Why do we study Cantor Set?

Question: For finding counter examples. That does not sound convincing enough, at least not always. Why as a object in its own right the study of Cantor Set has merit ?
Comment: Why are you not convinced that counterexamples are an important thing to study? (Re: "at least not always", the beauty of a counterexample is that only needs to be used once, in a particular key spot, to change our understanding forever.)
Answer:
in beginning real analysis: to counter the naive notion that a "closed set" is a union of closed intervals, plus a few single points.
In beginning Lebesgue measure: the easiest example of an uncountable set of measure zero
In general topology: sets homeomorphic to the Cantor set are useful in proofs
Fractal geometry: many fractals are homeomorphic to the Cantor set.  Mandelbrot calls such a thing a Cantor dust to suggest its appearance.
Comment: Can one claim and substantiate that the study of cantor set is something fundamental and not just a source of ready counterexample ?
Comment: Point (4) seems to address this.  The Cantor set is one of the simplest examples of a fractal, and is therefore studied in detail by students who are first learning fractal geometry or analysis (to the point that a colleague of mine, after writing a masters thesis on Cantor-like sets, started her PhD by stating that she never wanted to see the Cantor set again).
Answer: The Cantor set is quite useful in its own right. My preferred way to think of the Cantor set is as "the most general compact metrizable space." That it is the most general such space means that it is often good for counterexamples because it lacks the particulars. At the same time, one can construct things by specialization.
The formal expression of the Cantor set being the most general compact metrizable space is that every compact metrizable space is the image of the Cantor set under a continuous function. One can use this as a method of proof. For example, to show that a continuous surjection from $[0,1]$ onto $[0,1]^2$ exists, one can use the fact that $[0,1]^2$ is the image of the Cantor set under a continuous function and then extend this continuous function by linear interpolation (the complement of the Cantor set consists of open intervals) to all of $[0,1]$.
A lot more can be proven this way, including a number of very surprising results. A wonderful source for such arguments is the following paper, which (deservedly) won an award for mathematical exposition:

Benyamini, Yoav. "Applications of the universal surjectivity of the
  Cantor set." The American mathematical monthly 105.9 (1998): 832-839.

Among the results proven there is the entirely classical result of Banach and Mazur that every separable Banach space is linearly isometric to a subspace of $C[0,1]$; a result as far from being a counterexample as can be.
Answer: I think that you may be selling short the value of a counterexample! They are quite useful for making sure that you have not proven too much. i.e. When you have a plausible but only semi-formal argument, how do you tell if it is worth the effort in making it rigorous? Checking against counterexamples is often a useful step.
Still, there are relations to unbounded paths in an infinite binary tree without leaves. That is, suppose that you start at the root and write $0.$, thought of as a ternary number. As we travel to the left or right, write a $0$ or $2$, respectively, for the following digit. Continuing on we get a ternary expansion defining a real number in the Cantor set. Correspondingly, the ternary expansion of an element of the Cantor set gives rise to a path from the root in the tree.
Answer: Spaces that are homeomorpic to the Cantor set arise naturally in many mathematical settings, particularly in dynamical systems.
For one dynamical example, the Cantor set is homeomorphic to the phase space of any infinite Bernoulli process.
For another, the "nonescaping set" of many simple dynamical systems in the real line (or the complex plane) is homeomorphic to the Cantor set (this is a "Cantor dust" example as in the answer of @GeraldEdgar). Consider for example the dynamical system
$$z_n = (z_{n-1})^2 + 10
$$
(You can replace $10$ by any real or complex number of magnitude $>2$). One can prove that there is a subset $C \subset \mathbb C$ homeomorphic to the Cantor set such that if $z_0 \in C$ then the sequence $(z_n)$ is bounded (in fact it stays in $C$), whereas if $z_0 \not\in C$ then $\lim_{n \to \infty} |z_n| =\infty$. In short, points not in $C$ escape to infinity, points in $C$ do not.
Also, there are important theoretical descriptions/properties of the Cantor set, for example:

A topological space is homeomorphic to the Cantor set if and only if it is compact, metrizable, has no isolated points, and every component is a point.
Any compact zero-dimensional metrizable topological space is homeomorphic to a subspace of the Cantor set.

Cantor sets even occur naturally in number theory! The $p$-adic integers $\mathbb Z_p$ are homeomorphic to the Cantor set.
Comment: Do you mean some subset in $\mathbb{C}$?  I don't get the example; for any $z_{-1}$ real one has $z_n \geq 10^{2^n} \to +\infty$.
Comment: Oops, you're right, that was careless of me. Fixed.

Why do we study Cantor Set?


Any Yoga routine for Skinny Guys to Gain Muscles?

Question: As I know Yoga can give us many benefits in many ways to our body. I am a skinny guy and I think yoga will also help me to gain muscles. But is there any regular Yoga routine for Skinny Guy Rules to Gaining Muscles?
Comment: There are some aasanas and pranamyams like Kapalbhaati, if done properly, help in weight gain.
Answer: It can help only of you can do This kinds of exercices
 and that's not even true Yoga.
So to be honest with you, if you think that you're really skinny and you're not hitting bodybuilding at the same time, yoga won't give you the result you're looking for.
Yoga helps you add strength to the small muscles and parts of your body that are normally neglected by other sports. It also helps with flexibility and stress fighting, but flexibility and strength are DEFINITELY NOT mass.
Instead :

Try to go to a gym near you, try to focus on compound exercies
instead of isolation ones.
Favor good form over heavy weight.

If you are unable (for one of the reasons) to go to a gym, then you may look over on youtube for some body weight exercices that can be performed at home for mass building, it normally favors push-ups, squats, burpees and pull-ups (a bar may be necessary).
And I almost forgot, focus also on your DIET, which is a HUGE factor in bulking and adding mass to your body. Try to do a search on food and diets that can suit your reach and of course your pocket ;)
Hope this helps you, have a good day and good luck.
Answer: First of all, yoga never ment to be about body physique. As stated in yoga "bible" - "Yoga Sutras of Patanjali",

yoga is restraining the mind-stuff (Chitta) from taking various forms
(Vrittis)1

which is explained as

yoga essentially consists of meditative practices culminating in
attaining a state of consciousness free from all modes of active or
discursive thought, and of eventually attaining a state where
consciousness is unaware of any object external to itself, that is, is
only aware of its own nature as consciousness unmixed with any other
object 2

One can quite easily say that yoga is a form of religion or a philosophical school. Asanas or body postures are just instruments to attain required state of mind, just like prayers in Christianity or meditation in buddhism. Asanas were invented as a form of distraction for the mind, so it can let go of thinking process concentrating first on holding asana and then concentrating on itself.
With that in mind, if one to examine yoga from physical prospective, it is more about endurance and flexibility and some strength.

After analyzing the data, researchers found that the regular practice
of Hatha yoga significantly improved the subjects' flexibility,
muscular strength and endurance, and balance.
American Council on Exercise: "Does Yoga Really Do the Body Good?"
American Osteopathic Association

In order to grow muscle in size you have to basically damage them so they can repair themselves and when they do, they become bigger to be able to withstand the stress next time and not be damaged. In order to achieve visible growth in size you have to progressively increase the stress your muscles endure, which can only be achieved by so called "progressive overloading" i.e. gradually increasing the amount of weight your muscles have to work with.
In yoga, which is a body weight exercise (again, just from physical prospective), it is impossible to progressively overload. You can extend the amount of time you're exposed to stress, but in this case you'll be targeting muscle endurance, not size.
It needs to be mentioned, that if you're an absolute novice then ANY physical activity will result in muscle "toning" and muscles will appear bigger due to slight growth and reduction in body fat. But in order to achieve any significant gains in size, you have to start using heavy weights.
So, to sum it up:

Resistance training leads to trauma or injury of the cellular proteins
in muscle. This prompts cell-signaling messages to activate satellite
cells to begin a cascade of events leading to muscle repair and
growth... The most adaptable tissue in the human body is skeletal
muscle, and it is remarkably remodeled after continuous, and carefully
designed, resistance exercise training programs. 5
Answer: Practicing yoga is great for the body and mind, and yes, gaining muscle can be a nice perk. There are so many different types of yoga and practices, and you may have to experiment with a few to find one that suits your needs.
If you're interested in browsing around and trying a few different 'styles' of yoga, I would recommend either going to a yoga studio close to you, or checking out YogaGlo. It's a great website with lots of tutorial videos, different styles and classes of varied length. Most styles of yoga will help you gain muscle. Avoid a regular Yin practice as that is intense stretching (might be nice once a week on your 'off day'). Vinyasa-style classes are your best option for gaining muscle. You'll also build it in Hatha, but tends to be a less rigorous practice.
If you're interested in hot yoga, try finding a Moksha/ Modo studio, or a Bikram studio. These are set sequences, and can be as intense as you choose to make them (effort-wise), just keep in mind that if you fall in love with a hot yoga practice, make sure to up your calorie intake. The heat helps to loosen the muscle and get into poses and build lean muscle.
Alternatively, an intense, muscle-building practice would be Ashtanga. It is not done in a hot room, but with the intense breath work, you can work up a good sweat. This is a great practice to do at home, but I would strongly recommend going to a studio and practicing under the guidance of a teacher first (that goes for any style of yoga you choose).
Keep in mind that with yoga "no pain, no gain" does not exist. If it feels bad in your body, DON'T do it. With building muscle and practice, it is a process.


