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I might have to play with it and see. Free Power Perhaps you are part of that group of anti-intellectuals who don’t believe the broader established scientific community actually does know its stuff. Ever notice that no one has ever had Free Power paper published on Free Power working magnetic motor in Free Power reputable scientific journal? There are Free Power few patented magnetic motors that curiously have never made it to production. The US patent office no longer approves patents for these devices so scammers, oops I mean inventors have to get go overseas shopping for some patent Free Power silly enough to grant one. I suggest if anyone is trying to build one you make one with Free Power decent bearing system. The wobbly system being shown on these recent videos is rubbish. With decent bearings and no wobble you can take torque readings and you’ll see the static torque is the same clockwise and anticlockwise, therefore proof there is no net imbalance of rotational force.
A very simple understanding of how magnets work would clearly convince the average person that magnetic motors can’t (and don’t work). Pray tell where does the energy come from? The classic response is magnetic energy from when they were made. Or perhaps the magnets tap into zero point energy with the right configuration. What about they harness the earth’s gravitational field. Then there is “science doesn’t know all the answers” and “the laws of physics are outdated”. The list goes on with equally implausible rubbish. When I first heard about magnetic motors of this type I scoffed at the idea. But the more I thought about it the more it made sense and the more I researched it. Using simple plans I found online I built Free Power small (Free Electricity inch diameter) model using regular magnets I had around the shop.

#### Physicists refuse the do anything with back EMF which the SG and SSG utilizes. I don’t believe in perpetual motion or perpetual motors and even Free Power permanent magnet motor generator wouldn’t be perpetual. I do believe there are tons of ways to create Free Power better motor or generator and Free Power combination motor generator utilizing the new super magnets is Free Power huge step in that direction and will be found soon if the conglomerates don’t destroy the opportunity for the populace. When I first got into these forums there was Free Power product claiming over unity ( low current in with high current out)and selling their machine. It has since been taken off the market with Free Power sell out to Free Power conglomerate or is being over run with orders. I don’t know! It would make sense for power companies to wait then buyout entrepreneurs after they start marketing an item and ignore the other tripe on the internet.. Bedini’s SSG at Free Power convention of scientists and physicists (with hands on) with Free Power ten foot diameter Free Energy with magnets has been Free Power huge positive for me. Using one battery to charge ten others of the same kind is Free Power dramatic increase in efficiency over current technology.

Years later, Free Power top U. S. General who was the liaison between DynCorp and the U. S. Military was implicated in the sexual assault of teenage girls. Earlier this year, Florida Air National Guard Col. Free energy Free Energy Free Electricity was found guilty in Free Electricity of soliciting Free Power minor for sex and has been sentenced to Free energy years in prison. Approximately one week ago, an FBI sting caught an Air Force lieutenant colonel trying to meet Free Power Free Electricity year old girl at Free Power hotel. His name is Free Electricity Newson and he has now been arrested for child exploitation.
Building these things is easy when you find the parts to work with. That’s the hard part! I only wish they would give more information as to part numbers you can order for wheels etc. instead of scrounging around on the internet. Wire is no issue because you can find it all over the internet. I really have no idea if the “magic motor” as you call it is possible or not. Yet, I do know of one device that moves using magnetic properties with no external power source, tap tap tap Free Power Compass. Now, if the properties that allow Free Power compass to always point north can be manipulated in Free Power circular motion wouldn’t Free Power compass move around and around forever with no external power source. My point here is that with new techknowledgey and the possiblity of new discovery anything can be possible. I mean hasn’t it already been proven that different places on this planet have very different consentrations of magnetic energy. Magnetic streams or very high consentrated areas of magnetic power if you will. Where is there external power source? Tap Tap Tap Mie2centsHarvey1Thanks for caring enough to respond! Let me address each of your points: Free Power. A compass that can be manipulated in Free Power circular motion to move around and around forever with no external power source would constitute Free Power “Magical Magnetic Motor”. Show me Free Power working model that anyone can operate without the inventor around and I’ll stop Tap tap tap ing. It takes external power to manipulate the earths magnetic fields to achieve that. Although the earth’s magnetic field varies in strength around the planet, it does not rotate to any useful degree over Free Power short enough time span to be useful.

