think I'll have to address them in another question. Thanks.
Answer: I don't know if it is helpful for your problem, but I measured complexity in gen regulatory networks by parsing them into boolean networks. Then you can initiate all nodes with a random boolean value and update the network until it reaches an attractor. For a hand full of initial states you can compute the average entropy and mutual information. Additionally the topology (scale free, small world [Barabasi]) and count of reached and cycle lengths as well as robustness (reaching same attractor by flipping a randomly chosen node in the trajectory) can give you information about the complexity. To compute all attractors by all possible initial states of the boolean network can be done with a sat solver [Dubrova].
regards
Comment: The OP is asking about _organism_ complexity, not gene network complexity. One can argue that the two are connected but this is not an answer to the question.
Comment: Any references to published work defining or estimating these complexity measures for gene networks or the like?


Fermentation - only very small bubbles after 36 hours

Question: This is my first batch so I'm wondering if I am worrying over nothing or if I need to take action.  I brewed an all extract IPA kit from Williams Brewing.  I feel I was very careful with sanitation and pitched a liquid yeast pack just below 80 degrees.  I used an ice bath and was able to get the wort to that temp in just shy of an hour.  I am using a bucket and it appears that all the seals are good.  I did start the yeast packet and let it swell, but I did not have time to brew until close to three days after the pack was fully swollen.  I am getting very tiny bubbles in my airlock, but nothing that appears to really be moving and definitely not "vigorous" fermentation.  Do I need to pitch more yeast or should I wait it out?
Comment: I forgot to add that I am using a dark closet on the main floor of my home which has maintained a constant 64 to 65 degrees the entire time.
Answer: I don't think you have anything to worry about yet.
If you can see inside your bucket or carboy (without opening) you can see if the krausen has formed yet, and how thick it is. The krausen looks like this:

When it starts, it will just be a faint ring, then the entire surface is covered, and then it gets thicker.
Only when krausen has formed is fermentation really underway at it's most vigorous.
The quantities liquid yeast provided by the major manufacturers results in a slight underpitching, so there can be some lag time there. Also, 64F is on the cooler side, so things will proceed at a more moderate pace. (You'll also get cleaner tasting beer.)
With all that considered, 36 hours is still quite early.
Just wait it out - I'm sure you'll see it pick up in a day or so.
Answer: Given the fact that your yeast packet was activated and then left for 3 days, it's likely the yeast just went dormant again after consuming the nutrients in the packet (which caused all that swelling), which would account for a lag time in fermentation.
Wyeast's FAQs address a similar possible issue - pitching a yeast packet without activating it. Their answer is that the yeast is still perfectly viable, but to expect a possible lag time in fermentation.
As others have said, just be patient and have faith.
Answer: Welcome new brewer, you have asked every first time brewers favorite question. I would say just relax, you probably did nothing wrong, airlock bubbles are not a good sign of fermentation. But worrying during your first brew is part of the experience. If you don't have a hydrometer yet get one and measure the gravity is about a week. Google a bit for your brew's expected Final Gravity reading and see if you are coming close.
Cheers


collision resistant summarizer for long hash values

Question: how we can summarize SHA-256 into shorter values when we for example can't handle more than 128 bits on system? its better just use first 128bit and ignore the rest of bits or split full value and XOR them? or use another hash function that produce required bit long hashes to generate a hash from SHA-256's value (e.g SHA-1)? which way is more robust and unpredictable?
Answer: The answer is "any sane summarization function is about as good as any other; pick whichever is convenient"
To the best of our knowledge, SHA-256 acts pretty much like a random function (except for the length extension attack; that wouldn't apply here)
So, the output of SHA-256 is essentially a random 256 bit number; so, if we have a 128 bit hash function:

Hash(M) = Summarize(SHA256(M))

then, as long as the Summarize function has $2^{128}$ preimages (of 256 bits) for each possible output (of 128 bits), it should be good.  Both the "use the first 128 bits" and "split the full value and XOR them" functions have this property.
Now, since you talk about collision resistance, you can find a collision with a 128 bit hash function with about $O(2^{64})$ effort, by the simple expedient of generating $2^{64}$ random strings and hashing them, and then looking for two hashes with the same value (and, in case you're hoping the memory requirements of this would save you, there are clever ways to reduce the memory requirements drastically without requiring that much more computation time).  As long as those random strings are longer than 128 bits, there's nothing a hash function can do to avoid a collision.  If you need stronger collision resistance than that, you need to rethink the problem.
Also, since I mentioned the length extension attack, I'll summarize it here:
If you hash the value $SHA256(M)$, and the length of $M$, then you can compute a string $A$ such that, for any string $B$, you can compute the value:

SHA256(M | A | B)

You can compute this even though you know nothing more about $M$.
This wouldn't apply in this specific case, because for this to work, you need to know the entire $SHA256(M)$ output, and your summarize function will drop a lot (precisely 128 bits, actually) of that information.
Comment: Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/63526/discussion-on-answer-by-poncho-collision-resistant-summarizer-for-long-hash-valu).


Can I make a podcast out of any flash video to listen to on my iPod?

Question: Is it possible to make a podcast out of any flash video on the web that I can listen to on my iPod?
Comment: Why do you want it to be a podcast?
Answer: Maybe.
One of the reasons content providers choose Flash is that it makes grabbing the video for your own purposes nontrivial. Some sites provide links for downloading video, even if the embedded player is Flash. But I don't think that's what you're asking.
RealPlayer (http://www.real.com/realplayer/download) provides a free player as well as a program that will run in the background to try to grab Flash video. They also have a free converter program (RealPlayer Converter) for transcoding, if what you capture isn't something your iPod can play. If it's an iPod Touch, then you can play H.264/AAC or H.264/MP3 movies directly. Other formats you'd have to convert.
If you are talking an audio iPod, you need to pull the audio out of whatever movie you capture. I would do that using QuickTime 7 Pro, a $29 purchase. There are free alternatives, I'm sure. But taking just the audio from a Flash video is a bit obscure.


Text effect on any background color

Question: I want to display a text over a video. However, the video may include any color from white to black. What is the best text effect (color/stroke/shadow/glow ... etc) that can be applied on the text so that it will be very clear on any background color.
Comment: Sometimes unreadable text can be a feature: https://www.youtube.com/watch?v=SjnWOnJK-zw
Answer: The classic internet answer, at least for static images (and I imagine it would work for video as well), is "White text with black outline can be read on any color."
Answer: Another method would be to put an semi transparent background in white or black - depending on your text colour behind the text
although you may need to change the transparency of the background depending on the video's contrast
Answer: Instead you can try this method by adding following effects

Thin black outline
Heavy grey line
Hard black drop shadow
And soft grey shadow.

It appear very clear and readable
Answer: You need to make the background immediately behind your text contrast with your text color.
A very wide subtle soft white glow or spotlight will lighten the background adding contrast to make your black text (making it pop), same goes for a black glow behind white text (or any pair of contrasting colors). This method allows you to have a crisp/clean/sharp and minimal text style, however you can always add a dropshadow or stroke if you like to bring the text forward.
Personally I think adding a little bit of many effects looks better than a hard stroke or really dark dropshadow (unless you are doing for that more aggressive kind of look).
Answer: it's depend on Video subject, for example in a long time movie may choosing mostly white text with shadow is best choose (because of eye tiredness), but in a sport video choosing yellow text with shadow is best choose (because most sports have green field).
but in most case i prefer "mostly white text" and a "little amount of shadow" beneath it.

Text effect on any background color


Xray compression of secondary in hydrogen bomb

Question: Accounts of the "secret" of the hydrogen bomb describe  Xrays from a primary fission explosion reflecting off of the bomb case (occasionally passing through polystyrene foam) and compressing and heating the fusion fuel. No mention is made of gamma rays doing the same thing. Is there a physical explanation for why X-rays are able to compress the secondary, but gamma rays are not? Does the primary fission reaction produce gamma rays as well as x-rays?
Answer: The X-ray radiation is thermal i.e. it's generated by the very high temperatures as the primary goes off. The temperatures in an atomic bomb aren't high enough to generate thermal gamma rays. Presumably there must be some gamma radiation from nuclear reactions in the detonating primary, but this is small compared to the X-ray intensity.
Answer: In addition to John Rennie's answer, which is correct, gamma rays have a wavelength smaller than an atomic radius, and generally go through matter a ways inside before scattering or getting absorbed. To get an implosion of the secondary, you want the material on the surface of the secondary fusion clump to ablate away. This means it has to shoot out atoms from the surface very fast. This process only works if the rays shining down on it are absorbed at the surface, not deep inside. As explained on Wikipedia, the ablation pressure is the major process for compressing the secondary.