Fractional differential equation

Question: Does someone know how to solve this fractional differential equation? $$a\frac{d^2}{dx^2}u(x)+b\frac{d^\frac{1}{k}}{dx^\frac{1}{k}}u(x)+cu(x)=0$$ assuming $(a,b,c) =const$ and $k$ a parameter?
Thanks in advance
Answer: You can use the formula obtained via Laplace Transforms, which works as far as I know for $n \in \mathbb{Q}$
$$D^{-n}\{f(t)\} = f(t) *\frac{t^{n-1}}{\Gamma (n)} =\int_0^t \frac{{(t-u)}^{n-1}}{\Gamma(n)}f(u)du$$
So for example, setting $n= \dfrac{1}{2}$ and multiplying by $D$ you get an expression for the half derivative of $f(t)$
$$D^{1/2} f(t)= \frac{1}{{\sqrt \pi  }}\frac{d}{{dt}}\int\limits_0^t {\frac{{f\left( u \right)}}{{\sqrt {t - u} }}du} $$
So maybe you can multiply your equation by $D$ to get $D^{\frac{k+1}{k}}$ and then use the equation I give you with appropiate $n$.
Comment: Just wondering what does $D^.5$ means please? Thanks.
Comment: @Emmad: See [Fractional Calculus](http://en.wikipedia.org/wiki/Fractional_calculus)
Comment: @EmmadKareem You can try use the formula. You'll find that if you "half" differentiate $x^2$ twice you'll get $2x$.
Comment: Thanks to you and to @Simon.
Answer: it depends on the function $u(x)$. you can try Laplace Transform or Fourier Transform or some other approaches for sufficiently suitable functions to the chosen method.
Comment: This is not a very detailed answer. I suggest you remove this answer and place it as a comment instead ; you might get downvotes upon you.
Comment: @Patrick I guess, cem cannot put comments since he does not have enough reputation
Comment: Hm. I forgot about that! Then let's hope his answer does him some good..
Comment: @cem: the function u(x) is what we are going to find, so, a priori, we can't say anything about it.
Answer: Since some of the solutions (from integral transforms) are already listed I'll discuss some of the other approaches to this.
Looking at this:
$$
a\frac{d^2}{dx^2}u(x)+b\frac{d^\frac{1}{k}}{dx^\frac{1}{k}}u(x)+cu(x)=0
$$
One thing to note is that a solution is likely anylytic (since x is a real variable), so a polynomial series solution is an option.
Assume u(x) is an analytic function with series expansion (I'll use alpha in the series to avoid confusion with your a):
$$
u(x) = \sum_{i=0}^\infty \alpha_ix^i
$$
$$
u^{({{1}\over{k}})}(x) = \sum\alpha_i{{\Gamma(i+1)}\over{\Gamma(i-{{1}\over{k}}+1})}x^{i-{{1}\over{k}}}
$$
Which then gives an equation of the form:
$$
a\sum \alpha_i {\Gamma(i+1) \over \Gamma(i-2+1)}x^{i-2} + b\sum\alpha_i{{\Gamma(i+1)}\over{\Gamma(i-{{1}\over{k}}+1})}x^{i-{{1}\over{k}}} + c\sum\alpha_ix^i =0
$$
So something like:
$$
({bx^{-{{1}\over{k}}}  \over \Gamma( 1-{{1}\over{k}})} +c)\alpha_0+ ({2bx^{1-{{1}\over{k}}}  \over \Gamma( 2-{{1}\over{k}})} + cx)\alpha_1 + \sum_{i=2} \alpha_i ({a\over \Gamma(i-2+1)}+{bx^{-{{1}\over{k}}}  \over \Gamma( i-{{1}\over{k}}+1)} +{c\over\Gamma(1+i)}) \Gamma(i+1)x^i =0
$$
will give a recurrance relation of the second order for the $\alpha_i$ terms. This will be rather messy though (and will depend on your initial conditions).
Also, if you don't need an exact solution as k grows large the solution will be approximated by the solution to:
$$
a\frac{d^2}{dx^2}u(x)+(b+c)u(x)=0
$$
due to the order of differentiation approaching 0 (you can see this by taking the limit of the series representation of the fractional part for $ k \to \infty$.
and for k ~= 1:
$$
a\frac{d^2}{dx^2}u(x)+(b\frac{d}{dx}+c)u(x)=0
$$
The error will grow in factorial order from the points where these O.D.E. approximations exist, but for finding particular points (or a general form to start with variation of prarameters) they're useful.

Fractional differential equation


is Google map navigation only available to selected country

Question: I love using Google map to give me direction. however, I would like it to provide "turn left, turn right" voice while I am driving. I heard Google map navigation able to provide such feature. I search from android market but cannot find one. is it only available to selected country? is there any workaround, or other software I can try out?
Answer: It's limited to specific locations, and only available with a data connection.  See this question on the location limitations and this question on alternatives.
Answer: Google's official list of countries with Google Maps Navigation (Beta) available is here:
http://www.google.com/support/mobile/bin/answer.py?answer=172221
Currently 29 countries are listed there.
If your country is not in the list, the simplest workaround I found is brut.all ownhere modified changepn Google Maps apps on XDADev .


Sharepoint Search Box is requiring Authentication.. Help :)

Question: Seems like it should be relatively easy, but I'm having difficulties allowing anonymous users to enter a value in the search box without share point prompting that user for authentication. I have allowed anonymous users to access the site and that seems to be working fine. Any help would be appreciated
Comment: Have you also allowed anonymous access on your search center site?
Answer: Sounds to me like the Search Suggestions is requiring authentication. Try disabling them if you don't really need them. If you do need them, looks like you'll have to implement something yourself using a custom HTTP Handler and JavaScript, like in this blog: http://www.mavention.nl/blog/search-query-suggestions-anonymous-users-sharepoint-2013-mavention-query-suggestions
Answer: You might be interested to check this:
http://blogs.objectsharp.com/post/2008/02/20/SharePoint-search-and-anonymous-users.aspx
Or on the other hand you can also create your own custom search page and redirect the users to this custom page. This will not cause the anonymous users a problem is searching content on SharePoint Site. But if you are using the default OSSSearchResults.aspx page then the above article should solve your problem.


Network Connectivity problem in 12.04LTS installed on VMWare 10.0

Question: I installed Ubuntu 12.04LTS on Virtual Machines 10.0. My Host Operating System is Windows 8.1. My Laptop Hardware configuration is Intel 4th Generation i7 with 8GB RAM with an additional 2GB nvidia Graphic card. I have an internet connection with 20Mbps which is working with all other Operating systems which I already using in Virtual Machines 10.0 except with this newly installed Ubuntu 12.04LTS. In Virtual Machine settings the Network adopter is set to Host-only. Can any one help me in resolving this problem? Thanks for your help in advance.
Answer: choose an option of NAT which makes us connect to
 the host IP address in Network Adopter settings of VMWare for the corresponding operating system.


Ubuntu SDK for x86 tablets

Question: I was thinking about installing the regular Ubuntu on a Windows 8 tablet with x86 processor...so basically a tablet that is actually a PC. I want to make a program that supports all gestures and movement detection of a tablet. Windows 8 has full support on a variety of different programming languages and supports all mobile device sensors in its SDK. I am not sure about Ubuntu.
So my question is: Is there a Ubuntu SDK that supports all features of a x86 tablet? I am not interested in ARM tablets.
Thx for answering!
Answer: Ubuntu touch is not yet available for x86, But you can expect it in near future

Ubuntu SDK for x86 tablets


How do I fix the frequency of my audio?

Question: I am editing some audio files of an existing game.
The current frequency rate Im using is 6500hz and the music sounds pretty bad. P. D If I use 44100hz there will be a lot of noise.
I am using Audacity to fix the audio. The audio can play well in a computer but it plays horribly wrong on the mobile game I mentioned
Here is a link of the audio:
Comment: You need to sample at double the highest frequency you want to reproduce well. If you have already sampled at 6500 then you cannot make it sound better...
Comment: Sorry -- could you turn that into a question? I gather that you want some help, but... more info please, and as specific a request as possible.
Comment: I have just editted my question please help. Me
Comment: Is your question 'how to change the sampling frequency of an audio file from 6.5 kHz to 44.1 kHz without adding noise ?'
Comment: With 6.5 khz it already has noise. But Audacity nor any other program can detect thst noise.
Answer: Though you should always record at 44.1.KHz, I detect another issue with the sound. You can talk about analogue processing when you are recording, but when the audio is compiled it becomes encoded digitally. This is where bit rate comes into the mix. I believe the output of your recording was encoded into MP3, exactly what bit rate have you used to output the signal? It sounds incredibly low. As a general rule of thumb, you want to have your bit rate at either 128 kbps or above. I personally prefer 320 kbps as this ensures crisp clean audio.
In summary, record it at 44.1KHz, and make sure the output is at least 128kbps MP3. Alternatively you could use wav at 24-bit encoding.
Hope this helps
Comment: I had it at 16 bit. I will try that and let you know

How do I fix the frequency of my audio?


What is the value of sin(arcsin(4))?

Question: In this case arcsin() is the unrestricted sin inverse function. I know that it is either undefined or has the value of 4. It could be undefined because arcsin() has only a doman of -1...1 and 4 is out of the domain. On the other hand, it could be that since they are inverses the intermediary result does not matter and they will cancel to get back 4. Which one is it?
Answer: If you are working with real numbers, $\arcsin(4)$ doesn't exist.
Note that the formula
$$\sin(\arcsin(x))=x$$
only holds over the domain of $\arcsin(x)$ which is $x \in [-1,1]$.
Things are different when you work with complex numbers though....
Answer: If you are willing to allow complex numbers, $\arcsin(4)$ does exist: it is
 $\pi/2 - i \ln(\sqrt{15}+4)$ (using the principal branch of arcsin).  If not, as others have said, the expression is undefined.
Answer: Complex values aside, this expression cannot be evaluated since the $arcsin$ can only be taken from valus between -1 and 1 (inclusive) so the cancelation property of the inverses cannot be applied here. Cancelation property of inverses can be used for values that are in the respective domains of the functions. Think about $\sqrt(-2)^2=-2$ where I cancelled the squareroot against the square.

What is the value of sin(arcsin(4))?