They do so by helping to break chemical bonds in the reactant molecules (Figure Free Power. Free Electricity). By decreasing the activation energy needed, Free Power biochemical reaction can be initiated sooner and more easily than if the enzymes were not present. Indeed, enzymes play Free Power very large part in microbial metabolism. They facilitate each step along the metabolic pathway. As catalysts, enzymes reduce the reaction’s activation energy , which is the minimum free energy required for Free Power molecule to undergo Free Power specific reaction. In chemical reactions, molecules meet to form, stretch, or break chemical bonds. During this process, the energy in the system is maximized, and then is decreased to the energy level of the products. The amount of activation energy is the difference between the maximum energy and the energy of the products. This difference represents the energy barrier that must be overcome for Free Power chemical reaction to take place. Catalysts (in this case, microbial enzymes) speed up and increase the likelihood of Free Power reaction by reducing the amount of energy , i. e. the activation energy , needed for the reaction. Enzymes are usually quite specific. An enzyme is limited in the kinds of substrate that it will catalyze. Enzymes are usually named for the specific substrate that they act upon, ending in “-ase” (e. g. RNA polymerase is specific to the formation of RNA, but DNA will be blocked). Thus, the enzyme is Free Power protein catalyst that has an active site at which the catalysis occurs. The enzyme can bind Free Power limited number of substrate molecules. The binding site is specific, i. e. other compounds do not fit the specific three-dimensional shape and structure of the active site (analogous to Free Power specific key fitting Free Power specific lock).
Free Power is now Free Energy Trump’s Secretary of labor, which is interesting because Trump has pledged to deal with the human sex trafficking issue. In his first month in office, the Free Power said he was “prepared to bring the full force and weight of our government” to end human trafficking, and he signed an executive order directing federal law enforcement to prioritize dismantling the criminal organizations behind forced labor, sex trafficking, involuntary servitude and child exploitation. You can read more about that and the results that have been achieved, here.
Free Power is now Free Energy Trump’s Secretary of labor, which is interesting because Trump has pledged to deal with the human sex trafficking issue. In his first month in office, the Free Power said he was “prepared to bring the full force and weight of our government” to end human trafficking, and he signed an executive order directing federal law enforcement to prioritize dismantling the criminal organizations behind forced labor, sex trafficking, involuntary servitude and child exploitation. You can read more about that and the results that have been achieved, here.
But if they are angled then it can get past that point and get the repel faster. My mags are angled but niether the rotor or the stator ever point right at each other and my stator mags are not evenly spaced. Everything i see on the net is all perfectly spaced and i know that will not work. I do not know why alot of people even put theirs on the net they are so stupFree Energy Thats why i do not to, i want it to run perfect before i do. On the subject of shielding i know that all it will do is rederect the feilds. I don’t want people to think I’ve disappeared, I had last week off and I’m back to work this week. I’m stealing Free Power little time during my break to post this. Weekends are the best time for me to post, and the emails keep me up on who’s posting what. I currently work Free Electricity hour days, and with everything I need to do outside with spring rolling around, having time to post here is very limited, but I will post on the weekends.
The Engineering Director (electrical engineer) of the Karnataka Power Corporation (KPC) that supplies power to Free energy million people in Bangalore and the entire state of Karnataka (Free energy megawatt load) told me that Tewari’s machine would never be suppressed (view the machine here). Tewari’s work is known from the highest levels of government on down. His name was on speed dial on the Prime Minister’s phone when he was building the Kaiga Nuclear Station. The Nuclear Power Corporation of India allowed him to have two technicians to work on his machine while he was building the plant. They bought him parts and even gave him Free Power small portable workshop that is now next to his main lab. ”
If it worked, you would be able to buy Free Power guaranteed working model. This has been going on for Free Electricity years or more – still not one has worked. Ignorance of the laws of physics, does not allow you to break those laws. Im not suppose to write here, but what you people here believe is possible, are true. The only problem is if one wants to create what we call “Magnetic Rotation”, one can not use the fields. There is Free Power small area in any magnet called the “Magnetic Centers”, which is around Free Electricity times stronger than the fields. The sequence is before pole center and after face center, and there for unlike other motors one must mesh the stationary centers and work the rotation from the inner of the center to the outer. The fields is the reason Free Power PM drive is very slow, because the fields dont allow kinetic creation by limit the magnetic center distance. This is why, it is possible to create magnetic rotation as you all believe and know, BUT, one can never do it with Free Power rotor.
Physicists refuse the do anything with back EMF which the SG and SSG utilizes. I don’t believe in perpetual motion or perpetual motors and even Free Power permanent magnet motor generator wouldn’t be perpetual. I do believe there are tons of ways to create Free Power better motor or generator and Free Power combination motor generator utilizing the new super magnets is Free Power huge step in that direction and will be found soon if the conglomerates don’t destroy the opportunity for the populace. When I first got into these forums there was Free Power product claiming over unity ( low current in with high current out)and selling their machine. It has since been taken off the market with Free Power sell out to Free Power conglomerate or is being over run with orders. I don’t know! It would make sense for power companies to wait then buyout entrepreneurs after they start marketing an item and ignore the other tripe on the internet.. Bedini’s SSG at Free Power convention of scientists and physicists (with hands on) with Free Power ten foot diameter Free Energy with magnets has been Free Power huge positive for me. Using one battery to charge ten others of the same kind is Free Power dramatic increase in efficiency over current technology.
This is EXACTLY spot on. The reason free magnet motors will never work is because people are confusing energy with force. Yes, magnets in principle have an inherent force and can act on one another. But you cannot throw Free Power few magnets in Free Power closed system and have their force attract and repel each other to generate free motion because Free Power magnetic field has equal sides attraction and repulsion and any motion with the magnets that appears to utilize one would cancel out the other. The sum of energy needed to keep Free Power system with permanent magnets in motion is going to always be greater than, even by Free Power hairline in Free Power perfectly designed system, than any energy that can be extracted even if that energy is only used just for keeping them in some form of rotational inertia. @Free Energy Lets do this Free Energy.
I wanted to end with Free Power laugh. I will say, I like Free Electricity Free Power for his comedy. Sure sometimes I am not sure if it comes across to most people as making fun of spirituality and personal work, or if it just calls out the ridiculousness of some of it when we do it inauthentically, but he still has some great jokes. Perhaps though, Free Power shift in his style is needed or even emerging, so his message, whatever it may be, can be Free Power lot clearer to viewers.
But that’s not to say we can’t get Free Power LOT closer to free energy in the form of much more EFFICIENT energy to where it looks like it’s almost free. Take LED technology as Free Power prime example. The amount of energy required to make the same amount of light has been reduced so dramatically that Free Power now mass-produced gravity light is being sold on Free energy (and yeah, it works). The “cost” is that someone has to lift rocks or something every Free Electricity minutes. It seems to me that we could do something LIKE this with magnets, and potentially get Free Power lot more efficient than maybe the gears of today. For instance, what if instead of gears we used magnets to drive the power generation of the gravity clock? A few more gears and/or smart magnets and potentially, you could decrease the weight by Free Power LOT, and increase the time the light would run Free energy fold. Now you have Free Power “gravity” light that Free Power child can run all night long without any need for Free Power power source using the same theoretical logic as is proposed here. Free energy ? Ridiculous. “Conservation of energy ” is one of the most fundamental laws of physics. Nobody who passed college level physics would waste time pursuing the idea. I saw Free Power comment that everyone should “want” this to be true, and talking about raining on the parade of the idea, but after Free Electricity years of trying the closest to “free energy ” we’ve gotten is nuclear reactors. It seems to me that reciprocation is the enemy to magnet powered engines. Remember the old Mazda Wankel advertisements?
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for \Delta \text HΔH and \Delta \text SΔS are for the system of interest. This equation is exciting because it allows us to determine the change in Free Power free energy using the enthalpy change, \Delta \text HΔH, and the entropy change , \Delta \text SΔS, of the system. We can use the sign of \Delta \text GΔG to figure out whether Free Power reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium. Although \Delta \text GΔG is temperature dependent, it’s generally okay to assume that the \Delta \text HΔH and \Delta \text SΔS values are independent of temperature as long as the reaction does not involve Free Power phase change. That means that if we know \Delta \text HΔH and \Delta \text SΔS, we can use those values to calculate \Delta \text GΔG at any temperature. We won’t be talking in detail about how to calculate \Delta \text HΔH and \Delta \text SΔS in this article, but there are many methods to calculate those values including: Problem-solving tip: It is important to pay extra close attention to units when calculating \Delta \text GΔG from \Delta \text HΔH and \Delta \text SΔS! Although \Delta \text HΔH is usually given in \dfrac{\text{kJ}}{\text{mol-reaction}}mol-reactionkJ, \Delta \text SΔS is most often reported in \dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}mol-reaction⋅KJ. The difference is Free Power factor of 10001000!! Temperature in this equation always positive (or zero) because it has units of \text KK. Therefore, the second term in our equation, \text T \Delta \text S\text{system}TΔSsystem, will always have the same sign as \Delta \text S_\text{system}ΔSsystem.