Advice on doing mathematical research

Question: Please share any general tips or advice you have on doing mathematical research.
How do you identify good problems to work on or to think about? What do you do when you get stuck on a problem? Etc.
Answer: My 2c worth of personal perspective:
On problem selection - let the problems select themselves. In other words, when you have nothing to think about, talk to as many people as you can and read as much as you can in and out of your area of interest. Sooner or later a problem will emerge that doesn't let you go. It may not be an explicitly conscious choice but it will happen, and you'll just find yourself trying to solve it without ever having decided to do so. Realize that, and work on that problem.
On being stuck: One of the things that helped me most consistently is (my version of) Polya's dictum: inside every problem you can't solve lies a smaller problem you also can't solve. Find it!
Answer: One thing I find incredibly useful is the MathSciNet reviews. Often the reviews make the paper more clear. If the paper doesn't address the big picture, the review will often give context or, if the result fits into a larger program of research it will mention this. Sometimes the review highlights unique technical arguments.
Also, everything is hyperlinked, in that you can pull up all articles referenced by a paper or, better, all articles (or reviews) that reference a particular paper (this has often been useful when I didn't understand something in a paper or didn't know the point of a paper). I spent a lot of time reading reviews as a beginning grad student and it helped me get a sense of what questions people find interesting.
Comment: Is there a way to search for reviews directly? (rather than going through a paper that it reviews)
Answer: Some of the best advice comes from Richard Hamming's talk You and Your Research.
Answer: Terry Tao has some helpful ideas.
I can't say I've really mastered the practice of choosing problems.  I guess if there's an area of math that you're really familiar with, questions will just start coming to you, but I can't tell how much of that ability comes from some osmotic process from the examples of peers.
When I'm stuck, I try to write down where I am in a lot of detail, then I take a break.  If I'm lucky, I know what sort of knowledge might be able to crack my problem, and if I'm even luckier, I know who has that knowledge.
Answer: Another thing I do is go to a lot of conferences talk to a lot of people. Whenever people are idle I just ask them directly if some problem I know of is interesting and worth working on.
Comment: Just cold ask them for problems?
Answer: Often, I find that the good problems come from errors in my work -- I write a proof of something that (at the time) I think is trivial and later find a counter-example.  Suddenly, the problem is non-trivial and interesting.
When I get stuck on a problem?  I work on another problem, a related problem, a broader problem, a more specific problem, etc.  Anything to try to understand why it is difficult to solve this problem.
Answer: Some further advice and links can be found here.

Advice on doing mathematical research


How to plan routes for touring?

Question: I am a bit lost even to find the shortest safe way to the nearest city. How do you choose your routes for long journeys? Do you need to do compromises very often or change plans?
Comment: just found out that there are 3 different types of maps: topografical (shows the shapes of climbs etc), bicycle and and a general (shows high-ways etc). The book in my last reply (of a different post) mentioned that it is helpful to read different types of maps in different circumstances, sounds logical but not sure about practical issues, read my change history to find the references.
Answer: There's a variety of touring styles to consider. There are people who plot every single metre of the journey and then there are people who just ride until they run out of time or money. I assume there are people who do whatever seems interesting on the day or who just obey whatever directions they get from whoever they meet. I've kinda done all of those.
In outback Australia direction-finding is often very easy: there's one road, you follow it. After a few hundred kilometres you might reach a T intersection and have to make a decision (at Newman in WA, for instance). At the other extreme is navigating the old medieval parts of European cities where there's a maze of little corridor-sized streets.
My preference is to plan a route based on roughly 50km a day through an area with a reasonable density of interesting things. So if I fly into Christchurch, NZ (don't do that right now, ok) with a month to spend I might plan on riding down through Tekapo, out to the coast at Moeraki then back, for a nominal distance of about 1000km (30km/day). But when I actually ride I expect to find diversions - ride up to the Hermitage (Aoraki), through the Otago Rail Trail, and so on, probably riding more like 1500-2000km. My goal is that if I see a sign saying "historic dam 5km" I can ride up to see what that is. Or if someone says "you have to see the eccentric monolith" I can ride that way. It's a balance between planning and aimless wandering that suits me.
For CANC2 we spent a couple of months planning out where we'd ride, how long it will take (roughly) for each stage and what food and so on we'd need posted to each drop point. We also started contacting communities along the way to set up media and school visits, but that's pretty much what you expect for a political ride. I've never done a solo or friend-group ride with that level of planning.
The intense planning usually means looking not just at distance, but conditions - gradients, expected weather, road conditions and so on. Read tour reports from other people, look at nearby tourist attractions (for whatever type of tourist you are), work out where you want to stop each day. Often you start with the list of attractions and work from there. Friends of mine who've done that often start with an event (like PBP, SPEZI) or some other reason to travel, then work out how long they can afford to tour for and buy air tickets accordingly.
Answer: In the US I bought bicycle touring specific maps from Adventure Cycling Association. They're on the expensive side (I just paid about $50 for ~1,000 miles worth of maps), but I like knowing that someone has personally tested each route by bike and that food stops, bike shops and that sort of thing are marked.
I've met plenty of tourists who used GPS, road maps, or no maps at all, so in the end it really comes down to what your comfort level is.
Comment: +1 cool idea. Hopefully, I can find such things elsewhere, will simplify planning a lot.

How to plan routes for touring?


Where can I publicize a website I have developed for writers?

Question: I am not a writer, but I have developed a website for them. Where can I advertise and publicize it?
Comment: If you have a website for writers, why don't you have it listed on your profile on this site, which is for writers?
Answer: Why, wherever writers can be found. :)
Various writers' forums will have places for announcements. You could try approaching writing blogs - emailing blogger, asking for a link to your site if they find it useful. There's magazines like Writer's Digest where you can advertise. Here on Writers.SE, the chat room is the most appropriate place.
But really, beyond such rudimentary guidelines, you'll need to actually do some market research. I'm somewhat dubious about a site "developed for writers" if the development team doesn't know enough about writing to find some popular focal points. Find the places where people are talking about the problem your site solves, and advertise there. No?
Answer: Spread it at InkPop.com. We like spreading the word on new communities and sites for writers.


Why do we consider prime numbers  important, and what are their applications other than number theory in pure math?