Maintenance before trading in a car

Question: I'm planning to trade in my car on a newer model. My decision to go ahead and do it soon was based on the fact that I had several expenses and maintenance tasks coming up soon: registration, tires, flaky AC. For a trade-in, what kinds of maintenance and repairs generally make economic sense? Is it worth my trouble to remove my political bumper stickers? To get the interior professionally cleaned? To replace the tires or fix the AC? We're talking about a 2010 Honda Fit with a bluebook value of about $3-4k.
Comment: Most of the things you list are normal consumable items and you can add wipers, lights, additional floor mats - if you keep them in good condition then they do help the value of the car when you trade it in - if the mats are knackered, tires well-worn etc then that is why the dealer gives you the lowest price possible... But why run around on nerly bald tires - it is putting your safety in jeopardy and others as well...
Comment: I would say don't do anything extra, the dealer will likely undervalue it anyway.
Comment: For the flaky AC thing you can do a simple check yourself: with engine turned off and keys in your pocket, reach the compressor's clutch hub with your hands and turn it. You'll immediately know if your compressor is still good (smooth and uniform resistance across one whole turn).
Comment: @Al_ that assumes that the clutch is engaged. The pulley can spin quite nicely when disengaged from the compressor, even if the compressor is completely seized.
Comment: @3Dave You're wrong. The clutch hub (the plate in front of the pulley) is always mated to the compressor shaft, be the clutch engaged or not. If you're rotating the hub with your hands, you're rotating the compressor shaft. You need to rotate the pulley only if you want to check the pulley bearing. We're not talking about a clutch assembly for very large displacement bus AC compressors, where the pulley is often located in the front, and therefore you can only rotate the pulley with your hands (and never the "rotor" behind and thus the compressor's shaft), unless the coil is being powered.
Comment: @3Dave In fact, you can check for an automotive clutch-equipped AC compressor internal health even with the belt on. The belt needs to come off only if you want to check the internals of a clutch-less AC compressor. By the way, if you want to see an example of a bus AC compressor's clutch assembly, google for "Linnig LA16". You'll find it very different from the usual automotive one.
Answer: Honestly, don't do it. The ROI on doing maintenance before a trade in is negligible. Most dealers don't car about the tires on the car as long as there are tires on the car and they are not dangerously bald. Things like making the car look nice might add some additional value, but again the ROI would be low.
If you really wanted to see if you can negotiate a better deal for the car, bring it in as it and ask how much they would give you for it. Then ask how much would the give you for it if you did such and such. This would give you an idea which maintenance items are worth it.
Answer: The only time it's worth putting money into a car to trade it is if the money you get is more than the money your spend. If you have a car that's not running then getting it running makes the trade in value much higher, so can be worth it depending on how much it costs. Other than that, cleaning a car so it's not filthy makes sense but not paying for it to be professionally detailed. Putting new tires on costs you lots money but won't get you anything more on the trade in price, because the dealer can do it cheaper than you.
The most important thing is to know what your car is worth as a trade-in and be prepared to negotiate for it. Dealers will often give you a trade in far below value to maximize their profits, so know your stuff and be willing to stick to your guns.
Answer: Among everything you said, i would personally fix (and also tweak: things such as AC line soundproofing so you don't hear the compressor inside too much, and condenser external cleanup+radiator/condenser gap sealing with foam, things i've successfully done on my car myself) the AC, then show the compressor working correctly and then the cold and heat the AC system can provide inside at request. You don't know how many cars get sold with the "AC just needs regassing" trick that then turns out to be a serious leak (such as a leaking evaporator), a seized compressor or an inoperational compressor clutch, so keeping AC in top shape and then showing that to the buyer might definitely be added value.
Yes, your "flaky AC" might range from a simple cabin air filter replacement & regas to a seized compressor or evaporator leak, but in my own opinion it's definitely worth looking at.

Maintenance before trading in a car


How to wash Phil and Teds sports buggy?

Question: We used our Phil&Teds Sport buggy for our first two kids, and it got rather filthy in the process, and now there is the third one on the way! What is the best way to wash the buggy? The fabric parts don't seem to be detachable by design...
Answer: I am not familiar with this type of stroller, or, "buggy" as you call it, but a google search quickly brought me to the manufacturer's website where I found a "support" link. Here, you can search for your specific product's washing instructions (in case there are special considerations based on fabric type, etc). I'm my experience it is never apparent how to get the fabric off- you must consult the user manual, which I imagine is long gone. The website should have a downloadable version for you as long as you know the model #.
However, if you want some non- product specific advice from a mother of three boys it is this: hose it down.  Choose a nice, warm, dry spell (it takes a few days to dry depending on the fabric).  I used to fill a small bucket with a few drops of dish soap and warm water and then use a rag to gently spot clean the really nasty stains before the general hosing. I suggest doing this on a hard surface, such as asphalt, concrete, or the like so you don't get it even dirtier in a mud puddle (I made this mistake once- you don't need to make it also.) I would then wipe down the frame and soak up as much water as I could with some towels. If you don't have agreeable weather for a hosing, the spot clean alone should do the trick.
I used to focus on the straps-they tend to get the dirtiest, and kids can get them in their mouth, so they need to be petrified-jelly free.
UPDATE I recently started using oxygen bleach on my stroller straps because dishsoap was no longer adequate.  I spray it on then lightly agitate with an old toothbrush.  You can purchase it, or, you can make your own: in a plastic spray bottle mix 1/2 cup of baking soda with 1 cup of warm distilled water (you don't want any minerals contaminating your formula), shake to dissolve, than add 1/2 cup of 3% hydrogen peroxide (the kind you use for first aid).  I'm sorry, I don't know how to convert this to metric measurements. Please do a test to determine colorfastness before you start spraying this all over your stroller. It will need a rinse when you are done.

How to wash Phil and Teds sports buggy?


Free Multibody Dynamics Software?

Question: I'm working on modelling a feller buncher, its end effector consists of a saw to cut trees and claws to bundle them together. I'm looking for a free/open source software that can simulate dynamics of multiple constrained bodies, any recommendations?
Answer: If you are comfortable with programming, you can use a free physics engine, like Bullet or DART to create your model. There are also many other ones.
If you are less comfortable with programming (as in you will still need to code, but this softwear at least has a GUI) you can use a robotics framework for simulation like Gazebo.
You could take a look at MBDyn which has a declarative way of defining a simulation. May be simpler then the ones above.
I doubt that any automotive simulation framework would allow you to implement the mechanical linkages of yout vehicle.
If you have some CAD tools available, most of them have add-onds for simulating multi-body systems, but most-of them are not free...
If you do not want to code, I do not know any free solutions. If this is a hobby project, you could try the Matlab Home edition (Mechatronics option) for under 300 USD, (if it is a commercial project the matlab licenses will cost well above 15k). Students also get free/cheap Matlab versions. Other then Matlab, you can try MapleSim or other commercial software.
Answer: I think FreeCAD already has some capabilities in this regard. Also, if you are a hobbyist or a startup below $100K/year income, you are eligible to have Autodesk Fusion 360 for a year for free. It is the little brother of the Autodesk Inventor with more limited capabilities and using cloud to compute and store your models. It has dynamic simulation and constraining features. Personally, I am using the latter, but only because I used to Autodesk products already. Had a look on FreeCad too, it looks really promising for me considering that it is completely free.
If you don't want to create your 3D model, then Matlab has various really nice toolboxes, oh right, that would not be free... Hope this helps a bit.

Free Multibody Dynamics Software?


Nonzero elements of splitting field

Question: Let $F$ be a splitting field of $x^{p^{n}} - x \in \mathbb{Z}_p[x]$.
How is it that the nonzero elements multiply to $-1$ and sum to $0$? I don't get how we get that result.
Comment: That follows from Vieta's formulas (http://en.wikipedia.org/wiki/Vieta%27s_formulas).
Comment: Do you know that the elements of the field of size $p^n$ are exactly the zeros of that polynomial? If so, look [here](http://math.stackexchange.com/a/170614/11619) for the product. Also the usual argument works: if
$$x^r+p_{r-1}x^{r-1}+\cdots+p_0=(x-a_1)(x-a_2)\cdots(x-a_r),$$
then $p_{r-1}=-\sum_i a_i$ and $p_0=(-1)^r\prod_i a_i.$
Comment: What if $p=2$? The sum shouldn't be 0 in that case.
Comment: @ZhenLin I guess the trivial case $p=2,n=1$ should be excluded then.
Comment: @MihaHabič That's what I thought at first, and then I realised that there had to be some non-trivial cancellations happening.
Comment: The trivial case is $n=1$, in which case the polynomial above factors as $x(x^{p-1} - 1)$ *in* $\Bbb F_p$ (of which every element of $\Bbb F_p$ is a root). In which case, the theorem only holds for *odd* primes, which is why $2$ gums up what is otherwise an elegant argument.
Answer: Here's an answer that avoids using the polynomial $X^{p^n}-X$ and uses the fact that you are computing in a finite field.
So, you are asking why
$$
\sum_{x\in\Bbb F_q}x=0
\qquad\text{and}\qquad
\prod_{0\neq x\in\Bbb F_q}x=-1.
$$
in any finite field $\Bbb F_q$ with $q=p^n$ elements.
For the first one: let $s=\sum_{x\in\Bbb F_q}x$. Pick any $z\in\Bbb F_q$, $z\notin\{0,1\}$. Multiplication by $z$ defines a bijection $\Bbb F_q\rightarrow\Bbb F_q$. Thus
$$
zs=\sum_{x\in\Bbb F_q}zx=\sum_{y\in\Bbb F_q}y=s,
$$
from which $s=0$.
For the other one, consider first the case $p\neq2$. Then you have an even number of $x$'s different from $0$. Pair them together considering pairs $(x,x^{-1})$ where $x\neq x^{-1}$. What is left out is the elements such that $x^2=1$ of which there are exactly two: the roots of the polynomial $X^2-1$, i.e. $1$ and $-1$ (distinct!). Now the product of each previous pair is $1$ and of the last two is $-1$.
Finally, consider the case $p=2$. It goes as above, except that now there are an odd number of elements to be multiplied together but also there's only one repeated root of $X^2-1=(X-1)^2$. Thus, $1$ is the only element left out from the pairings and $\prod_{0\neq x\in\Bbb F_q}x=1=-1$ (last equality holds since $p=2=0$ in $\Bbb F_q$).
WARNING : All of the above works with one exception, namely $q=2$. In this case $\sum_{x\in\Bbb F_2}x=0+1\neq0$. The above arguments breaks down because there's no element $z$ than can be chosen as requested.
Answer: Let $\theta_1,\theta_2,...,\theta_{p^{n-1}}$ be the non-zero roots of the polynomial in a splitting field.  Then, in the splitting field, we have
$$x(x^{p^n-1} - 1) = x\prod_{k=1}^{p^n-1}(x-\theta_k) = x(x^{p^n-1} + ... + x(\theta_1 + ... + \theta_{p^n-1}) + \theta_1\theta_2...\theta_{p^n-1}).$$
Hence mathching up coefficients on the left and right we find that the product of the non-zero roots is $-1$ while the sum is $0$.
Answer: We appeal to the structure theorem for finite fields: the field $\mathbb{F}_{p^n}$ is precisely the splitting field of the polynomial $x^{p^n} - x$ over the prime field $\mathbb{F}_p$. We thus have a factorisation in $\mathbb{F}_{p^n}$
$$x^{p^n} - x = x (x-a_1) \cdots (x-a_{p^n - 1})$$
and expanding the right hand side and comparing coefficients, assuming $p^n > 2$, we find
\begin{align}
a_1 + \cdots + a_{p^n - 1} & = 0 \\
a_1 \cdots a_{p^n - 1} & = (-1)^{p^n}
\end{align}
which is what we wanted. (Note that the polynomial $x^{p^n} - x$ is separable because its derivative is the constant $1$, so its zeros are all distinct and constitute all the elements of $\mathbb{F}_{p^n}$.)
Comment: Your answer does not answer the OP's question if you don't show $\,a_1,...,a_{p^n-1}\,$ are *all* the nonzero elements of the splitting field $\,\Bbb F\,$, and for that I think you need the structure theorem of finite fiels.
Answer: There is another approach. For note that the only elements which are their own multiplicative inverses are $1$ and $-1$ (if $p =2,$ these are the same, but the argument still applies). Hence when you multiply all non-zero elements of the field together, the terms other than $1$ and $-1$ cancel out in mutually inverse pairs, leaving a product $-1$ (ok even if $p=2$!). Similarly, when you add all elements of the field, when $p \neq 2$, every element except zero is different from its additive inverses, the mutually inverse pairs cancel each other out, and  only zero is left. When $p =2,$ the case $n = 1$ is a genuine exception to the additive property, for the sum of the element of the field is $1$ in that case. It is slightly unsatisfactory, but one way to deal with the case $p = 2, n > 1$ is to use the fact that the multiplicative group of the field of $p^{n}$ elements is cyclic. Let $\alpha$ be an element of multiplicative order $p^{n}-1.$ Then the non-zero elements of the field are $1, \alpha, \alpha^{2}, \ldots, \alpha^{p^{n}-2}.$ Their sum is $\frac{1 - \alpha^{p^{n}-1}}{1-\alpha}$ which is zero since $\alpha \neq 1.$