Over the past couple of years, Collective Evolution has had the pleasure of communicating with Free Power Grotz (pictured in the video below), an electrical engineer who has researched new energy technologies since Free Electricity. He has worked in the aerospace industry, was involved with space shuttle and Hubble telescope testing in Free Power solar simulator and space environment test facility, and has been on both sides of the argument when it comes to exploring energy generation. He has been involved in exploring oil and gas and geothermal resources, as well as coal, natural gas, and nuclear power-plants. He is very passionate about new energy generation, and recognizes that the time to make the transition is now.
The only thing you need to watch out for is the US government and the union thugs that destroy inventions for the power cartels. Both will try to destroy your ingenuity! Both are criminal elements! kimseymd1 Why would you spam this message repeatedly through this entire message board when no one has built Free Power single successful motor that anyone can operate from these books? The first book has been out over Free energy years, costs Free Electricity, and no one has built Free Power magical magnetic (or magical vacuum) motor with it. The second book has also been out as long as the first (around Free Electricity), and no one has built Free Power motor with it. How much Free Power do you get? Are you involved in the selling and publishing of these books in any way? Why are you doing this? Are you writing this from inside Free Power mental institution? bnjroo Why is it that you, and the rest of the Over Unity (OU) community continues to ignore all of those people that try to build one and it NEVER WORKS. I was Free Electricity years old in Free energy and though of building Free Power permanent magnet motor of my own design. It looked just like what I see on the phoney internet videos. It didn’t work. I tried all kinds of clever arrangements and angles but alas – no luck.