Question: Why do we consider prime numbers  important, and what are their applications other than number theory in pure math? I know that Number theory is devoted to studying prime numbers, but there must be other reason to study the properties of prime numbers, right?
Answer: I would start by saying that much of number theory, to the extent that you can really describe number theory in a single sentence, is devoted to solving equations with integer solutions. And it turns out that understanding how these equations behave with respect to primes is often the key to understanding how they behave with respect to all integers.
Prime numbers are also important in understanding a great many concepts in abstract algebra, which generalize way beyond number theory. For example, if we look at an object with a finite number of symmetries, primes play key large role in understanding such symmetries (what I'm alluding to here is finite group theory).
Also, much of the abstract machinery developed to understand primes has found applications in other areas, particularly geometry. When we look at the set of zeros of a polynomial or complex analytic function in space, we do so by trying to understand the irreducible components of these sets, which correspond to something called "prime ideals."
I won't even get into the Riemann Hypothesis, except to say that a single conjecture (which we are unable to prove at present) is simultaneously a statement about primes, complex functions, convergence of series, and random matrices, to name just a few of the myriad formulations.
But perhaps the best reason to study primes is that they are simultaneously elementary and mysterious. It's remarkable just how little we know about these numbers after pondering them for millennia. For many mathematicians, that alone is sufficient motivation.
Comment: I think the second paragraph nails it.
Answer: The additive structure of the integers is trivial: it's generated by 1. This is in essence the Peano axioms.
The multiplicative structure of the integers is not trivial: it's generated by prime numbers. In other words, prime numbers are the multiplicative building blocks of the integers in the sense that every nonzero number is either a prime or a product of primes (the empty product gives 1).
This gives a rich ring structure to the integers. Number theory is about that structure. Add to this the amazing regular irregularity of the primes and you get the richness of analytic methods.
Answer: One reason is that prime numbers are the basis of RSA cryptography, which is based entirely on using large prime numbers in clever ways. Studying the primes directly can change how secure we believe the RSA cryptographic algorithm to be. Currently we believe it to be very secure, but an unexpected advancement in the study of primes could lead to a way to break it, meaning we would have to change to something else.
This certainly has real-world application in that RSA cryptography is (to my knowledge) the standard of many important and sensitive information (think banks, credit cards online, etc).
Comment: "An unexpected advancement in the study in of primes" is not the only threat to RSA-based security solutions. Much more likely (perhaps inevitable?) is the development of [quantum computers](http://en.wikipedia.org/wiki/Quantum_computer), which will theoretically be able to crack any RSA-algorithm in a split second (a remarkable achievement considering that doing this with today's computers would take even hundreds of years).
Answer: Relative primes are also useful when building gear trains. It helps to reduce wear. If one gear has a number of teeth that is a factor of the other gear the same teeth always come into contact with one another.
For instance if one gear has 10 teeth and the other 20, the first tooth on the 10t gear always comes into contact with the 1st and 11th teeth on the 20t gear.
If, on the other hand, one gear has 11 teeth and the second has 20 teeth (20 is not prime but 11 and 20 have no common factors) the first rotation the first tooth contacts the 1st and 12th teeth on the first revolution and the 3rd and the 14th on the second, the 5th and 16th on the third, and it will take 21 revolutions to get back to the first tooth.
In nature prime numbers show up time to time. Cicadas live under ground for prime numbers of year before coming up and making lots of racket (here is a blog post about it http://www.factodiem.com/2010/04/prime-years-of-life.html)
Of course lots of thing are important to study, not because of what they HAVE been used for but for the things they WILL be used for.
Hope this helps.
Answer: The prime numbers have connections to pseudo random numbers. They might also have connections to "true randomness", but I'm not aware if there has been much progress on the conjectures which point in this direction. I wonder whether the Riemann Hypothesis is equivalent to a statement about the relation of the prime numbers to "true randomness", or whether it would at least imply such a statement.


Proving ${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2}$

Question: While simplifying an inequality, this inequality was derived:
$${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2},\quad\quad\quad\quad n\in \mathbb{N}$$
Do you have any idea to prove it? It is very close for small $n$.
Comment: For zero it is false :)
Comment: Please don't use display math in titles; it takes up a lot of unnecessary space on the front page.
Comment: I am confused...you are asking for *any* hint to solve it, however you labelled your question with the "induction" tag.
Comment: @MatemáticosChibchas: hint or proof with or without induction is welcome :) (However a direct proof is better)
Comment: $n\in \mathbb{N}$
Answer: Useful bounds on $n!$ (rather than asymptotics) are given here; in this case, you can use the fact that
$$
n! \le e n^{n+1/2}e^{-n}
$$
for all $n\ge 1$.  Taking the $n$-th root of both sides,
$$
\sqrt[n]{n!} \le \left(\frac{n}{e}\right)\left(e \sqrt{n}\right)^{1/n}.
$$
Now,
$$
\left(e\sqrt{n}\right)^{1/n}=\exp\left({\frac{1}{n}\ln (e\sqrt{n})}\right)=\exp\left({\frac{2+\ln n}{2n}}\right) \rightarrow 1,
$$
decreasing monotonically from above; in particular it's less than $9/8$ for $n\ge 22$ (by a direct calculation).
You want to show that
$$
\sqrt[n]{n!}<(\sqrt{2}-1)(n-2)+\sqrt{2}=(\sqrt{2}-1)n+(2-\sqrt{2})
$$
for all $n\ge 3$.  The above calculation shows the stronger inequality
$$
\sqrt[n]{n!}<\frac{9}{8e}n<(\sqrt{2}-1)n$$
for $n \ge 22$.  The remaining cases, $n=3,4,\ldots,21,$ can then be shown by hand to satisfy the original inequality.  As you point out, for $n=3$ and $n=4$ it is close (and of course for $n=1$ and $n=2$ you have equality).
Comment: thanks. still a lot of calculations for n=3,...,21 is needed! but mathematically logical.
Answer: Using Stirling,
$$n!\approx n^ne^{-n}\sqrt {2\pi n}$$
we can approximate the left hand side with
$$ (n+2)!\approx (n+2)^{n+2}e^{-n-2}\sqrt{2\pi (n+2)}$$
and the right hand side with
$$ (n(\sqrt 2-1)+\sqrt2)^{n+2}>n^{n+2}(\sqrt2-1)^{n+2}=(n+2)^{n+2}\left(1+\frac2n\right)^{-(n+2)}(\sqrt2-1)^{n+2}.$$
If we drop the common factor $(n+2)^{n+2}$ and note that $\left(1+\frac2n\right)^{-(n+2)}=\left(1-\frac2{n+1}\right)^{n+2}\to e^{-2}$, we need only compare the growth of
$\sqrt{2\pi (n+2)}$ and $((\sqrt 2-1)e)^{n+2}$ . Since $(\sqrt 2-1)e>1$, the exponential "wins" for $n$ big enough. To turn this into a proper proof, you need to take a precise look at the error hidden in the "$\approx$" of Stirling and tthus determine, what "$n$ big enough" means. It turns out that the relative error is below $\frac 1{12n}$ and therefore $n=22$ can be taken as "big enough".
EDIT: Thanks to CutieKrait's comment, I noted that I had $(1+\frac2n)^{n+2}$ instead of its reciprocal. Thus one needs to have a closer look at its value as well (it's simply no longer "$>1$", but at least bounded from below ...)
Comment: why $$n^{n+2}(\sqrt2-1)^{n+2}=(n+2)^{n+2}\left(1+\frac2n\right)^{n+2}(\sqrt2-1)^{n+2}$$? And you used a few approximations. finding a large enough $n$ and proving for other $n$s may be harder.
Comment: It's funny that my approach also hit on $n=22$ as a cutoff... a coincidence, I think, but I'm not sure.
Answer: A completely different approach, that's why I don't simply edit my first (lousy) attempt:
We want to show that
$$\tag1 n!<(\alpha n+\beta)^n\quad\text{for }n\ge 3,$$
where $\alpha=\sqrt 2-1\approx0.414$ and $\beta=\sqrt 2-2\alpha=2-\sqrt 2\approx0.586$.
Assume we findan inequality
$$ \tag2 n!\le (\gamma n)^n\quad\text{for }n\ge n_0.$$
Then we have shown $(1)$ for all $n$ with $n_0\le n\le \frac\beta{\gamma-\alpha}$  if $\gamma>\alpha$. Of course, if $(2)$ holds even for some $\gamma <\alpha$, then we have shown $(1)$ for all $n\ge n_0$.
We can try to find good such $\gamma$.
By the AGM inequality, we have
$$\prod_{k=r}^s k\le \left(\frac{r+s}2\right)^{s-r+1}.$$
With $r=1$ this gives us immediatyly $(2)$ with $\gamma=\frac12$, hence $(1)$ for $n\le 6$ because $\frac\beta{\frac12-\alpha}\approx6.8$.
Let $a$ be a number with $0<a<1$ such that $an$ is an integer. Then
$$\tag3n!=\prod_{k=1}^{an-1}k\cdot \prod_{k=an}^nk\le \left(\frac{an}{2}\right)^{an}\left(\frac{(1+a)n}{2}\right)^{n-an}=\left(\frac n2a^a(1+a)^{1-a}\right)^n.$$
Let $\gamma=\frac{a^a(1+a)^{1-a}}2$.
Numerically, if $0.19<a<0.19+\frac17$ (which can definitely be achieved if $n\ge 7$), then $\gamma<0.42$, which gives us $(1)$ for $n\le 101$ (because $\frac{\beta}{0.42-\alpha}\approx101.2$).
These results alone may already be helpful for other methods that establish $(1)$ for $n$ "big enough". However, we could not possibly proceed significantly further up.
We can refine $(3)$: If $an-1\ge 7$, we have already seen that we can estimate $(an-1)!\le (0.42an)^{an}$ and obtain
$$\tag4 n!\le (0.42an)^{an}\left(\frac{(1+a)n}2\right)^{n-an}=(\gamma n)^n $$
with
$$\gamma = (0.42a)^a\left(\frac{1+a}2\right)^{1-a}.$$
If $0.3<a<0.6$ we find numerically that $\gamma<0.41<\alpha$.
This shows $(1)$ for all $n\ge 27$ (so that $an-1\ge 7$).
In summary, $(1)$ holds for all $n\ge 3$.
Remark: Note that a bit of work is hidden in what appears above a "numerical findings": You need to show that the expressions for $\gamma$ have a unique local minimum for $a\in(0,1)$, which then allows us to conclude $\gamma(a)\le\max\{\gamma(x_1),\gamma(x_2)\}$ for all $a\in[x_1,x_2]$.
Edit: It seems like I have made an off-by-one error somewhere, but don't see where exactly right now ...
Answer: Let
$$a_k=\sqrt[k] {k!}$$
By Prove elementarily that $\sqrt[n+1] {(n+1)!} - \sqrt[n] {n!}$ is strictly decreasing
we have
$$a_{k+2}-a_{k+1}< a_{k+1}-a_k$$
taking sum we get this inequality:
$$\sum_{k=1}^{m}{a_{k+2}-a_{k+1}}< \sum_{k=1}^{m}{a_{k+1}-a_k}$$
that is
$$a_{m+2}-a_{2}< a_{m+1}-a_1$$
that is
$$a_{m+2}-a_{m+1}< a_{2}-a_1=\sqrt{2}-1$$
taking sum again we get this inequality:
$$\sum_{m=1}^{n}{a_{m+2}-a_{m+1}}< n(\sqrt{2}-1)$$
that is:
$$a_{n+2}-a_{2}<n(\sqrt{2}-1)$$
or
$$a_{n+2}<n(\sqrt{2}-1)+\sqrt{2}$$
or
$$\sqrt[n+2] {(n+2)!}< n(\sqrt{2}-1)+\sqrt{2}$$
or
$${(n+2)!}< \left(n(\sqrt{2}-1)+\sqrt{2}\right)^{n+2}$$