Nonzero elements of splitting field


Cempasúchitl vs maravilla

Question: I was reading about the Day of the Dead festival in Mexico and learned that the yellow marigold flower is a central theme in its celebrations. And I also read that it's traditionally called by its Classical Nahuatl name, cempasúchitl. But that sounds like such a difficult word to pronounce! I don't mind learning it but I am really curious to know if it's really the most commonly used name for this flower in Mexico or do they prefer the regular Spanish name, maravilla? I just want to know which of the two words am I more likely to hear in Mexico out of the context of this festival, say at a flower shop. Also, some websites spell it as zempasúchitl; which of the two is "more correct"?
Comment: *cempasúchil* is the spelling that the RAE recognizes, but Wikipedia lists a number of minor variations on it.
Answer: This flower in Mexico is called Cempasúchil, Cempasúchitl is the nahuatl name. If there are other names, I've never heard them before, I didn't know that was called maravilla in Spain.
Words written with final "tl" have two options of pronunciation:

-tl: yes, hard to pronounce.

and

-l: just an l or lt, but in this case only for extremely common words, for example "Xóchitl" is a female name but I've never heard someone saying it as "Sochitl"[the X sounds S], I've allways heard and said "Sóchil" or "Sóchilt"[Names can be written sightly different from the pronunciation]. Remember, this is not a correct pronunciation, it appears in some words and is ok but if you see a word "tl" say it as it written.

And the Zempasúchitl, is a misspelling, I found a couple of English sites that have that Z but it can be an archaism, old writing, all the antiques "ze" and "zi" now is "ce" and "ci".
Comment: @fedorqui'SOstopharming' this is indeed based on frequently misspelled words (though I edit the odd HNQ post as well)
Comment: @Glor just for curiosity: are you editing these old posts following some patterns (like fixing old URLs and stuff you did in the past) or just randomly?
Comment: @Glorfindel OK, I was curious about this one: https://spanish.stackexchange.com/review/suggested-edits/20786

Cempasúchitl vs maravilla


Am i suppose to feel a burn?

Question: So I have started working out for the first time, using 10 pound dumbbells to build muscle on my arms and forearm for the time being. I do all the usual exercises, for example curls and hammers, etc. at decent reps (10-15) and 3-4 sets, yet I still dont get this burning feeling that people tell me I should be getting. I do feel exhausted during the exercise, but after completion I don't feel any pain or soreness in my arms. So should I increase the weight or the number of reps per set, or is that just normal to feel no soreness after a workout? I'm 16 years of age, weigh 113 pounds and am rather skinny.
Answer: In my opinion, it sounds like you need to increase the weight you are using and add some variety to the kinds of curls you are doing. Like Jaydee stated, you might want to also decrease the number of reps you are doing. Personally, I like to do about 8 reps and use a weight that only allows me to get those 8 reps (so I'm working until failure on the 8th). However, increasing weight may cause you to sacrifice your form, which is NOT a good thing. You can find hundreds of videos/tutorials/guides on proper form (here is one example - Proper Form Dumbbell Bicep Curls).
Overview

Increase Weight
Decrease Reps (to stimulate growth)
Maintain Proper Form
Add Variety
Answer: If you want to build muscle you should probably reduce the reps and increase the weight.
I am underweight. How do I gain weight and muscle?
Secondly, it is probably worth getting somebody to put a full body workout program together for you and teach you some lifting technique. Some of the best weight exercises are things like squats and deadlifts which work pretty much everything, but require good technique.
Answer: Burn doesn't really indicate a positive or negative stimulus. You'll often get the burning sensation if you do an exercise where the time under tension for your muscles is constant. For example, cable curls may cause more of a burning feel than dumbell curls because your muscles are contracted to some degree throughout the rep (so long as you don't deload the cable but dropping the weight all the way back to the stack).
But again, burn itself is not really much of an indicator of training effectiveness. When you first start out, there are really two things to key on: central nervous system strain and delayed onset muscle soreness. While these two are also not clearcut indicators of positive results, they do often correlate with intensity. So if you are stressing your muscles to a large degree, you will find that motor control degrades over the course of our workout. It's not just because your muscles are sore but also because your body itself is firing signals like crazy to coordinate muscle contractions to meet the demands you're placing on it. You're straining your CNS to train your body to recruit more muscle efficiently for future workouts. If you did a really heavy compound lift such as a squat or deadlift the systemic strain on your entire body can leave you quivering not because of muscular exhaustion but rather because your nervous system itself needs to recover.
Muscles soreness can be a bit scary, especially early in your training where frankly everything can hurt early and often. But it's fairly common for a very taxing workout to cause the muscles you loaded to be sore 1 to 3 days after. This again is often a good indicator that you're providing a heavy adaptive stimulus. It's not that the pain is good (and by all means, DO NOT train for pain), so much as your body is telling you it was overloaded and needs to repair and adapt. As you continue to train, the delayed onset muscle soreness (DOMS) will become less shocking and easier to cope with if you follow a progressive loading pattern (i.e. up your weights by 5 lbs a week).
The burn itself doesn't really factor into you reaching your training goals. It's not something that signals a whole lot compared to you making note of your recover between workouts (CNS and muscle strain).
TL;DR:

Lift heavier weights for less reps. You want to stimulate your body
to grow not just to feel tired.
Concentrate on monitoring your
recovery. The higher your intensity, the much longer you need to rest
and heal between workouts.
Burn really does not indicate a whole lot
in comparison.
Answer: Sounds like you might be using weights that are too light for you, or too heavy for you and you are just swinging the weights up (assuming you just do curls), not actually engaging your muscles.

Am i suppose to feel a burn?


Importance of Priests in temples

Question: Most of the Hindu temples are maintained by Priests. What are the qualifications that one need to become a priest? What are the duties of the priests towards the temple and god (if any)?
Answer: Now-a-days Archakatva course is offered by different universities and there are specific Veda schools run by many individuals and temple bodies. The basic qualifications include he should be able to do all services starting from Suprabhatam (hymns to wakeup God ) to Pavalimpu/Unjal Seva (hymns to make God sleep) including various activities like Archana, Abhishekam, Alankara, different Upacharas.
Priest should see god in the idol and should accept HIM as omnipotent. He should be capable of performing different kinds of Homas and Vratas specific to the deity. Each temple follows different Aagama Saastra and priest should have thorough knowledge of them. He should do Prokshana (sprinkling of holy water reciting vedic hymns) in case of any eclipse or any Dosha (impurity caused to temple or god by any reason like some being dies in temple or someone comes to visit temple when he is not supposed to like early days of someone's birth or death in the family).
He should co-ordinate regularly with the Dharmakarthas (noble men entitled to take care of temple activities) and with the Matt with which the temple is associated. If someone equal to the God like Sanyasi or any other who has self-realization visits the temple, they should be welcomed with Purnakumbam to the temple by the priest.
The list of responsibilities of priest is big or just cant be explained much here.

Importance of Priests in temples


Yeast for dubbel

Question: I was planning on making 35L of 1.070 ish OG dubbel on Friday.  I don't have any yeast to hand and was wondering whether the smack packs of http://www.brewuk.co.uk/wyeast-activator-3787-trappist.html would be appropriate at, if so, what pitch rate would be best?  I use brewtoad for recipes and it's suggesting >300 bn cells are needed, which would make this pretty pricey yeast order.  Better to leave for another time and make a starter, or would under-pitching this be OK?
Answer: When can you get the yeast? I would definitely not go for it if you're not going to be able to let it get started at all. If you can get the yeast a day or two in advance and get it partially started then you're more likely to avoid off flavors. Even if you have to cut it a little short. You might try buying a couple packs and giving those a short start.
However, let's be clear the best beer will come from waiting for a proper starter. Looking at the BJCP style guide for the dubbel doesn't look good. Under pitching you're likely to end up with more fruity flavors and less of many other characteristics, like body and malt, according to this article at Science Brewer.
Comment: As I feared :(.  I'll have to go have a look for other high ABV beers (looking to get the best use out of a conical that I want to leave alone for 6 weeks).
Comment: What about pitching the yeast 3 days after brewday - or is the risk of infection significantly increased this way?
Comment: I think that it's a question of how much and what kind of risk you want to take. Under pitch and you have a low risk of contamination, but a higher risk of mediocre beer. Waiting to pitch you have a higher risk of contamination.

Yeast for dubbel


"Оттесченный" or "оттестенный"?