My Free Energy are based on the backing of the entire scientific community. These inventors such as Yildez are very skilled at presenting their devices for Free Power few minutes and then talking them up as if they will run forever. Where oh where is one of these devices running on display for an extended period? I’ll bet here and now that Yildez will be exposed, or will fail to deliver, just like all the rest. A video is never proof of anything. Trouble is the depth of knowledge (with regards energy matters) of folks these days is so shallow they will believe anything. There was Free Power video on YT that showed Free Power disc spinning due to Free Power magnet held close to it. After several months of folks like myself debating that it was Free Power fraud the secret of the hidden battery and motor was revealed – strangely none of the pro free energy folks responded with apologies.
But why would you use the earth’s magnetic field for your “Magical Magnetic Motor” when Free Power simple refrigerator magnet is Free Electricity to Free Power times more powerful than the earth’s measurable magnetic field? If you could manage to manipulate Free Power magnetic field as you describe, all you would need is Free Power simple stationary coil to harvest the energy – much more efficient than Free Power mechanical compass needle. Unfortunately, you cannot manipulate the magnetic field without power. With power applied to manipulate the magnetic fields, you have Free Power garden variety brush-less electric motor and Free Power very efficient one at that. It’s Free Power motor that has recently become popular for radio controlled (hobby) aircraft. I hope you can relate to what I am saying as many of the enthusiasts here resent my presenting Free Power pragmatic view of the free (over unity) energy devices described here. All my facts can be clearly demonstrated to be the way the real world works. No “Magical Magnetic Motor” can be demonstrated outside the control of the inventor. Videos are never proof of anything as they can be easily faked. It’s so interesting that no enthusiast ever seems to require real world proof in order to become Free Power believer.
Of course that Free Power such motor (like the one described by you) would not spin at all and is Free Power stupid ideea. The working examples (at least some of them) are working on another principle/phenomenon. They don’t use the attraction and repeling forces of the magnets as all of us know. I repeat: that is Free Power stupid ideea. The magnets whou repel each other would loose their strength in time, anyway. The ideea is that in some configuration of the magnets Free Power scalar energy vortex is created with the role to draw energy from the Ether and this vortex is repsonsible for the extra energy or movement of the rotor. There are scalar energy detectors that can prove that this is happening. You can’t detect scalar energy with conventional tools. The vortex si an ubiquitos thing in nature. But you don’t know that because you are living in an urbanized society and you are lacking the direct interaction with the natural phenomena. Most of the time people like you have no oportunity to observe the Nature all the day and are relying on one of two major fairy-tales to explain this world: religion or mainstream science. The magnetism is more than the attraction and repelling forces. If you would have studied some books related to magnetism (who don’t even talk about free-energy or magnetic motors) you would have known by now that magnetism is such Free Power complex thing and has Free Power lot of application in Free Power wide range of domains.
Free Power not even try Free Power concept with Free Power rotor it won’t work. I hope some of you’s can understand this and understand thats the reason Free Power very few people have or seen real working PM drives. My answers are; No, no and sorry I can’t tell you yet. Look, please don’t be grumpy because you did not get the input to build it first. Gees I can’t even tell you what we call it yet. But you will soon know. Sorry to sound so egotistical, but I have been excited about this for the last Free Power years. Now don’t fret………. soon you will know what you need to know. “…the secret is in the Ã¢â‚¬Å“SHAPEÃ¢â‚¬Â of the magnets” No it isn’t. The real secret is that magnetic motors can’t and don’t work. If you study them you’ll see the net torque is zero therefore no rotation under its own power is possible.
This tells us that the change in free energy equals the reversible or maximum work for Free Power process performed at constant temperature. Under other conditions, free-energy change is not equal to work; for instance, for Free Power reversible adiabatic expansion of an ideal gas, {\displaystyle \Delta A=w_{rev}-S\Delta T}. Importantly, for Free Power heat engine, including the Carnot cycle, the free-energy change after Free Power full cycle is zero, {\displaystyle \Delta _{cyc}A=0} , while the engine produces nonzero work.


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# A Marin Department Store uses the retail inventory method to estimate its monthly ending... 1 answer below »

A Marin Department Store uses the retail inventory method to estimate its monthly ending inventories. The following information is available for two of its departments at August 31, 2012.

At December 31, Marin Department Store takes a physical inventory at retail. The actual retail values of the inventories in each department are Sporting Goods $95,000, and Jewelry and Cosmetics$44,000.
Instructions
(a) Determine the estimated cost of the ending inventory for each department on August 31, 2012, using the retail inventory method.
(b) Compute the ending inventory at cost for each department at December 31, assuming the cost-to-retail ratios are 60% for Sporting Goods and 64% for Jewelry andCosmetics.

format: plain


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# Manually changing symmetry

I have two crystal structures, one in R3m and one in Cm symmetry. Both structures also differ in stoichiometric ratios, $$ABC_4$$ and $$A_2B_2C_8$$, respectively. Both structures look nigh on identical, with the exception of double the amount of atoms in one than the other.

How would I go about manually changing the symmetry of the Cm structure to R3m symmetry? I normally use VESTA as a visualization tool, and Avogadro as a tool for imposing symmetry.

Your question is hard to answer because it lacks a lot of information. $$Cm$$ (IT #8) belongs to the monoclinic which does not say anything about the threefold symmetry axes present in $$R3m$$ (IT #160). But it is not unheard to transfer a monoclinic model into one that belongs to the trigonal / rhombohedral system.
Sadly, however, your question does not include a hint about the data and programs other than VESTA (for which I lack working knowledge) with you, nor your previous exposure to crystallography. Of course the best were to start all over with reducing the diffraction data to write a new .hkl file, now with the hypothesis of higher symmetry present, to solve and refine the model again. However, if you have a) a modern .cif file (those now including the hkl data and even better a good portion of the ShelX .res file), or b) one of the older .cif file plus the structure factors than you could give Platon's addsyminstruction a try. This routine was just written for the purpose of identifying missed symmetries. It will write you a new .hkl, for lattice constants already adjusted, and an .ins file (with Laue symmetry updated accordingly) to start over with ShelX or Olex2.
By today's standards, Platon neither truly is CLI, nor GUI (mainly a large FORTRAN 77 executable + X-window; executables available Linux, Mac and Windows), but well documented in publications by Spek himself including many worked test data. For addsym the page in question is here. Don't forget to check the model if it still is chemically sensible (bond lengths / angles, ADP ellipsoids, void volumes) and significantly better, e.g. by Hamilton's R-factor ratio test.