Sets of equations

Question: Let $I,J,K$ be three non-void sets, and let $\gamma$:$I\times J\times K\rightarrow\mathbb{N}$.
Is there some nonempty set $X$, together with some functions {$\{ f_{i}:X\rightarrow X;i\in I\} $},
some subsets {$\{ \Omega_{j}\subset X;j\in J\} $}, and some
points {$\{p_{k}\in X;k\in K} $} s.t. $\mid f_{i}^{-1}\left(p_{k}\right)\cap\Omega_{j}\mid=\gamma\left(i,j,k\right)$
$\left(i\in I,j\in J,k\in K\right)$, and $\mid f_{i}^{-1}\left(p\right)\mid\leq\mid\mathbb{R\mid}$$\left(i\in I,p\in X\right)$
? In other words, is $\gamma$ ''representable'' as the number of
solutions of some ''reasonable'' equations? [An elementary problem,
indeed.]
Comment: Try first for I of size 1. Put the (generally countable) preimage of p_k as a row, so you have a Kx\omega array.  Now pick \gamma(j,k) elements from the kth row and put them in \Omega_j.  This gets you f_1 as defined on the first array.  For larger I and the ith f, use a disjoint by K x \omega array and repeat. Define f_h(x) for x in the ith array (for h not equal i) to be the 1st element that is not any p_k and is in the same row as x.  All preimages are countable, although K and \Omega_j may not be.
Gerhard "Ask Me About System Design" Paseman, 2011.04.17
Answer: Here is a second attempt (see edit history for previous version).
For each $t\in\mathbb{N}$, let
$$P_{i,j,k,t}=\{1_{i,j,k,t},\ldots,n_{i,j,k,t},\ldots,\gamma(i,j,k)_{i,j,k,t}\}$$
 (so that for each choice of $i\in I$, $j\in J$, $k\in K$, and $t\in\mathbb{N}$, we have a disjoint set of size $\gamma(i,j,k)$).
For each $t\in\mathbb{N}$, let
$$Q_t=\{a_{k,t}\mid k\in K\}$$
(so for each $t\in\mathbb{N}$, this is just a copy of $K$, up to relabeling).
Let
$$X=\coprod_{t\in\mathbb{N}}\left(Q_t\coprod_{\substack{i\in I,j\in J\\\k\in K}}P_{i,j,k,t}\right).$$
Define
$$\Omega_j=\coprod_{i\in I,k\in K}P_{i,j,k,1}\subset X,$$
and $f_i:X\rightarrow X$ by
$$f_{i_0}(n_{i,j,k,t})=\begin{cases}a_{k,1}\text{ if }i=i_0,t=1\\\ n_{i,j,k,t+1}\text{ otherwise}\end{cases}$$
$$f_i(a_{k,t})=a_{k,t+1}$$
Thus
$$f_{i}^{-1}(n_{i,j,k,t})=\begin{cases}\emptyset\text{ if }t=1,2\\\ \{n_{i,j,k,t-1}\}\text{ if }t>2\end{cases}$$
$$f_i^{-1}(a_{k,t})=\begin{cases}\coprod_{j\in J}P_{i,j,k,1}\text{ if }t=1\\\ \{a_{k,t-1}\}\text{ if }t>1\end{cases}$$
We choose $p_k=a_{k,1}$.
Thus $f_i^{-1}(p_k)\cap \Omega_j=P_{i,j,k,1}$, so $|f_i^{-1}(p_k)\cap\Omega_j|=\gamma(i,j,k)$.
Unfortunately this still doesn't address your size concerns, i.e. the preimage of any element of $X$ being countable, because if $J$ is uncountable then $f_i^{-1}(a_{k,1})$ is uncountable (I added the whole mess with the $t$'s to make the preimages of all the other elements countable). I'll leave this as a community wiki, and if anyone sees a way of fixing it they are welcome to edit this.
Comment: @Zev Thank you. This is interesting, yet what about the second condition? I mean [as you've already remarked], you are not controlling the size of the preimages [not necessarily countable!].
Comment: @Ady - Yes, my first attempt was just to write down what seemed to be the most natural sets and functions, without controlling the sizes. I've changed things a bit to reduce the problem somewhat, but it is still there. Perhaps we need to mod out by an appropriate equivalence relation to force things to get smaller?
Comment: He is asking for each preimage under f_i to be countable, right? Or is he asking for the union (over I) of the preimages of each p to be countable?  Gerhard "Ask Me About System Design" Paseman, 2011.04.17
Comment: No. He is asking for each preimage under f_i to be not larger [in size] than the power of the continuum.
Comment: Good.  I hope he (or perhaps other than he, forgive my presumption) likes the ideas in the comment I made to the question.  I think that keeps the preimage size down.  Gerhard "Ask Me About System Design" Paseman, 2011.04.17
Answer: This is basically a detailed description of a solution, based on Gerhard's answer.
Let $X=\mathbb{N}_0 \times I_0 \times K $, where $\mathbb{N}_0$ includes $0$ and similarly $I_0=I\cup 0$ with $0\not \in I$.
Let $p_k =(0,0,k)$, and let $\displaystyle \Omega_j=\bigcup_i \bigcup_k \bigcup_{n=1}^{\gamma_{ijk}} (n,i,k)$.
Define $f_i(n,i,k)=(0,0,k)=p_k$ and $f_i(n,i',k)=(n+1,i',k)$ for $i'\neq i$.  We add $1$ to $n$ so that $f_i(p_k)\neq p_k$.
Note that for $x\neq p_k$, $|f_i^{-1}(x)|\leq 1$.  On the other hand, $f_i^{-1}(p_k)=\mathbb{N} \times  \{i\}\times \{k\} $, and then $f_i^{-1}(p_k)\cap \Omega_j =\{(n,i,k)\mid 1\leq n\leq \gamma_{ijk} \}$, which has the desired cardinality $\gamma_{ijk} $.
Remark:  In a comment to Gerhard's answer, Ady says "I think you're underestimating the size of J."  The size of $J$ actually plays no role in this problem as stated, as the sets $\Omega_j$ may overlap (and will overlap a lot in my construction).  If you want to require the $\Omega_j$ to be disjoint, note that the other conditions force $|J|\leq |\mathbb{R}|$, as $\left|\bigcup_j \left( f_i^{-1}(p_k)\cap \Omega_j \right)\right| \leq |f_i^{-1}(p_k)|\leq |\mathbb R|$.  We can modify the construction above by replacing $\mathbb N$ with $\mathbb R$, and can assure that the $\Omega_j$ are disjoint if we are more careful in choosing which $\gamma_{ijk}$ elements of $\mathbb{R}\times\{i\}\times\{k\}$ to include in $\Omega_j$ (above I chose the points $(n,i,k)$ for $n$ from $1$ to $\gamma_{ijk}$).
Comment: Thanks for the clarification.  Not only does adding 1 to n keep f_i(p_k) from being p_k, it keeps it from being p_l for any l in K.  I hope this clears things up for Ady.  Gerhard "Ask Me About System Design" Paseman, 2011.04.18
Comment: +1 I hereby give you my second upvote.  (I would give you+ 7, but then others might think me strange in a bad, unfriendly way.)  Gerhard "Ask Me About System Design" Paseman, 2011.04.19
Answer: Consider the following construction.  Let $Y$ be a subset of
$X$ such that $Y$ is (equipollent to) $I \times K \times \omega$.
I think of it as $I$-many copies of an array with
$K$-many rows and each row
has countably many elements.  The $k$th row in the $i$th array
is the preimage of $p_k$ under $f_i$.  (For $h$ not equal to $i$,
let $f_i$ send the $k$th row in the $h$th array to, say, the
first element in that row, or perhaps instead to some subset
of elements in that row, under the condition that
those images are disjoint from the set of $p_k$.)  For the sets
$\Omega_j$, pick precisely $\gamma(i,j,k)$ elements from the $k$th
row in the $i$th array and put them into $\Omega_j$.  So far, we
have achieved that the preimage of every point in the range of
$f_i$ is at most countably infinite, for every $i$.  We also have
the desired condition on the intersection of the preimage
of $p_k$ under $f_i$ with the set $\Omega_j$.
Now everything is done except for deciding where to put the
$p_k$.  As long as you avoid sending $f_i(x)$ to a $p_k$ for $x$
outside the ith array, you can label some of the array elements
with $p_k$; this should be doable because you have control of
how $f_i$ acts outside the $i$th array.
Alternatively, let the $p_k$ be disjoint from $Y$ and the
$\Omega_j$, and let $f_i$ send the $p_k$ to themselves, or to some
other set disjoint from the $\Omega_j$.
Gerhard "Ask Me About System Design" Paseman, 2011.04.17
Comment: @Gerhard Sounds promising, yet too vague. E.g., may I ask who is X? Thx anyway. P.S. I think you're underestimating the size of J.