Question: There is a word тестировать in modern Russian which differs from проверять/испытывать (both of these has wider meaning). This word has also colloquial form тестить.
I want to ask, regarding de-facto existing rules of word formation in Russian, which derived adjective is more appropriate: оттесченный (or even  оттесщенный compare to обесчестить/обесчещенный) or оттестенный? Or may be some other form? Why?
Comment: The verb _to test_ in this sense better correspond to the russian verb _испытывать_. Not a single inspection, as _проверка_ implies, but a series of thorough tests, or _испытаний_.
Comment: @theUg, valid note, I've edited the question.
Answer: The word you are looking for is протестированный. If you want to sound a little more formal, прошедший тестирование is for you.
Comment: @shabunc: you meant "how do I derive adjective from the colloquial "тестить"".
Comment: You can delete and retype the whole comment if nobody answered yet :-).
Comment: Not exactly, since you are talking about adjective derived from "full" term  - "тестировать". And I wonder how do I derive adjective from the colloquial "тестировать". Because, well, such ajdective is used real life )))
Comment: @shabunc But you can say "I have already can" - is it OK?
Comment: no, guys, hard day's night for poor non-native speaker, sorry for annoying you with such harsh grammar errors.
Comment: Thoroughly formal would be _прошедший испытания_.
Answer: At work I can say: Те́стить(imperfect) → Проте́стить(perfect) → Проте́щенный(derived adjective) and coworkers will understand me.
Probably, grammatically more correct is протеСЩенный or протеСТЩенный, but it's hard to pronounce, in colloquial speech it shorten to single "щ".
Answer: I think that "оттестированный" (compare to "ремонтировать"/"отремонтированный") would be a better choice.
Answer: Because the word тестить is colloquial and kind of funny it would be logical to use оттестенный.
Saying "logical" I mean that the word оттестенный needs to break some rules to be funny.
Оттесченный may be more proper by the rules, but it sounds to me as if it is derived from the verb чесать somehow. As if somebody wanted to say отчесанный but misspoke. :-)
Comment: I think that «оттесченный» is good enough because «сч» is read as «щ».

"Оттесченный" or "оттестенный"?


average distance in a graph

Question: Having a graph of $n$ vertices in Euclidean $m$-dimensional space, is it possible to find average (Euclidean) distance between the vertices in $O(n)$ steps? Is there a deterministic algorithm for this?
Answer: This is not a proof, but maybe an indication that a positive answer is unlikely.
In $2$ dimensions, consider the $2N+2$ points $(j,0)$ and $(0,j)$ for $0 \le j \le N$.  The average distance $\overline{D}$ will be a linear combination over $\mathbb Q$ of $\sqrt{a^2 + b^2}$ for all pairs of positive integers $(a,b)$ with $a,b \le N$.  In particular note that each prime $p < N^2$ with $p \equiv 1 \mod 4$ can be expressed as $a^2 + b^2$, so $\overline{D}$
will include terms in $\sqrt{p}$ for such primes, of which there are on the order of $N^2/\log N$.  Now it is known that these $\sqrt{p}$ are linearly independent over the rationals, and indeed that $\sqrt{p}$ is not in the field generated by the square roots of the other primes.  I suspect (but again, I have no proof) that any algorithm to calculate $\overline{D}$ using integers, $\sqrt{},+,-,\cdot, /$ would require at least $\Omega(N^2/\log N)$ square root operations.
Comment: This is a nice intuition!

average distance in a graph


Windows Live Mesh: "It is already being synced, contains folders that are already being synced, or is in a folder that is already being synced."

Question: On my Windows 7 x64 laptop, I'm trying to sync My Documents using Mesh (version 15.4.3502.922) but I keep getting this error:

This folder can't be synced.  It is
  already being synced, contains folders
  that are already being synced, or is
  in a folder that is already being
  synced. Please select a different
  folder or move this folder to another
  location.

But this folder ISN'T being synced.  If I go to devices.live.com and browse the node for my laptop, it says:

You aren't syncing any folders. To
  start syncing a folder, make sure
  Windows Live Mesh is installed on this
  computer. Then start Windows Live Mesh
  and click Sync a folder.

I get the same error when trying to sync "My Videos" and "My Photos".  These 3 folders were synced using a different hotmail account on my old laptop.  These folders were copied from that old laptop to my new laptop.  On this new laptop, I'm using a completely different Windows Live Id.  And this is the first time I'm using Mesh on this laptop.
How can I resolve this issue?
Answer: I deleted pretty much everything under "C:\users\\AppData\Local\Microsoft\Windows Live Mesh" and things worked as expected.


Are there downsides to using a high shutter speed when it's not required?

Question: Is there any detrimental effects to using a high shutter speed for let's say a still picture, provided there is abundance of light. Does it produce better pictures or makes them worse?
Comment: I would start with the basics of what shutter speed is. Take a look at http://www.cambridgeincolour.com/tutorials/camera-exposure.htm
Comment: Existing questions wondering about specific potential problems: [Do very short shutter speeds result in diffraction?](http://photo.stackexchange.com/questions/19758/do-very-short-shutter-speeds-result-in-diffraction) and [Does taking photo using the camera fastest shutter speed shorten the life of a camera?](http://photo.stackexchange.com/q/24375/1943)
Answer: The key is that shutter-speed is not independent of anything and is related to other exposure parameters.
To increase shutter-speed without under-exposing one must either open up the aperture or raise the ISO:

A wider aperture is often softer and will therefore reduce  image quality. It can also be sharper if you were past the diffraction limit and improve image quality.
A higher ISO generally adds more image noise thus reducing image quality. Except if you were using an Expanded ISO setting below the sensor's native ISO, in which case you will get better dynamic-range.

Increasing shutter-speed without changing other parameters results in darker image. Darker images show more noise because the signal is lower and therefore so is the Signal-to-Noise (S/N) ratio.
Answer: Well, assuming that the lens can open up enough to compensate for the faster shutter speed, the only thing that happens is that you get a shallower depth of field than you might ideally have wanted to use. (I am assuming here that you balance the equation by opening up the lens, rather than by upping the ISO, which is an altogether more expensive operation in terms of image quality!) If you have not focused on the correct point of interest, or if your autofocus is a bit off, this could in turn mean that your subject falls out of focus entirely.
Plus, as Itai mentions, most lenses are noticeably less sharp at their widest-open aperture than stopped down a bit, which would affect picture quality.
Answer: Put another way, you generally set the shutter speed when you need to capture movement.  The general rule is you set your aperture according to your desired depth of field, your shutter as fast as it needs to be based on the amount of movement and the ISO (sensitivity) to whatever is needed to make the shot work.
The higher the ISO, the noisier the image, so if the shutter can be slower, it is generally going to result in a better image quality at a fixed aperture.  If you are shooting free-hand or something that is moving however, there will be practical limits on how slow you can make your shutter.
Generally, when taking a photo in the real world, you have to make decisions to balance the depth of field you want, the amount of noise that is acceptable and the amount of sharpness you want on things that are moving.  Thus, you balance those three parameters accordingly.
Answer: Simply put......No.
A high shutter speed unto itself does not have any detrimental effect on your pictures.
Comment: Does a low shutter speed affects the image quality then? Or will this essentially mean less noise? I'm shooting stills in dark surroundings with a tripod and a trigger, and have done that for a while. But I never really want to bring the shutter speed way down because I have an idea that it affects my picture quality wise. Is this true?? Or can I go way more crazy than I thought?


What is hyperbolic eccentric anomaly F?

Question:
I have no idea what this is and have been searching on the net but all I get is eccentric anomaly. The equation that relates hyperbolic eccentric anomaly F to its true anomaly $\theta$ is exactly the same as that between eccentric anomaly E and its true anomaly. I am guessing it is probably the hyperbolic version of the ellipse.
Comment: A question about the definition of eccentric anomaly for hyperbolic Keplerian orbits is most certainly on-topic here. I think the unexplained close vote for *off-topic* is both *eccentric*, and an *anomaly!*
Answer:
The equation that relates hyperbolic eccentric anomaly F to its true anomaly θ is exactly the same as that between eccentric anomaly E and its true anomaly.

It looks like they are not the same.
From this handy table I've reconstructed the following equation sets. $E$, $D$, and $F$ are the eccentric anomalies for elliptical (and circular), parabolic, and hyperbolic Keplerian orbits.
The source of that table is in turn:

Reference: "Fundamentals of Astrodynamics"  by R.R.Bate, D.D.Mueller, and J.E. White, Dover Publications (1971)

You might be able to find pdf versions online, or better yet, physical versions in a library.
note: I have borrowed much of this from this answer.

Ellipse, Circle $(0 \le \epsilon < 1)$:
$$\tan \frac E2 = \sqrt{\frac{1-e}{1+e}} \tan\frac\theta2$$
$$\text{or}$$
$$\cos E = \frac{e+\cos\theta}{1+e\cos\theta} \tan\frac\theta2,$$
$$M = E - e\sin E,$$
$$\Delta M = M_2 - M_1,$$
$$\Delta t = \sqrt{\frac{a^3}{\mu}} \Delta M,$$
Hyperbola ($\epsilon > 1)$:
$$\tanh \frac F2 = \sqrt{\frac{e-1}{e+1}} \tan\frac\theta2$$
$$\text{or}$$
$$\cosh F = \frac{e+\cos\theta}{1+e\cos\theta} \tan\frac\theta2,$$
$$M = e\sinh F-F,$$
$$\Delta M = M_2 - M_1,$$
$$\Delta t = \sqrt{\frac{(-a)^3}{\mu}} \Delta M,$$
Parabola ($\epsilon = 1)$:
$$D = \tan\frac\theta2,$$
$$M = D + \frac{D^3}{3},$$
$$\Delta M = M_2 - M_1,$$
$$\Delta t = \sqrt{\frac{q^3}{\mu}} \Delta M,$$

To get the semi-major axis $a$ or to get $q$, use the following (don't worry that $a$ is negative for the hyperbola):
Ellipse, Hyperbola:
$$a=\frac{r_{peri}}{1-e}$$
Ellipse:
$$a=\frac{r_{peri}+r_{apo}}{2}$$
Circle:
$$a=r$$
Parabola:
$$q=r_{peri}$$
A quick check with $\mu=1$ and $r_{peri}=1$:
<code> e      theta      a     v_peri    E/D/F       M         t
1.5   90.000000  -2.0   1.581139  55.14281  40.94513  2.021271
1.0   90.000000   n/a   1.414214  57.29578  76.39437  1.885618
0.5   90.000000   2.0   1.224745  60.00000  35.19020  1.737177
0.0   90.000000   1.0   1.000000  90.00000  90.00000  1.570796
</code>
If you want to try it in Python:
<code>def deriv(X, t):
    x, v = X.reshape(2, -1)
    acc  = -mu * x * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

def get_D(theta, e):
    if e == 1.0:
        D    = np.tan(0.5*theta)
    else:
        D    = np.nan
    return D

def get_E(theta, e):
    if e < 1.0:
        term = np.sqrt((1.-e)/(1.+e)) * np.tan(0.5*theta)
        E    = 2.*np.arctan(term)
    else:
        E    = np.nan
    return E

def get_E_alt(theta, e):
    if e < 1.0:
        term = (e + np.cos(theta)) / (1. + e*np.cos(theta))
        E    = np.arccos(term)
    else:
        E    = np.nan
    return E

def get_F(theta, e):
    if e > 1.0:
        term = np.sqrt((e-1.)/(e+1.)) * np.tan(0.5*theta)
        F    = 2.*np.arctanh(term)
    else:
        F    = np.nan
    return F

def get_F_alt(theta, e):
    if e > 1.0:
        term = (e + np.cos(theta)) / (1. + e*np.cos(theta))
        F    = np.arccosh(term)
    else:
        F    = np.nan
    return F

def get_M_from_E(E, e):
    if e < 1.0:
        M = E - e*np.sin(E)
    else:
        M = np.nan
    return M

def get_M_from_F(F, e):
    if e > 1.0:
        M = e*np.sinh(F) - F
    else:
        M = np.nan
    return M

def get_M_from_D(D, e):
    if e == 1.0:
        M = D + D**3/3.
    else:
        M = np.nan
    return M