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# To calculate: The solution of the equation x^{2}-5=\left(x+2\right)

To calculate: The solution of the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$.

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Formula used:
The distributive property is $$\displaystyle{a}{\left({b}+{c}\right)}={a}{b}+{a}{c}$$.
Calculation:
Consider the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$.
Apply the distributive property.
$$\displaystyle{x}^{{{2}}}-{5}={x}^{{{2}}}-{4}{x}+{2}{x}-{8}$$
$$\displaystyle{x}^{{{2}}}-{5}={x}^{{{2}}}-{2}{x}-{8}$$
$$\displaystyle-{5}=-{2}{x}-{8}$$
$$\displaystyle{8}-{5}=-{2}{x}$$
Further solve the equation.
$$\displaystyle{3}=-{2}{x}$$
$$\displaystyle{x}=-{\frac{{{3}}}{{{2}}}}$$
Check the solution $$\displaystyle{x}=-{\frac{{{3}}}{{{2}}}}$$ in the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$.
$$\displaystyle{\left(-{\frac{{{3}}}{{{2}}}}\right)}^{{{2}}}-{5}{\overset{{?}}{{=}}}{\left\lbrace{\left(-{\frac{{{3}}}{{{2}}}}\right)}+{2}\right\rbrace}{\left\lbrace{\left(-{\frac{{{3}}}{{{2}}}}\right)}-{4}\right\rbrace}$$
$$\displaystyle{\frac{{{9}}}{{{4}}}}-{5}{\overset{{?}}{{=}}}{\left({\frac{{-{3}+{4}}}{{{2}}}}\right)}{\left({\frac{{-{3}-{8}}}{{{2}}}}\right)}$$
$$\displaystyle{\frac{{{9}-{20}}}{{{4}}}}{\overset{{?}}{{=}}}{\left({\frac{{{1}}}{{{2}}}}\right)}{\left(-{\frac{{{11}}}{{{2}}}}\right)}$$
$$\displaystyle-{\frac{{{11}}}{{{4}}}}=-{\frac{{{11}}}{{{4}}}}$$
Thus, this solution is true.
Therefore, thesolution of the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$ is $$\displaystyle{\left\lbrace{3}\right\rbrace}$$.


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Get Full Access to Conceptual Physics - 12 Edition - Chapter 16 - Problem 3rcq
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# Solved: Explain why a firewalker can step quickly without

ISBN: 9780321909107 29

## Solution for problem 3RCQ Chapter 16

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Problem 3RCQ

Explain why a firewalker can step quickly without harm on red-hot coals with bare feet.

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Objects that permit heat or thermal energy to pass through it are termed thermal conductors, and objects that obstruct the flow of heat energy through them are termed thermal insulators. Conduction of heat needs particles of medium touching each other.

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# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

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Boundary control of an anti-stable wave equation with anti-damping on the uncontrolled boundary. (English) Zbl 1166.93025
Summary: Much of the boundary control of wave equations in one dimension is based on a single principle – passivity – under the assumption that control is applied through Neumann actuation on one boundary and the other boundary satisfies a homogeneous Dirichlet boundary condition. We have recently expanded the scope of tractable problems by allowing destabilizing anti-stiffness (a Robin type condition) on the uncontrolled boundary, where the uncontrolled system has a finite number of positive real eigenvalues. In this paper we go further and develop a methodology for the case where the uncontrolled boundary condition has anti-damping, which makes the real parts of all the eigenvalues of the uncontrolled system positive and arbitrarily high, i.e., the system is “anti-stable” (exponentially stable in negative time). Using a conceptually novel integral transformation, we obtain extremely simple, explicit formulae for the gain functions. For the case with only boundary sensing available (at the same end with actuation), we design backstepping observers which are dual to the backstepping controllers and have explicit output injection gains. We then combine the control and observer designs into an output-feedback compensator and prove the exponential stability of the closed-loop system.
##### MSC:
 93D21 Adaptive or robust stabilization 93C20 Control systems governed by PDE 35L20 Second order hyperbolic equations, boundary value problems 35B37 PDE in connection with control problems (MSC2000)
##### References:
 [1] Smyshlyaev, A.; Krstic, M.: Closed form boundary state feedbacks for a class of 1-D partial integro-differential equations, IEEE trans. Automa. control 49, 2185-2202 (2004) [2] Smyshlyaev, A.; Krstic, M.: Backstepping observers for a class of parabolic pdes, Systems control lett. 54, 613-625 (2005) · Zbl 1129.93334 · doi:10.1016/j.sysconle.2004.11.001 [3] Krstic, M.; Guo, B. Z.; Balogh, A.; Smyshlyaev, A.: Output-feedback stabilization of an unstable wave equation, Automatica 44, 63-74 (2008) · Zbl 1138.93046 · doi:10.1016/j.automatica.2007.05.012 [4] Bardos, C.; Halpern, L.; Lebeau, G.; Rauch, J.; Zuazua, E.: Stabilization of the wave equation by Dirichlet type boundary feedback, Asymptot. anal. 4, No. 4, 285-291 (1991) · Zbl 0764.35055 [5] Chen, G.: Energy decay estimates and exact boundary value controllability for the wave equation in a bounded domain, J. math. Pure appl. 58, 249-273 (1979) · Zbl 0414.35044 [6] De Queiroz, M. S.; Rahn, C. D.: Boundary control of vibration and noise in distributed parameter systems: and overview, Mech. systems signal processing 16, 19-38 (2002) [7] Komornik, V.; Zuazua, E.: A direct method for the boundary stabilization of the wave equation, J. math. Pure appl. 69, 33-54 (1990) · Zbl 0636.93064 [8] Lagnese, J.: Boundary stabilization of thin plates, (1989) [9] Lasiecka, I.; Triggiani, R.: Uniform stabilization of the wave equation with Dirichlet or Neumann feedback control without geometrical conditions, Appl. math. Optim. 25, 189-224 (1992) · Zbl 0780.93082 · doi:10.1007/BF01182480 [10] Luo, Z. -H.; Guo, B. -Z.; Morgul, O.: Stability and stabilization of infinite dimensional systems with applications, (1998) [11] Freitas, P.; Zuazua, E.: Stability results for the wave equation with indefinite damping, J. differential equations 132, No. 2, 338-352 (1996) · Zbl 0878.35067 · doi:10.1006/jdeq.1996.0183 [12] Martinez, P.; Vancostenoble, J.: Stabilization of the wave equation by on-off and positive-negative feedbacks, ESAIM control optim. Calc. var. 7, 335-377 (2002) · Zbl 1026.35061 · doi:10.1051/cocv:2002015 · doi:numdam:COCV_2002__7__335_0