Sets of equations


What is the best way to hold a sphere and for how long?

Question: I have been playing around with the spherification process. My questions are how long can I hold a sphere and what is the best way to do so??
Answer: In my experience, spheres done with the reverse method (calcium solution dipped in alginate) can last up to four days, well refrigerated, and kept submerged in vegetable (soy or canola) oil. I do not know the science behind it, only that it works. For reference, these were spheres made of a carrot-ginger consomme, not certain of specific amounts of calcium chlorate (in the carrot mix) and sodium alginate (bath) used. Note that acidity may play a role in how long spheres will last, and (often) the chemical reaction will continue, albeit slower; on Day 1 we'd have thin membranes with very liquid interiors. By Day 4, the membrane would be perceptibly thicker and the contents more glutinous.


Prime conditional problem in modulo

Question: Any odd prime p can be written in the form of $x^2 + y^2$ when $p \equiv 1 \pmod{ 4}$. explain the proof.  Is there any such condition equations and satisfying the primes? if there, how to construct them? Also, is there any function, which will give a prime.
Comment: You are not using LaTeX correctly. You need to place dollar signs round complete formulae, and not just small parts of them. Please have a look at the edit I made.
Answer: Regarding the proof: Here's an elementary argument. It can be modified to the following similar problems:

$x^2 + 2y^2 =p \leftrightarrow p \equiv 1,3 \mod 8$ or $p=2$.
$x^2+ 3y^2 = p \leftrightarrow p \equiv 1 \mod 3 $ or $p=3$.

Neccesity: Note that for any $x,y$, we have $x^2+y^2=0,1,2 \mod 4$, so for it to be a prime $p$ we need either $p=2$, or $p= 1 \mod 4$.
Sufficiencty: If $p=2$, $x=y=1$ will work. If $p \equiv 1 \mod 4$, then $t^2 + 1 = 0 \mod p$ is solvable, as it is equivalent to finding an element of order 4 in $(\mathbb{F}_{p})^{\times}$, which is a cyclic group of order $p-1$, i.e. divisible by $4$, so you can just take $t=g^{\frac{p-1}{4}}$ for any generator $g$.
By the pigeonhole principle, two of the elements $\{ ta-b \mod p| 0 \le a,b < \sqrt{p} \}$ are the same mod $p$, say: $ta-b=ta'-b' \mod p$, i.e.
$$t(a-a')=(b-b') \mod p$$
Square both sides of the last equation and move both terms to same side to get: $(b-b')^2 + (a-a')^2 \equiv 0 \mod p$. Note that $|a-a'|,|b-b'| < \sqrt {p}$, so $0< (b-b')^2 + (a-a')^2 < 2p$, implying:
$$(b-b')^2 + (a-a')^2 = p$$
Regarding more such conditions - I suggest the wonderful book "Primes of the form $x^2 + ny^2$" by David A. Cox. Another related book is "Introduction to the Construction of Class Fields" by Harvey Cohn.
Regarding a prime-generating function: consult this Wikipedia page. There are many formulas, most of them impractical. But there's a system of Diophantine equations "defining" the primes, which is maybe what you're looking for.
Answer: This is a famous theorem of Fermat.
Answer: Here is (perhaps) a simplification of the proof by Euler given in the wikipedia link. If $p$ is a prime which is congruent to 1 (mod $4$), then $-1$ is a square in the field $\mathbb{Z}/p\mathbb{Z}.$ Hence there is an integer $x$ such that $p$ divides $x^{2}+1.$ We may replace $x$ by $-x$ if necessary, and suppose that $|x| \leq \frac{p-1}{2}.$ Then $x^{2}+1 \leq 1+ (\frac{p-1}{2})^{2} \leq \frac{p^{2}-p}{4}.$ If there is a prime congruent to 1 (mod $4$), which is not the sum of two squares, there is a smallest such prime, so suppose that that prime is $p.$ Choose integers $y$ and $z$ (not both zero) so that $y^{2}+z^{2} = mp$ where $m$ is an integer, but as small as possible. We know that $m \leq \frac{p-1}{4}.$ We claim that $m = 1.$  Note first that $y$ and $z$ are relatively prime.
For suppose that $h = {\rm gcd}(y,z) \neq 1.$ Now $|h| <p$ as $y^{2} + z^{2} <p^{2}.$ Since $h^{2} | mp,$ we have $m = h^{2}n$ for some integer $n.$ Then $np = (\frac{y}{h})^{2}+ (\frac{z}{h})^{2}$ contrary to the choice of $m.$
If $m = 2n$ for some integer $n$, then $y$ and $z$ are both odd, (they can't both be even), and we use the standard trick $np = (\frac{y-z}{2})^{2} + (\frac{y+z}{2})^{2},$ which dates to antiquity. But this contradicts the choice of $m,$ so $m$ is odd. Hence if $m >1,$ then $m$ is divisible by an odd prime $q < p.$ We have excluded the case that both $y$ and $z$ are divisible by $q,$ so neither is divisible by $q,$ since $y^{2}+z^{2}$ is divisible by $q.$ Both $y$ and $z$ have non-zero images in the field $\mathbb{Z}/q\mathbb{Z},$ so $-1$ is a square in that field, ie $q \equiv 1$ (mod 4) as $4 | (q-1)$. By the minimal choice of $p,$  we know that $q$ is a sum of two integer squares, say $q = u^{2} +v^{2}.$ Now we can argue as Euler did that $\frac{m}{q}$ is a sum of two integer squares, and $\frac{m}{q}$ is still divisible by $p,$ contrary to the choice of $m.$
Answer: I do not know the prime function. It is my out of expectation. We can create your cited type problems. For example, primes can be written as primes $x^2$ + 2$y^2$ if and only if p = 1 or 3 (mod 8). Regarding your first question:
Let L = (1, 0, 1) and L has discriminant = -4. If (a, b, c) is reduced representative of some equivalence class with |b| < or = a < or = |c| => $b^2$ = 4(ac -1). Since |ac| > or = $b^2$, we should have ac -1 = 0. The only form of L is (1, 0, 1). Thus h (-4) = 1 and p can be expressible by (1, 0, 1) if and only if -4 = () (mod 4p). I hope you can do the necessary things to complete the solution of  $x^2$ + 2$y^2$ if and only if p = 1 or 3 (mod 8).
All the best!
Comment: You are not using LaTeX correctly. You need to place dollar signs round complete formulae, and not just small parts of them. Please have a look at the edit I made to the question.