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

# http://www.bogan.ca/orbits/kepler/orbteqtn.html

quarterpi, halfpi, pi, twopi = [f*np.pi for f in [0.25, 0.5, 1, 2]]
rads, degs = pi/180, 180/pi

mu = 1.0

th0, th1 = 0.0, halfpi
print "th0, th1 (degs): ", degs*th0, degs*th1

eccs = [1.5, 1.0, 0.5, 0.0]

for e in eccs:

    print "e: ", e

    rp =  1.0  # periapsis

    if e < 1.0:
        print "     is ellipse!"

        ra = rp * (1+e)/(1-e)
        print "rp, ra: ", rp, ra

        a0 = 0.5*(rp + ra)
        v0 = np.sqrt(mu * (2./rp - 1./a0))
        print "a0, v0: ", a0, v0

        E0,  E1  = [get_E(th, e) for th in [th0, th1]]
        M0,  M1  = [get_M_from_E(E, e)  for E  in [E0,  E1 ]]
        print "E0, E1 (degs): ", degs*E0, degs*E1
        print "M0, M1 (degs): ", degs*M0, degs*M1

        print "E0, E1: ", E0, E1
        print "M0, M1: ", M0, M1

        dt = np.sqrt(a0**3/mu) * (M1-M0)

        print "dt (sec): ", dt

    elif e > 1.0:
        print "     is hyperbola!"

        ra = rp * (1+e)/(1-e)
        print "rp, ra: ", rp, ra

        a0 = 0.5*(rp + ra)
        v0 = np.sqrt(mu * (2./rp - 1./a0))
        print "a0, v0: ", a0, v0

        F0,  F1  = [get_F(th, e) for th in [th0, th1]]
        M0,  M1  = [get_M_from_F(F, e)  for F  in [F0,  F1 ]]
        print "F0, F1 (degs): ", degs*F0, degs*F1
        print "M0, M1 (degs): ", degs*M0, degs*M1

        print "F0, F1: ", F0, F1
        print "M0, M1: ", M0, M1

        dt = np.sqrt((-a0)**3/mu) * (M1-M0)

        print "dt (sec): ", dt

    elif e == 1.0:
        print "     is parabola!"

        print "rp: ", rp

        v0 = np.sqrt(mu * (2./rp))
        print "v0: ", v0

        D0,  D1  = [get_D(th, e) for th in [th0, th1]]
        M0,  M1  = [get_M_from_D(D, e)  for D  in [D0,  D1 ]]
        print "D0, D1 (degs): ", degs*D0, degs*D1
        print "M0, M1 (degs): ", degs*M0, degs*M1

        print "D0, D1: ", D0, D1
        print "M0, M1: ", M0, M1

        q = rp

        dt = np.sqrt(2.*q**3/mu) * (M1-M0)

        print "dt (sec): ", dt

    time = np.array([0, dt])
    X0   = np.array([rp, 0, 0, v0])

    answer, info = ODEint(deriv, X0, time, atol=1E-13, rtol=1E-13, full_output=True)

    x, y, vx, vy = answer.T
    theta = np.arctan2(y, x)

    print degs*theta[0], degs*theta[-1], " should be ", degs*th0, degs*th1
</code>
Comment: Thanks for the help! However I have no idea what is this hyperbolic eccentric anomaly F. For the eccentric anomaly E, it is the angle from the center to the Auxiliary circle. But what about F? F is not visually represented on the diagram and thats why I am really confused
Comment: @newbie125 oh I see, you'd really like a geometrical explanation or really an illustration. Hmm... I don't know beyond how it's used mathematically, but I will try to find out for you. Thanks for your comment.
Comment: @newbie125 I see there is a [new and much better](https://space.stackexchange.com/a/27869/12102) answer to your question!
Answer: Hyperbolic anomaly is the hyperbolic equivalent of eccentric anomaly.
As you mentioned in a comment above, eccentric anomaly is the angle from the central body to the auxiliary circle of the orbit.  Because a hyperbolic orbit does not have an auxiliary circle, we need a different formulation.
For hyperbolic anomaly, we use an equilateral hyperbola, which has an eccentricity of $\sqrt2$.
The tricky bit with hyperbolic anomaly is that instead of an angle, it is defined as an area.
$ F = 2*Area / a^2 $
Where $Area$ is defined as the area between the X axis, the equilateral hyperbola from periapsis until the vertical projection of the spacecraft location, and the line between this projection point and the origin.
In your image, this is the shaded area.
It's a non-intuative parameter that doesn't really have a physical meaning that anyone uses.  Primarily, it is just used as a mathematical quantity.
Comment: Is it possible to calculate it from the mean anomaly?

What is hyperbolic eccentric anomaly F?


Can one obtain free energy from the vacuum?

Question: It is known that from the vacuum of a quantum field theory, virtual particle pairs are created and destroyed; is it possible to capture these particles thus obtaining free energy from the vacuum?
Comment: We earnestly hope not. If so, this would imply the "vacuums" of our universe are not true vacuums, and the universe could spontaneously tunnel to its "true", lower energy vacuum, wiping out everything in the process, at any time!
Comment: See also, for interest and amusement http://www.google.com/patents/US7379286 although mainstream physics with considerable justification does not take such patents seriously. This particular patent has "slicked" itself up over the years and looks respectable: the original provisional patent bore a photograph (very seldom in patents) of a putative vacuum energy extraction device (a small, plastic box) with a regular, everyday 240V mains lit lightglobe dangling from the box by a cable! It's one thing to be genuinely sucked in by misguided physics, but such a photo borders on outright fraud.
Comment: @WetSavannaAnimalakaRodVance not slick enough.  There is a typo in the abstract.  These sorts of patents and claims are harmful to the less educated that take them as evidence that this sort of physics is real.
Comment: Related: http://physics.stackexchange.com/q/11800/2451 and links therein.
Comment: @BrandonEnright You're right. I must say I never looked at this one closely after the box and lightbulb - I stopped looking at the physics when the question of the writer's honesty was decisively answered for me by the box and lightbulb. It seems someone's hoping to build a pyramid on this one - even the name "Jovion" seems to be making a joke at the expense of anyone who puts money into this (it reminds me of Enron at the height of its integrity - they had shelf companies and fictitious individuals to hide debt in and they named them things like M. Yass (permute trivially to read "My Ass").
Comment: @SeleneRoutley Why would the universe spontaneously do that? It seems to me like that means it would spontaneously decrease its energy level, which would break conservation of energy.
Answer: The energy is borrowed from the Heisenberg Uncertainty Principle to create virtual particles and has to be paid back in a very short time.
$\Delta{t} \geq \frac{\hbar}{2\Delta{E}}$
This is why virtual particles live for very short times (i.e pop in and out of existence). We cannot manipulate this energy.
Comment: For additional information, I recommend Prof. Tong's notes: http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdf. The very first section on "why quantum field theory?" discusses virtual particles.
Comment: Why is this accepted? It doesn't answer the question.
Comment: These notes are good QFT notes but I think they are better suited for grad students! I'm wondering are there any undergrad level QFT notes?
Comment: A. Zee's "Quantum Field Theory in a Nutshell" is perhaps better suited for undergraduates, as it focuses on the physical intuition/understanding.
Answer: Whether you can extract energy from this or not (and I strongly suspect not) the Casimir effect is a consequence of vacuum fluctuations.
Essentially when two metallic plates are very close to each other, the wavelengths of virtual particles that can be created between the plates is restricted and hence there are fewer particles between the plates and outside, where no restriction occurs. This creates a pressure on the plates and pushes them together. Not a source of free energy, but an interesting (and experimentally verified) result of vacuum fluctuations.
Another well-known effect due to virtual particles is Hawking radiation. This says that when virtual particles created across the event horizon of a black hole, one can escape and the other fall in, essentially turning a virtual particle into a real particle, since it's antiparticle is inside the black hole. This is not free energy however, as the energy required comes from the mass of the black hole, causing it to (very, very) slowly evaporate over time.
So in short, no, we can't get free energy from vacuum fluctuations, but that's not to say it it doesn't have some very interesting effects.
Answer: No.
Just like in Chemistry and Thermodynamics, we never get anything for free.
On a mechanistic level, it's important to recognize that zero-point (vacuum) energy represents the lowest energy state waveform. I remember thinking that because the EM fields are everywhere and quantized, that there was some sort of magic taking place. Realistically, zero-point energy is more like a spring between objects (very good analogy Slereah) than something unknown. It is a mathematical consequence of very well understood relations.
The second matter of virtual particles requires a little more thought. When we expand (any) Schroedinger equation into its particle form (using the number operator (two opposing ladder operators) which changes the differential equation to a sum of a series of integrals), we get an infinite number of these terms (the virtual particles). They are a distraction at this level.
Finally, Nuclear_Wizard explained the Casimir effect brilliantly, and I won't add to that except to say that this effect represents the primary experiment validating the theory. To my knowledge and at this time, the other experiments have had a great deal of background noise to wade through (they're looking for the LOWEST energy level) and have not given nearly the experimental validation that the two plates experiment has.
One critical point that is fairly unrelated to your question (but somehow the accepted "right" answer), there is no time operator. The derivation of that relationship depends on somewhat specific circumstances relating to the decay of wavestates. If we look at a graph of the energies of incoming particles, there will be some sort of distribution - the particles don't all have the exact same energy level. If we repeat the experiment and instead look at the time of arrival (which gets us to the lifetime of the wavestate by subtraction) of particles (different setup, identically prepared source), I will also get a distribution. The Heisenberg equation (in that form) relates the uncertainty (width) of those distributions where the location of the peak represents the quantity of those distributions. There is no borrowing energy but only for a short time - the Heisenberg inequality represents a fundamental property of the Fourier Transform and NOT a vacuum energy lender.
Answer: The answer kinda is "You can, but why would you".
It is indeed possible to extract energy from the vacuum. It has been studied, both theoretically and experimentally, using a variety of metal plates and other Casimiresque gizmos.
The problem is just that it basically acts like a spring. To put the Casimir effect in action, you must first approach together the two metal plates, working against the pressure of the Casimir effect. You then let the plates go, giving back said energy (minus whatever loss in the process).
There's a number of other methods you can use for equivalent effects, such as oscillating metal plates to emit a (very weak) radiation from the Casimir effect. But none of them break energy conservation in the end.
Although I am sure that such devices will be plenty useful for various nanotechnology things.
Edit : Oh, by the way :
1) While the quantum vacuum is a nice explanation for it, it is (probably) not true. You can derive Casimir effects just from higher order QED effects (you can't really have a Casimir effect with metal plates because they are not perfect conductors. Though you do have the topological Casimir effect in compact spaces).
2) For some sources on various experimental gadgets using the Casimir effects, you can try "Frontiers in Propulsion Physics". It is quite a neat book that contains a lot of experimental results on such crazy topics (usually negative unfortunately!), and it contains quite a lot of references on applications of the Casimir effect.
Answer: No...but it would certainly be neat if true. Vacuum energy is a measure of "tension" in space time.  Over time, as the universe expands, tension diminishes.  Interestingly, the more tension, the faster light travels.  Early in the universe, after the big bang, tension was very high and the universe expanded faster than what we would consider the speed of light. In any case, tension is an expression of potential against a rest state, defined by the rest state of the initial universe.  Trying to extract energy from this potential would be analogous to trying to extract energy by connecting a wire to one pole of a battery.  Unless you can access the initial rest state of the universe, there is no way to extract energy from the vacuum.
Answer: Yes, but no, as any energy "gained" would not only be undetectable, but would never last. For every virtual particle, there is a virtual antiparticle. Even if you did manage to "extract" energy through somehow -impossibly- harvesting virtual particles, the antiparticles would just neutralize it. Even in a hypothetical -and very impossible- scenario where one could somehow "contain" the antiparticles, you wouldn't be able to draw any energy from it. There are several reasons for this.
1. Matter and energy can neither be created nor destroyed, only conserved, and converted into another form (e=mc². The only reason virtual particles don't violate this law, is because virtual particles and antiparticles always eradicate each other soon after forming via quantum tunnelling.
2. Virtual particles remain virtual unless their virtual antiparticle counterpart meets certain conditions: It must be inaccessible, it must not be in the same worldline, and/or must not be able to be observed. This means that in order to make a virtual particle "real", so one could harvest it, you would need to get rid of its antiparticle; which can only be done through interaction and annihilation with another particle, or through crossing an event horizon.
3. Even if one could contain a virtual antiparticle long enough to harvest its virtual particle counterpart's energy, there would be no net gain of energy, as soon as the antiparticle would become real, it would "destroy" any matter and energy equivalent to the amount harvested from its positive energy counterpart.
It would be not only unconventional, and redundant, but would also be nothing more than a means of "destroying an object in its entirety just to harvest a small amount of equivalent energy". It wouldn't be free energy, merely a means of "quantum tunnelling shenanigins".