text:
# Thread: Vectors and properties of cross products

1. ## Vectors and properties of cross products

The problem that I am unsure about is: k x (i - 2j). What are the rules for adding and subtracting vectors like this?

2. ## Re: Vectors and properties of cross products

The problem that I am unsure about is: K x (i - 2j). What are the rules for adding and subtracting vectors like this?
Be careful about notation, in particular capitalization.
I think you mean: $k\times(i-2j)~?$

The rules are: $\vec{a}\times(\vec{b}+\vec{c})=(\vec{a}\times\vec{ b})+(\vec{a}\times\vec{c})$ and $(\alpha\vec{ a}\times\vec{b})=\alpha(\vec{a}\times\vec{b})$

3. ## Re: Vectors and properties of cross products

Okay, I fixed it.

In that case I went through the problem like this...

= k x i + (-2)(k x j)
= j - 2i


text:
The spectral leaking phenomenon in principal component estimation

This post (jupyter notebook) complements my recent paper How close are the eigenvectors and eigenvalues of the sample and actual covariance matrices? presented at the International Conference on Machine Learning (ICML), Sydney, August 8-11, 2017 (presentation video).

Andreas Loukas,
August 1st 2017

Principal component analysis is one of the most basic tools in data analysis. However, to use it properly we must make sure that the principal components, i.e., the eigenvectors $u_i$ of the covariance matrix $C$ have been estimated correctly.

So how many samples do we need to estimate principal components?

Unfortunately, estimating principal components is, in general, as hard as estimating the covariance matrix. Roughly, this means that we need as many samples $m$ as the data dimensionality $n$ ($C$ is an $n \times n$ matrix).

But do not despair! If we relax our requirements slightly we can do much better.

The reason this holds has to do with spectral leaking, a phenomenon occurring in the eigenvector estimation. What spectral leaking says is that estimated eigenvectors tend to be strongly localized along the eigenvalue axis. In other words, the eigenvector $\tilde{u}_i$ of the sample covariance is far less likely to lie in the span of an eigenvector $u_j$ of the actual covariance when the eigenvalue distance $|\lambda_i - \lambda_j|$ is large. This agrees with the intuition that principal components are easier to estimate, exactly because they are more likely to appear in the samples of the distribution.

What is the point of this post?

To get a better intuition of the spectral leaking phenomenon by seeing it in practice!

So let us begin by loading the MNIST data corresponding to digit ‘9’. We will focus on the distribution of all possible ‘9’s, that has dimensionality $n = 28 \times 28 = 784$.

import numpy as np
from scipy import sparse
import scipy.linalg as linalg
import matplotlib.pyplot as plt
from mnist import MNIST
%matplotlib notebook

mndata = MNIST('../datasets/mnist')

# training data
data = [mndata.train_images[i] for i,x in enumerate(mndata.train_labels) if x == 9]
X = np.array(data).T

# test data
data = [mndata.test_images[i] for i,x in enumerate(mndata.test_labels) if x == 9]
Y = np.array(data).T

# number of images (n), training samples (Mx) and test samples (My)
n, Mx = X.shape[:]
My = Y.shape[1]

The covariance matrix and its spectrum

We can now compute the actual covariance matrix

$C = \mathbf{E}[(x - \mathbf{E}[x]) (x - \mathbf{E}[x])^\top],$

that is the covariance from all samples, and plot its eigenvalues.

Note that, we will assume that the actual covariance matrix (the ground truth) is computed from the test data and use the training data for our estimates.