Prime conditional problem in modulo


Would an elastic IP affect SEO?

Question: I am considering moving my site to Amazon EC2 instances. Would an elastic IP affect SEO?
Answer: No.
The reason this won't negatively impact your SEO is because of something called DNS. Basically DNS resolves your Domain Name back to your IP address. When you're setting up accounts on various sites for Webmaster tools, they're only concerned with the domain name. When they attempt to crawl the site, they will perform a DNS lookup to find the IP. If the site has moved, then they would see that in DNS and go to the site's new location to check if the content has changed.
For most sites on the web, the servers are managed by companies where the actual owners of the site have no control over what their hosts do with their IP addresses. For most larger sites (Amazon, Microsoft, Google, Etc), they work from a pool of addresses, either through VPS pools, or through load balancing between servers around the world. Because of this a company like Google would be unwise to limit search results to any content based on the IP address alone.
In regard to SEO, in the past they looked at sites based on their IP address and IP ranges to see if you were cross linking between your own sites to gain search engine influence, but now they focus more on established social media.
Answer: Plenty of websites that rely on search traffic are hosted on Amazon.   Using their elastic IP is no more risky than getting an IP address from any other hosting company.
Google doesn't usually target penalties to an IP address.  Sometimes lots of sites with different authors (some good and some bad) share an IP address.
Answer: Not negatively.  If anything, the use of an elastic IP to bypass any failures of any one instance would slightly improve SEO as there would be far fewer outages.
Answer: Generally anytime you get a new IP there is a risk that the IP has a bad reputation. A good indication (but not guarantee) is to check the email anti-spam services if the elastic IP is blacklisted somewhere.
You can check the health of the IP here: http://whatismyipaddress.com/blacklist-check
Keep in mind if you don't find the IP blacklisted it doesn't mean it is clean. It just means that the IP wasn't used for sending email spam.
Comment: Something to keep in mind if your webserver sends email. Amazon has their own email services, pass the traffic on to that.

Would an elastic IP affect SEO?


Delete a purchased file from ITunes Download Queue without downloading file

Question: This is related to an earlier question, but the advice there doesn't work for me. My disk space is full. Whenever I open iTunes, it tries to download some purchases that I don't want anymore (old Lost episodes) that I haven't wanted for years. All that I want to do is cancel these downloads. Deleting them in itunes doesn't work. The recommended advice on an earlier question is to finish downloading them, then delete them, but I can't do this for diskspace reasons (there are about seventeen episodes, so it's not practical for me to sit there and slowly deleted enough diskjunk to download an episode at a time, then delete the episode - and even if it were, it seems rude to hog this much bandwidth.) Any other solution?
Answer: In iTunes Downloads folder (open Terminal, type: open ~/Music/iTunes/iTunes\ Media/Downloads/), delete list.plist file to remove the queued downloads.
Then restart iTunes.
You may also want to delete other files in there too since they are certainly not needed.
(From my answer, https://discussions.apple.com/message/24479839#24479839)
Comment: I have iTune 12.3 (Windows) and deleting contents of `Media Downloads` doesn't work. iTune create that `list.plist` magically again and start downloading.
Comment: @Emadpres Just closed iTunes before deleting plist file
Answer: I had a download (a Lost episode in fact) that wouldn't stop downloading.  Even though it was already in my library, it would download every time I checked for purchases and then error out at the end of the download.
I don't remember everything I tried, but a tried a LOT of things to get it to stop.  The only thing that eventually worked for me was contacting iTunes Support: http://www.apple.com/support/itunes/contact/
They had the item removed from my queue within a couple hours.  It will probably expedite things if you give them the exact episode names and numbers as they appear in the iTunes Store.
Answer: Remove Them from Store=>Purchased Playlist
iTunes 10.7.0.21 on my Windows 7 machine seems to automatically download only those items in  the Store=>Purchased "playlist".  In my case, deleting a movie from Store=>Purchased stopped it from being added to the download queue automatically.
To download purchased items that have been removed from Store=>Purchased, go to Store=>Purchased=>Download Previous Purchases and find the item there.


Scale invariance of power law distribution.

Question: On wikipedia i have find this statement:

...it is scale invariant, and the only continuous distribution that fits this (scale invariance) is one whose logarithm is uniformly distributed.

how can be proven?
Answer: This is the argument as it's sketched by Julian Havil in Gamma: Exploring Euler's Constant; some fiddly details (e.g., involving domains of definition) are ignored.
Let $X$ be a scale-invariant continuous random variable with probability density function $f(x)$ and cumulative density function $F(x)$. Fix a lower bound $u$ on $X$. Scale invariance means that for any $a>0$, $$P(u<X<x)=P\left(u<\frac1aX<x\right)=P(au<X<ax)\;,$$ and hence $F(ax)-F(au)=F(x)-F(u)$. Differentiating with respect to $x$, we see that $af(ax)=f(x)$, so $$f(ax)=\frac1af(x)\;.\tag{1}$$
Now fix a base $b$ for logarithms and let $Y=\log_b X$, with pdf $g(y)$ and cdf $G(y)$. Then $$G(y)=P(Y\le y)=P(\log_bX\le y)=P(X\le b^y)=F(b^y)=F(x)\;,$$ and hence
$$\begin{align*}
g(y)&=\frac{d}{dy}G(y)=\frac{d}{dy}F(x)\\\\
&=\frac{d}{dx}F(x)\frac{dx}{dy}=f(x)\frac{dx}{dy}\\\\
&=xf(x)\ln b\;.\tag{2}
\end{align*}$$
Substitution of $ax$ for $x$ and using $(1)$ and $(2)$ yields
$$\begin{align*}g(\log_b ax)&=axf(ax)\ln b\\
&=ax\cdot\frac1af(x)\ln b\\\\
&=xf(x)\ln b\\\\
&=g(\log_b x)\;,
\end{align*}$$
which can be rewritten as $g(\log_b x+\log_b a)=g(\log_b x)$. But $a$ was an arbitrary positive scaling factor, so $g(y+c)=g(y)$ for all $c\in\Bbb R$, and therefore $g$ must be a constant function.
This shows that if $X$ is continuous and scale-invariant, then $\log_bX$ is uniformly distributed.
Comment: a comment: in a lot of books i have seen scale invariance defined as $f(ax)=a^k f(x)$. In this case $\log(x)$ does not distribute uniformly i suppose.
Comment: @emanuele: I'd have to work through the algebra, but it doesn't look like it.
Answer: If you start by assuming that $X\sim\operatorname{Unif}[0,1]$ and then examine the leading digit of $Y=b^X$ base $b$, I think you will be able to derive the law. Let $B$ be the leading digit base $b$ (B for Benford). Then $B = \lfloor Y\rfloor$ and $$P\left(B=k\right)=\log_b(k+1)-\log_b k=\log_b\left(1+\frac1k\right)$$
follows from observing that
$$k\le b^X\lt k+1\qquad\iff\qquad\log_bk\le X\lt \log_b(k+1)$$
and $X$ is uniform.