What was the earliest concept of a robot?

Question: I'm a highschool student studying electronics and for an assessment task on the history of electronics I have decided to focus on the history of robotics. I want to begin with the earliest possible concept of a robot and progress through major developments in robotics to the current day. Where should I begin my research?
Comment: Props for asking what would in most instances be a list or subjective question (which would get closed pretty quickly) in a clear and answerable way.
Answer: I've found the wikipedia article (as well as its linked articles) on the history of robots to be enlightening:

The history of robots has its roots as far back as ancient myths and legends. Modern concepts were begun to be developed when the Industrial Revolution allowed the use of more complex mechanics and the subsequent introduction of electricity made it possible to power machines with small compact motors. After the 1920s the modern formulation of a humanoid machine was developed to the stage where it was possible to envisage human sized robots with the capacity for near human thoughts and movements, first envisaged millennia before.

Also, you might mention R.U.R., a play in the 1920s that popularized the term robot:

R.U.R. is a 1920 science fiction play in the Czech language by Karel Čapek. R.U.R. stands for Rossum's Universal Robots, an English phrase used as the subtitle in the Czech original.1 It premiered in 1921 and introduced the word "robot" to the English language and to science fiction as a whole.
Comment: Indeed, you can't talk about the history of robotics without mentioning where the term comes from.
Answer: If I recall correctly Leonardo da Vinci is credited with the earliest design of a robot. My personal favorite is generally referred to as Leonardo's robot. However some accounts place Leonardo's cart as having come first and is generally considered a robot because it could be "reprogrammed" by replacing its cogs.
Answer: If you are interested in real and fictional robots (for concepts), the CMU Robot Hall of Fame may be useful.
Answer: This site gives a great time line of robotics. Keep in mind the origins of robots come well before electricity. As the article describes, the early egyptians designed and constructed simple automatons, which are considered the earliest type of robot. Another starting point worth mentioning is the materials that early robots were made from and continue to present the evolution of carbon fibre and other materials that make robots as amazing as they are today.
Answer: You can find some nice presentations on prezi about robotic history (and about many other topics).
For example this or this or this presentations mention the ancient greek Archytas's robotic pigeon well before Leonardo's work around 350 B.C. and the klepsydra from around 300 B.C. having a feedback control system.
Comment: You are not new to stackexchange websites, but I need to tell you that unfortunately the internet changes all the time. [Links rot](http://en.wikipedia.org/wiki/Linkrot) and [answers with only links are not considered good answers](http://meta.stackexchange.com/a/8259/169090). It would be nice if you quoted some of the more important/interesting parts of the articles you linked, just in case that particular website goes down.
Comment: Ok, updated. My intention was to show a possible source "where he should begin his research".
Answer: I started where you are but found lots of misconceptions and omissions so I started a small page on History making robots at http://davidbuckley.net/DB/HistoryMakers.htm
Reuben Hogget contributed a lot of information and after our trip round Europe talking to researchers who had built robots in the 1950s he started http://cyberneticzoo.com - not everything about the history of robotics is there yet. Humans have talked about robots (by other names) for a long time.
In the Greek myths Hephaestus creates 'walking' tripods and golden ladies to help him. The Greeks built large animated sculptures such as a 12 foot high statue of Nysa mounted on a huge cart. Heron of Alexandria (~50AD) created animated figures for theatre shows and also described mechanically programmable devices. There was nothing particularly new about Leonardo's drawings of a mechanical statue of a knight, it is just that his drawings have survived. The first 'modern' 'robot' was probably Dederick's 1868 steam man (Boilerplate is fictional). From early in the last century there have been many drawings of electromechanical men and several working ones (albeit of limited funtionality and really precursors of Disney's animatronic figures). Capek's RUR really only went along with people's fascination with robots (the robots he describes though are biological). A major advance came in the 1940s with Grey Walter's tortoises Elmer and Elsie which had simple electronic brains. The mobile robot Shakey (1966) was another 'milestone'. After that you have the walking robots from Waseda University, the subsumption (a dead end) robots from the MIT Mobile Robot Lab, and then Honda's P2 humanoid which led to a veritable explosion of humanoid robots from Japan followed by Korea and others. Whatever you read, all the writers will have their own slant on things and it is a huge field. Reuben has well over 500 posts on robotics history and PlastiPals.com is the site to read for more modern robots.
Answer: One major development milestone was the first functioning robot to use electricity. As mentioned in Jordan Brown's link, Nikola Tesla demonstrated a radio-controlled robotic boat (more of an ROV I suppose) in a specially-built pool in Madison Square Garden in 1898. He even implemented a crude logic gate designed to prevent other transmitters from taking control from him.
Radio being a not exactly well-known technology in 1898 (Marconi had not even filed for any patents in the US yet), disbelieving onlookers gave such explanations as witchcraft, telepathy, and a tiny trained monkey hidden inside the vehicle.
Related US patents:
Patent 613,809: Method of and Apparatus for Controlling Mechanism of Moving Vessels or Vehicles
Patent 725,605: System of Signaling
Answer: When you start talking about the first robots, you quickly run into the question: "What is a robot?".
Before we can say whether or not some machine is the first robot, we need to define what exactly we mean by the word 'robot'. The problem is, that there's really no single definition, and it depends who you ask. For example, I had a massive argument with someone on the Wikipedia robots talk page about the definition of robot. And you can forget about looking up the word in a dictionary, all of the dictionary definitions I have seen are hopelessly naive. The best definition I have seen is on the Wikipedia page:

A robot is a mechanical or virtual artificial agent.

The most important word here is 'agent'. A machine which seems to have a sense of agency is more likely to be called a robot. I can imagine that if Leonardo had built his 'robot' then people would really have thought it has real agency of its own, and been practically alive, and would therefore have qualified to be included in that definition of robot. Today, however, roboticists have stricter definitions, and require robots to be more than just clockwork, and have some degree of autonomy and interaction with the environment.
Answer: I think it is better to start with explaining about an automation. And then go for robotics and their milestones.


What happens when I browse items in my Lastpass vault on Chrome while running other extensions that can "Access all your data on all websites"?

Question: The way I use Lastpass with Chrome, I often open and browse my vault items, including the passwords. I just noticed with horror that I have several other extensions enabled that have permission to  "Access all your data on all websites." Does this mean these other extensions "see" whatever I see in my vault, and that they can learn logins and passwords I've saved? If not, I'd love to understand why not, you know, just so I can sleep at night.
Comment: I don't think so, because when you open your LastPass vault, you're actually launching a local Chrome Extension.
Answer: The only way for a Chrome extension to gain access to the HTML code inside a tab, or in your case your password, is to inject a content script in the form of a javascript file.
I've created a quick Chrome extension to try this out myself. Luckily, it is not possible to inject content scripts into other chrome extension tabs - only in tabs that match http://*/* and https://*/*. It's therefore impossible for a Chrome extension to be able to get your passwords from your LastPass vault.
However, this doesn't prevent extensions with the ""Access all your data on all websites" permission to just simply get your password when you log in on any website, including your online banking account for example.
Only install extensions from sources you trust. Check the reviews when downloading an extension from the Chrome web store. It's likely that someone will notice a malicious extension quickly.