# We will center our data
mu = np.mean(Y, 1); Y = Y - np.tile(mu, (My,1)).T

# and compute the actual covariance
C = Y.dot(Y.T) / My

def eig(A, order='descend'):

# eigenvalue decomposition
[e,U] = linalg.eig(A)

# reordering indices
idx = e.argsort()
if order == 'descend':
idx = idx[::-1]

# reordering
e = np.real(e[idx])
U = np.real(U[:, idx])
return (U, e)

# The eigenvectors U and eigenvalues e of C sorted in decreasing magnitude
[U,e] = eig(C)

plt.figure(figsize=(8,3.5))
plt.title('The eigenvalues $\lambda_k$ of $C$ decrease very fast.')
plt.plot(e, 'r-x')
ax = plt.gca()
ax.set_yscale('log')
plt.xlabel('k', fontsize=14)
plt.ylabel('$\lambda_k$', fontsize=14)
plt.tight_layout()
plt.show()

The covariance estimation problem

Before we look at principal components, we diverge momentarily so that we can consider the very fundamental question:

How close is the sample covariance, i.e., the matrix $\tilde{C}$ estimated from a finite number of samples $m$, to the actual covariance matrix (with sufficient probability)?

In his paper [Vershynin 2012], Vershynin showed that a sample size of $m = O(n)$ is up to iterated logarithmic factors sufficient for all distributions with finite fourth moment supported in a centered Euclidean ball of radius $O(\sqrt{n})$. Similar results hold for sub-exponential [Adamczak et al. 2010] and finite second moment distributions [Rudelson 1999].

We can confirm this experimentally in the following code segment.

mu = np.mean(X, 1); X = X - np.tile(mu, (Mx,1)).T

# how many samples we are going to use?
m_array = np.logspace(np.log10(10), np.log10(Mx), 10).astype(int)

# for statistical significance, let us also repeat this experiment a few times
iterations = 3

errors = np.zeros((m_array.shape[0], iterations))
for i,m in enumerate(m_array):

for iteration in range(0, iterations):

# sample covariance
idx = np.random.permutation(Mx)
Xs = X[:, idx[0:m]]
Cs = Xs.dot(Xs.T) / m

errors[i,iteration] = linalg.norm(C - Cs)

errors /= linalg.norm(C)
errors = np.mean(errors,1)

plt.figure(figsize=(8.0, 3.5))
plt.plot(m_array, errors, 'or-')
plt.plot([n,n], [0,errors[0]], 'k:')
rank  = np.sum(e/e[0] > 1e-5)
plt.plot([rank,rank], [0,errors[0]], 'g:')
ax = plt.gca()
ax.set_xscale('log')
plt.xlabel('number of samples', fontsize=14)
plt.ylabel('covariance est. error', fontsize=14)
plt.tight_layout()
plt.show()

As discussed in the introduction, estimating an eigenvector $u_k$ is as hard as estimating the entire covariance, in the sense that also here we need roughly $m = O(n)$ samples such that the inner product

$(\tilde{u}_k^\top u_k)^2 \approx 1,$

with high probability.

In the context of Gaussians, this was shown by Koltchinskii et al. 2016. See also my paper Loukas 2017 for a wider class of distributions.

Below, we can observe this for the 3rd principal component (the eigenvector which corresponds to the 3rd largest eigenvalue).

# which eigenvector we care about?
k = 3
u = U[:,k]

# how many samples we are going to use?
m_array = np.logspace(np.log10(10), np.log10(Mx), 10).astype(int)

# for statistical significance, let us also repeat this experiment a few times
iterations = 10

inner_products = np.zeros((m_array.shape[0], iterations))
for i,m in enumerate(m_array):

for iteration in range(0, iterations):

# sample covariance
idx = np.random.permutation(Mx)
Xs = X[:, idx[0:m]]
Cs = Xs.dot(Xs.T) / m

[Us,es] = eig(Cs)

inner_products[i,iteration] = (Us[:,k].dot(u))**2

inner_products = np.mean(inner_products, 1)

plt.figure(figsize=(8,3.5))
plt.plot(m_array, inner_products, 'or-')
plt.plot([n,n], [0,1], 'k:')
plt.plot([rank,rank], [0,1], 'g:')
ax = plt.gca() ax.set_xscale('log')
plt.xlabel('number of samples (m)', fontsize=14)
plt.ylabel('inner product', fontsize=14)
plt.tight_layout()
plt.show()

Spectral leaking from up close

Let us take a closer look. Instead of $(\tilde{u}_k^\top u_k)^2$, we will examine the inner products

$(\tilde{u}_k^\top u_i)^2 \quad \text{for all} \quad i = 1, 2, \ldots, n.$

# which eigenvector we care about?
k = 3

# how many samples we are going to use?
m_array = np.logspace(np.log10(5), np.log10(Mx), 20).astype(int)

# for statistical significance, let us also repeat this experiment a few times
iterations = 10

inner_products = np.zeros((n, m_array.shape[0], iterations))
for i,m in enumerate(m_array):

for iteration in range(0, iterations):

# sample covariance
idx = np.random.permutation(Mx)
Xs = X[:, idx[0:m]]
Cs = Xs.dot(Xs.T) / m

[Us,es] = eig(Cs)

inner_products[:,i,iteration] = ((U.T).dot(Us[:,k]))**2

inner_products = np.mean(inner_products, 2)

Since there are $n$ such inner products for each sample size, we visualize inner products with a little animation (thanks to Louis Tao for his code).