Can not connect wirelessly to internet in CrunchBang 10

Question: I am trying to install Crunch Bang 10 on a clients computers. The problem is, wireless internet is not working. Internet on the computer has worked before when it had Lubuntu on it. I have wired works. The computer could temporaily be connected with a wire to do the fix,  but it can not be permanent. The output from lspci (I saw at http://www.cyberciti.biz/faq/linux-find-wireless-driver-chipset/ it told you some wireless info.)
00:00.0 Host bridge: Intel Corporation 82845 845 [Brookdale] Chipset Host Bridge (rev 04)
00:01.0 PCI bridge: Intel Corporation 82845 845 [Brookdale] Chipset AGP Bridge (rev 04)
00:1d.0 USB Controller: Intel Corporation 82801CA/CAM USB Controller #1 (rev 02)
00:1e.0 PCI bridge: Intel Corporation 82801 Mobile PCI Bridge (rev 42)
00:1f.0 ISA bridge: Intel Corporation 82801CAM ISA Bridge (LPC) (rev 02)
00:1f.1 IDE interface: Intel Corporation 82801CAM IDE U100 Controller (rev 02)
00:1f.5 Multimedia audio controller: Intel Corporation 82801CA/CAM AC'97 Audio Controller (rev 02)
00:1f.6 Modem: Intel Corporation 82801CA/CAM AC'97 Modem Controller (rev 02)
01:00.0 VGA compatible controller: ATI Technologies Inc Radeon Mobility M7 LW [Radeon Mobility 7500]
02:00.0 Ethernet controller: 3Com Corporation 3c905C-TX/TX-M [Tornado] (rev 78)
02:01.0 CardBus bridge: Texas Instruments PCI1420 PC card Cardbus Controller
02:01.1 CardBus bridge: Texas Instruments PCI1420 PC card Cardbus Controller
02:03.0 CardBus bridge: Texas Instruments PCI1410 PC card Cardbus Controller (rev 01)

Answer: If you haven't got one of the 3.2 kernel backports, you might need to do that to get the necessary drivers.


With iLife 11, considerations of slide shows

Question: I have to turn a big stack of images into one of those somewhat sappy slide shows with music.
Google shows various advice from various points in time. The last advice I found was iDVD. Is that the right path to launch off on?
Answer: Nope, not at all. iPhoto has great slideshow capabilities.
Lots of examples on YouTube.
No need to involve DVDs; or slow, aging software like iDVD.
Answer: iMovie has the "Ken Burns" effect (the actual name in the application!), designed specifically to do slow panning across images that dissolve into others, etc. It's very flexible and quite powerful.
iDVD is definitely not what you want. If nothing else, it hasn't actually been updated in years, and both iPhoto and iMovie significantly outshine it.


Does naming sensitive things discreetly provide protection?

Question: I had a job where a manager said he liked naming servers after places because if they were named descriptively e.g. "passwords database server" they become obvious targets. I sort of generalized this concept and when I'm writing code I avoid naming variables things like "password". Does this help security or does this confuse people who I should be working with?
Answer: Giving non-obvious names to things is akin to security through obscurity which is usually frowned upon in these parts. Problem with that kind of security is not that it does not work; indeed, it has some value, which was demonstrated many times through History (e.g. that's why a tank is called "tank" and not "armored chariot"). But you cannot quantify security through obscurity. You cannot know "how much" security it buys you; and, in particular, you cannot know whether it works at all (and though it often works well, it also often fails spectacularly).
Therefore, when thinking about risk management, the value of non-standard names must be considered low. This would not be a problem if it could be done "for free"; but non-standard names have a huge hidden cost: if they confuse attackers, they are likely to confuse sysadmins too. It will siphon out thinking time and resources, and will increase reaction latency if trouble arises.
Really, in IT security and in IT in general, descriptive names are a must-have.
Answer: This would be a prime example of Security Through Obscurity.  While this does provide some security value, the gain is generally very minimal and it should never be relied upon as an ultimate defense.
As with any security measure, how to balance security and usability here is one which your organization needs to decide.  If you can come up with a convenient naming convention that would be reasonably obscure to outsiders, but easy to learn for you and your team, it may be a worthwhile consideration.  If you're constantly running into significant problems because people can't remember, or can't easily find a list of, essential server or variable names then it might be time to re-consider whether or not this particular solution is right for you.
Answer: If you're writing a security-sensitive piece of code, it behooves you to use names which clearly spell out what all of the sensitive pieces of information are.
If a piece of text is a password, but you call it string1, then the next programmer who comes in after you might write code that does not treat the variable with proper care.
For instance, he or she might add a debugging statement which spits out the contents of string1 into a log, making passwords available to someone who can turn on debugging and gain access to the logs.
Anyway, the source code does not actually contain passwords, only variables called passwords, unless it is horribly insecure. Nobody is going to gain any passwords by looking for variables called password in source code.
Comment: +1 for noting the risk of confused developers further down the line misunderstanding the need to handle security sensibly.
Answer: There is a big difference between selecting names for servers which do not identify their functionality, possibly making them more obvious targets for hackers and using obscure names in program code.
The names you use for variables in your code do not end up in the final binary/executable of the program. Those names only exist in your source code. If someone has access to the source code, obscure names will provide little protection. However, for those poor maintainers who follow you it will be really inconvenient. In fact, in the long term, it may even decrease the security of the program as later maintainers are more likely to misunderstand or misinterpret what the variable is used for.
When writing code, the first and most important objective is correctness. After that, it is communication of the algorithm and intent of the code. Using variable names which do not acurately reflect the purpose of the variable will detract from this and make maintenance harder. It isthe algorithm which will determine how secure your solution is, not what names you use for variables.
Comment: Thanks, I was thinking more about web scripting languages which are not compiled so the variables names would be apparent. "It isthe algorithm which will determine how secure your solution is, not what names you use for variables." that sums it up nicely!
Answer: The primary gain you can get from server naming is increasing the need to guess and makes it easier to deploy something like a honeypot to try and identify attackers.  It's still security through obscurity and doesn't provide any security in it's own right.
As for using obfuscation in programming code, it's important to consider that depending on the language, this may already be done for you.  Many compiled languages will discard the names of methods and variables in releases and code obfuscation tools are available for many languages that either aren't compiled or keep track of variable names (such as .Net).
The main problem with obfuscating names in code is that it decreases security by decreasing readability.  Complex, unmaintainable code is almost always less secure since added complexity increases errors and errors can lead to security holes.  The best bet is to write very simple, very maintainable code when it is security critical and then if you want to, obfuscate it with a commercial package to make it harder for an attacker to analyze what they need to do to try and attack it.


What is the best way to create a slideshow in iphoto across multiple events?

Question: I want to put together a slideshow but i want to use pictures across multiple events.  I see you can click on Slideshow on a particular event.  Do I have to copy these picture all into a new "event" just to do a slideshow on it ?
Answer: Even easier: Make an Album of the photos you want to use in the Slideshow, and base it on that. This has a couple of advantages:

It simply references the photos in the Events, and uses no ezra disk space, while keeping your Events intact
You can sort the photos manually, into any order you like, which you can't do with Events.

To make a new Album: select a couple of the pics and go Command-n; then drag the rest of the photos to that
Or, shift-command-n to make a new empty album
Answer: To add to the previous post, once you've created your album and it shows in your source list on the left-hand side, you would then select the album for which you want to create the slideshow and click on the Create button.


Is there a limit to the length of an infinitely scrolling page of content where readers are willing to read?