Has anyone tried Stripe payments through OmniPay?

Question: The Stripe payment processor extension works okay for the most part, but when there is an issue, it seems hard to get any answers or feedback. I am curious if anyone has tried integrating Stripe payments through the OmniPay extension, and if so how that went?
Comment: Yes, I think if you could highlight the issues you've been having that might also help. We've done quite a bit of work on the stripe extension recently and are pushing a few code changes back to the maintainers. If you could share your issues we might be able to confirm if they're being addressed.
Answer: I think Stripe through the Omnipay extension is not working at the moment. We are looking into Stripe as well and will wait for the regular extension to be updated like Parvez mentioned.


Resource Monitor not showing all memory in use

Question: I have a server currently sitting at 97% memory utilization (Just over 30GB on a 32GB server), yet only show about 1gb of memory being used when I add up the values in Task Manager or Resource Monitor.  Any thoughts on how to find out what is consuming all of my memory resources?
Comment: Why do you care? If you want memory not to be used, take it out of the server and sit it on your desk.
Comment: Because SQL Server is being starved at less than 1gb, but thanks for the smartass answer without actually offering anything constructive.  Happy holidays.
Comment: If you describe your actual problem, I'll do my best to help you solve it. How did you determine that SQL Server was memory starved?
Answer: There are more memory classifications than are shown in task manager.  In particular, the Standby List.  RAMMap from SysInternals would be the defacto tool for providing that information.

Active: (also called Valid) The page is part of a working set (either a process working set, a session working set, or a system working set), or it's not in any working set (for example, nonpaged kernel page) and a valid PTE usually points to it.
Transition: A temporary state for a page that isn't owned by a working set and isn't on any paging list. A page is in this state when an I/O to the page is in progress. The PTE is encoded so that collided page faults can be recognized and handled properly. (Note that this use of the term "transition" differs from the use of the word in the section on invalid PTEs; an invalid transition PTE refers to a page on the standby or modified list.)
Standby: The page previously belonged to a working set but was removed (or was prefetched/clustered directly into the standby list). The page wasn't modified since it was last written to disk. The PTE still refers to the physical page but is marked invalid and in transition.
Modified: The page previously belonged to a working set but was removed. However, the page was modified while it was in use and its current contents haven't yet been written to disk or remote storage. The PTE still refers to the physical page but is marked invalid and in transition. It must be written to the backing store before the physical page can be reused.
Modified no-write: Same as a modified page, except that the page has been marked so that the memory manager's modified page writer won't write it to disk. The cache manager marks pages as modified no-write at the request of file system drivers. For example, NTFS uses this state for pages containing file system metadata so that it can first ensure that transaction log entries are flushed to disk before the pages they are protecting are written to disk.
Free: The page is free but has unspecified dirty data in it. (These pages can't be given as a user page to a user process without being initialized with zeros, for security reasons.)
Zeroed: The page is free and has been initialized with zeros by the zero page thread (or was determined to already contain zeros).
Rom: The page represents read-only memory.
Bad: The page has generated parity or other hardware errors and can't be used.
Comment: Thanks!  This was exactly what I was looking for.  Turns out it was a VM config issue that had 30GB driver locked - let the ops team know, and after a reboot I'm no longer suffering from resource starvation for my SQL process.
Answer: Modern operating systems only make memory free if they have some reason to do so. It takes resources to make memory free and that effort is just wasted when it has to make the memory used again. So they only bother to make memory free if they have absolutely no choice.
The memory is just still being used by whatever used it last. Most likely, it contains copies of data on disk. If that data is read again, it will save the system a disk operation. So this is preferable to having it free.
You're thinking you want memory to be free now so you can use it later. But memory doesn't have to be free now to be used later. In fact, memory that's being used now is more likely to be used later.
Imagine you walk into a factory where all the employees are doing things. Maybe some are doing unimportant things like sweeping the floor or sitting near the telephone in case it rings. But nobody is on the couch doing nothing at all. Do you wonder why more employees aren't sitting on the couch doing nothing -- ready to work should there be something super-important to do? There's always something you can do. Maybe it's not the most useful thing in the world, but it's better than sitting on the couch doing absolutely nothing.
So some of your employees are doing important things. Some are just waiting for the phone to ring. But you don't want them on the couch doing absolutely nothing.


Recording Multiple Kinds Of Microphones In Reaper

Question: I've had a AT2020 USB mic for a while but I recently bought a Scarlett 2i2 and a MXL 990. Is it possible for me to record both microphones in Reaper? If so how? I know it'd be ideal to just get another xlr microphone and I will eventually, but for now I'd like to utilize what I have if at all possible.
Answer: You can do this, I think, using Mac OS X's aggregate audio devices. Some advice on how these work and how you can set one up is found at the following links:
http://support.apple.com/kb/HT1215?viewlocale=en_US&locale=en_US
http://www.loopblog.net/tutorials/studio-and-gear/studio-techniques/use-multiple-audio-interfaces-on-mac-os-x/
The Reaper Blog says this is 'ok overall but not reliable enough for professional studio work' due to some latency and timing issues. Fair enough.
http://reaperblog.net/2013/01/using-multiple-audio-devices-in-reaper-osx/
Answer: If you're on Windows, ASIO4all might do it.
You can select ASIO4all as your audio device within Reaper, and then select multiple devices in ASIO4all's configuration, which will then look like a single device to Reaper. You'll then need to close and reopen Reaper's audio device configuration before you can choose which inputs / outputs Reaper will use.
I haven't tested this, so it's possible you might run into sync problems or something - worth a try though...


How do you prevent being logged out of the control panel?

Question: One of my clients' irritations is being logged out of the control panel when he has left his session idle for a few hours. How can I prevent this?
Comment: I'd be remiss if I didn't point out that it's a _good_ thing your client is logged out after being idle (especially for several _hours_). It prevents baddies from accessing the site's control panel if, for example, your client's laptop is stolen. Site security should always be a top priority, and I'd recommend pointing that out to your client.
Answer: If for some reason you need to use sessions, you can change the session time with a config variable
http://ellislab.com/expressionengine/user-guide/general/hidden_configuration_variables.html#cp-session-ttl
<code>$config['cp_session_ttl'] = 86400; // Keep admins logged in for one day!
</code>
Comment: Tried this, cookies-only, login_reminder set to `'n'`, and am still getting logged out after some time (and getting the reminder). Could be b/c of having many tabs open for so long, trying in incognito mode right now to see if that's the issue.
Comment: private browsing didn't help, found answer in this comment: http://expressionengine.stackexchange.com/a/19140/61
Answer: Try changing the Control Panel Session type to Cookies Only:
http://ellislab.com/expressionengine/user-guide/cp/admin/security_settings.html
Answer: I use the following code within the config.php file that is located at /system/expressionengine/config/config.php.
And you have to set "cookies" in your security preferences.
<code>// Cookie Settings
$config['cookie_prefix'] = "";
$config['cookie_domain'] = '.yourdomain.com';
$config['admin_session_type'] = 'c'; // if c does not work try 's' (session)

// logout time controlling: it's best to set only cookies in security prefs
// be sure to set cookies domain as well
$config['user_session_ttl'] = 14400; // user session
$config['cp_session_ttl'] = 14400;  // control panel session
</code>
This works quite well!
Cheers
outline4
Answer: I'd prefer to just up vote MediaGirl's comment, but since I don't have enought rep yet to do that I'll just say that we did the two things recommended here: 1) switch to cookies only and 2) up the cp_session_ttl and that seems to have worked best for us. Probably didn't need to set both, but it worked regardless.
Comment: Adam ask a question and you'll get some points action to start up-voting.
Comment: Thanks Frank. I think I actually have a few that are sitting unanswered in the EE forums that maybe I'll move over here. ;)
Comment: Yeah I'm asking stock standard ones as well just to get this thing out of private beta. Although even the easy ones are actually producing very interesting answers...
Answer: I was under the impression the staying logged-in problem was a known bug. As recently as 3 months ago complaints were still showing up on Ellis Labs official support forums, and that seemed to be the standard response.
Comment: I have had intermittent problems with this as a bug, with CP sessions as cookies.  In one instance it had to do with accessing two different EE installations on the same domain name (different subdomains). Logging into one would log the user out of the other.
Answer: John D Wells mentioned this on his site, not me, but "as of EE 2.2.somethingorother, there are two new hidden configuration variables that allow you to control the session length of both CP users and front-end visitors. See User Session Length and Control Panel Session Length for more."
Here's a link to the "Hidden Configuration Variables" page:
http://ellislab.com/expressionengine/user-guide/general/hidden_configuration_variables.html
Comment: I see Richard Frank said basically the same thing, checking the "hidden config variables." Sorry for the duplicate info, but you may have luck trying both of the settings mentioned in John D Wells' post.
Answer: This is somewhat tangential, but I highly recommend your client use something like 1password to manage credentials instead of staying logged in for a long time,


How to change the Linux Localization?

Question: How to do change the Linux or ubuntu   Localization?  is there easy steps which I can convert Linux or ubuntu  English into Tami language localization operating system ? or is there any free available Tamil Linux or ubuntu ?
Answer: Have you check in :
System > Administration > Language Support ?

If there is your language, you can easily change it without re-install.
Answer: Ubuntu comes with Tamil.  Try @Iterrier's answer first.  It's the easy way.  This will change your locale to a supported one, if it's in the list.
If you're having problems, make sure /var/lib/locales/supported.d/local contains a line starting with ta_IN.
If it's missing, add ta_IN UTF-8 and run sudo dpkg-reconfigure locales
otherwise, continue:
In /etc/default/locale, change

LANG="ta_IN"
LANGUAGE="ta_IN:ta"

Honestly, I can't confirm if you should be using ta_IN or just ta, but just make this entry match what's in /var/lib/locales/supported.d/local.
If it's not already on your system, you'll need the ttf-tamil-fonts package - it's in the universe or multiverse.
You'll also want to change your keyboard settings as well.   Also see The Ubunutu Tamil team wiki.
Answer: Mandriva comes with Tamil fonts already pre-installed. For other distributions you can download the fonts. for changing your locale, follow this answer.
There is also a user group called tamilinix dedicated to this, you may want to check them out.

How to change the Linux Localization?