Question: I have done some research on this subject and find much about the "myth of the fold," why infinite scrolling is not great for ecommerce, and how users benefit from a button to click to load more content. I have not found that magic answer for how infinite is too infinite?
Answer: My personal experience leads me to believe that you have indeed found the answer already, namely that there is no hard-and-fast rule to follow.  It depends entirely on the type of content being presented and the target audience.
The general rule of thumb (just guidance mind you) is that users tend to like to have both options available to them so their experience can be tailored to their browsing style.  In many of the online journals/blogs/magazines/etc. the content is often presented in web-paginated views, meaning I have to click some sort of "Next" or page-numbered button to move to the next page.
However they also often provide a "printable view" or (even better) some kind of "see all pages" view.  That way if I don't want my reading experience artificially interrupted, I can skip it altogether.  I find that I enjoy this most when reading content that requires focus, because I don't have to break my concentration to pay attention to navigation controls.
The key is this: know your content and know your audience.  That will give you a firm idea of how you should (or should not) segregate content.
Answer: There isn't "a" correct answer. Depends on what your site's goals are. Depends on what goals the people using your site have.
Think about what you want to achieve. Think about the assumptions you are making that are pushing you towards a infinite scrolling solution. Test those assumptions. If they work out well - go for infinite scrolling. If they don't - don't.
There's a nice presentation from the folk at Etsy that takes apart their introduction of infinite scrolling and why it didn't work. Might be worth a read http://danwin.com/2013/01/infinite-scroll-fail-etsy/
Answer:
how infinite is too infinite?

There's no hard limit.  As the page grows it takes more memory and that can cause the system to slow (virtual memory swapping). You don't know what type of system the user is viewing your site on, so a user with a small memory system (an old desktop or a low end tablet) might reach the point of performance problems quickly.  This can be a major issue on an ecommerce site if you can't limit page size.
Even if the memory limits aren't stressed to the point of performance problems, an extremely large page is unwieldy.


SQL Server 2008 R2 Express Edition - generating database create script

Question: I need to script a database schema out to T-SQL for inclusion source control. The Express version of SSMS doesn't seem to let you script the entire database, just individual tables. Are there any free tools (scripts or otherwise) that can do this ?
Answer: This SQL script is a SQL Server 2005 table scripting tool that I wrote to reverse engineer a database schema to SQL.  It might do what you want.  Load it in SSMS and set 'Results to Text' from the menu.  Then run the script.  It will reverse engineer most of the major database objects.
Answer: Not sure if this will work in Express or not, but you could look into SQL Powershell scripting...  Check this link out.  It's a rough transcript of a SQL Powershell video (the link to that on the page is broken, unfortunately).  Specifically, you're looking for the command that looks like this:
<code>gci | % {$_.script()}
</code>
gci is Get-ChildItem, | just pipes that output to the next command, % is a short hand for foreach, $_ holds the current item, and we're calling the .script() function.
You should be able to save or copy that info out into a file.
For data scripting, I recommend the SSMS Tools Pack.
Comment: Thanks, I already had the SSMS Tools Pack installed and while it will create CRUD scripts it won't just dump the entire schema to T-SQL.
Answer: You can do this with the SMO framework and something like Powershell.
There are a number of tutorials available online.
In brief, you will want to create an SMO UrnCollection for all objects in the DB, then pass that UrnCollection to the SMO.Scripter class to script out all the objects.


How can I add ventilation to my basement office?

Question: I have a finished basement with an office, guest bedroom and bathroom. All about 700 sq ft.
I work 12+ hours in the office and really need to get some fresh air circulating by extracting stale air. I'm looking for a good method of doing this. I had considered running an air vent with a slow/continuations fan or a high volume fan that I can turn on a time for 15 minutes. The office is adjacent to a utility room and I'm concerned that too much of an exhaust could extinguish the water heater pilot light and cause other problem.
I do have an air-in vents (heat/cool), but nothing going back out (circulating).
The office is next to a window which opens to a window well.
Any advice would be appreciated.
Comment: yep, I have 2 jobs.:)
Comment: "I work 12+ hours in the office" each day!?
Answer: Energy Recovery Ventilation (ERV)  You may want to consider either spot ERV or whole house ERV.
These devices both exhaust and intake fresh air, doing some energy exchange to reduce the impact on your air when the temperature differences between inside and outside are great.
Some ERV discussions here
As a spot unit, you could just run the unit in your office.  The Panasonic unit I've read about has automatic controls, but I would supplement them with a countdown timer to limit use even further as necessary.
Leviton 12 Hour countdown timer

The divided wall cap means only 1 penetration is needed and it incorporates a pre filter for the incoming air.

Declaration:  No financial interest in Panasonic or Leviton.  I have used (personally) and installed (for customers) Panasonic fans and Leviton switches.
Comment: Its the right size for a bath, to clear moisture (rule of thumb 1cfm / SF ), perhaps too much for circulation.  The ERV was 40/20 or 20/10 (max / min).  An adjustable speed fan would be perfect.
Comment: great suggestion... however, expensive at $350 just for the fan. I suppose I'll have to bite the bullet.
Comment: a cheaper alternative was discussed in the other link.. A passive, filtered inlet that will balance an exhaust only fan.
Comment: Aldes Airlet 100 was the device mentioned
Comment: I do like the fresh air in, but I may just go with a outlet fan with a countdown timer. I think 80CFM would be fine for a 100 sqf room?
Answer: If you have forced air for heating and cooling, a thermostat that would also run just the fan once in a while to provide circulation might be sufficient. Honeywell's VisonPro TH8000 is one model that has that feature. Look for "Circ" in the fan settings in the manual, which is available online at Honeywell's site here: http://yourhome.honeywell.com/home/products/thermostats/7-day-programmable/visionpro+8000.htm
You may find better options, but it seems to me that feature is the easiest remedy to your requirements. I would consider making sure that the door into this area allows enough airflow. Either with a sufficient gap at the bottom, or with a grate.
Answer: Making a few assumptions here inferred from the details in your questions so please comment to correct any mistakes in my assumptions. The basement is already vented from your A/C unit so you don't really want to open the window unless you have to. I'm guessing that the room does not have any ceiling fan or anything, that would help circulate the air in the room a bit but it could still be an extreme hot or cold area and feel stale. If the basement opens into the house and you don't mind leaving the door open, a simple solution would be a Door Frame Fan or Doorway Booster Fan (same thing different names) this mounts in the corner of the door frame and just helps circulate the air from room to room. A more full solution would be to install an exchange/return back to your A/C handler in the basement but that's probably not practical.
As for the pilot light concern, it would be near impossible to have an ambient flow in an adjacent room or hall to blow out a pilot light. A small amount of flow would actually provide more oxygen to the fire anyways. A large enough pull to create vacuum (in effect) in a large space would be nearly impossible as well as give you a headache and probably make you pass out before extinguishing a flame, a.k.a it's not a practical concern without an industrial fan in a small enclosed space.
Comment: Thanks @Monso. The basement is vented for IN air only, not out. I'm looking for a way to pull stale air out.


How do I perform directional inputs for Street Fighter 4 on a PC Keyboard?

Question: I play SF4 on a Windows PC using the keyboard. How do I perform moves related to circle movements (ex:- quarter circle fwd, half circle fwd) using the keyboard keys?
Comment: Off-topic, but please buy a controller. You can get a decent one for as little as $5, and a good usb adapter for a snes controller for under $25. That's probably about 1-2% of what your PC cost you.
Comment: thank you for the suggestion .even i think controller is good choice .
Answer: I've managed to figure out all the moves of ken (after properly mapping the controls) and it turned out to be the same (almost at-least) for many others. It wasn't really easy figuring them out.
Make sure you have the movement controls as wasd, uio as punches, jkl as kicks, y as triple punch and h as triple kick(or whatever its called). The rest where ever you find convenient.
The ultra combo for Ryu is sd sd and y. Check how the movement is in the command list and cross check for a similar movement in the below mentioned characters.
Here's a list of players having a similar controls(i.e.combos) scheme -
Sagat,Ken,Ryu,Sakura,Dan,Abel,Viper & Dhalsim.
Note: some characters may have a slightly different movement for certain combos.
Thanks to that, I've managed to finish the game as Sagat,Ryu,Ken,Dan and Sakura in medium difficulty and yes, I am aware of every move of those characters.
Edit: You can pull of El Fuerte's ultra combo as well.
Also, for shoryuken and similar moves in other characters - s(hold) tap d twice and press any punch/kick move (depending on the character).
Answer: Okay, got it now.
This guide explains how to do perform these moves using the arrow keys.
An even better guide can be found here.