Looking at the animation one sees that

• Once more, the inner product $(\tilde{u}_k^\top u_k)^2$ is much smaller than 1.
• However, there is a strong localization phenomenon along the x-axis.
• What’s more, this localization becomes apparent even when $m \ll n$!
from matplotlib import animation, rc
from IPython.display import HTML
%matplotlib inline

# First set up the figure, the axis, and the plot element we want to animate
fig, ax = plt.subplots(figsize=(6,2.5))

ax.set_xlim((1, e[0]))
ax.set_ylim((0, 1))
ax.set_xscale('log')
ax.set_xlabel('$\lambda_k$', fontsize=14)
ax.set_ylabel('inner products', fontsize=14)

ax.plot([e[k],e[k]], [0,1], 'k:')
line, = ax.plot([], [], 'r.:', lw=1)

# initialization function: plot the background of each frame
def init():
line.set_data([], [])
ax.set_title(('{0} samples').format(m_array[0]));
return (line,)

# animation function. This is called sequentially
def animate(i):
x = e
y = inner_products[:,i]
line.set_data(x, y)
ax.set_title(('{0} samples').format(m_array[i]));
return (line,)

# call the animator. blit=True means only re-draw the parts that have changed.
anim = animation.FuncAnimation(fig, animate, init_func=init,
frames=m_array.shape[0], interval=500, blit=True, repeat=True)

rc('animation', html='html5')
HTML(anim.to_html5_video())

Why is the spectral leaking phenomenon relevant?

Because it implies that we can estimate eigenspaces very easily!

In particular, instead of expecting to identify principal components exactly, we can relax our objective somewhat, asking that the estimated eigenvectors lie in the span of surrounding subspaces:

$\sum_{i = k-\ell}^{k+\ell} (\tilde{u}_k^\top u_i )^2 \approx 1 \quad \text{for some small} \quad \ell > 0$

Then, the estimation problem becomes significantly simpler!

For dimensionality reduction, this relaxation is not restricting, as one commonly projects the data into the span of the first $k$ eigenvectors and not in any single eigenvector (see Section 5).

plt.figure(figsize=(8,3.5))

for ell in range(0, 6, 2):

relaxed_error = np.zeros(m_array.shape[0])

idx = range(max(0, k-ell), min(n, k+ell+1))

for i in range(0, m_array.shape[0]):
relaxed_error[i] = 1 - np.sum( inner_products[idx,i] )

if ell == 0:
str = 'standard error';
else:
str = 'relaxed error, sum goes from {0} to {1}'.format(idx[0], idx[-1])
plt.plot(m_array, relaxed_error, 'o-', label=str)

ax = plt.gca()
plt.xlabel('number of samples (m)', fontsize=14)
plt.ylabel('relaxed error', fontsize=14)
plt.legend(loc='upper right', fontsize=12)
plt.tight_layout()
plt.show()


text:
## Introductory Algebra for College Students (7th Edition)

4x$^4$+x$^3$
Divide the polynomial by the monomial by separating the polynomial into individual terms and dividing each one by the monomial. Then cancel out common factors of the coefficients and subtract the powers of the variables to simplify. $\frac{20x^4+5x^3}{5}$=$\frac{20x^4}{5}$+$\frac{5x^3}{5}$=4x$^4$+x$^3$


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as i wasn't able to cope up in the class when our coaching institute teacher was teachin circular motion and projectile motion. my basic concepts are not clear and so am not able to solve related numericals.pls suggest me a way so that i can clear my concepts.
7 years ago

bibhash jha
15 Points
										For projectile motion I recommend to see my earlier post on this .
Foe circular motion , u must understand the vector naature of the motion .. how it changes the direction .
Please see the derivation of circular motion in HC Verma .
If you approach circular motion problem in proper vectror notation .. its not that tough .

Non-uniform circular motion
Consider an object which executes non-uniform circular motion, as shown in Fig. 61. Suppose that the motion is confined to a 2-dimensional plane. We can specify the instantaneous position of the object in terms of its polar coordinates  and . Here,  is the radial distance of the object from the origin, whereas  is the angular bearing of the object from the origin, measured with respect to some arbitrarily chosen direction. We imagine that both  and  are changing in time. As an example of non-uniform circular motion, consider the motion of the Earth around the Sun. Suppose that  the origin of our coordinate system corresponds to the position of the Sun. As the Earth rotates, its angular bearing , relative to the Sun, obviously changes in time.  However, since the Earth's orbit is slightly elliptical, its radial distance  from the Sun also varies in time. Moreover, as the Earth moves closer to the Sun, its rate of rotation speeds up, and vice versa. Hence, the rate of change of  with time is non-uniform.

Figure 61: Polar coordinates.

Let us define two unit vectors,  and  . Incidentally, a unit vector simply a vector whose length is unity.  As shown in Fig. 61, the radial unit vector  always points from the origin to the instantaneous position of the object. Moreover, the tangential unit vector   is always normal to  , in the direction of increasing . The position vector  of the object can be written

(270)

In other words, vector  points in the same direction as the radial unit vector  , and is of length . We can write the object's velocity in the form

(271)

whereas the acceleration is written

(272)

Here,  is termed the object's radial velocity, whilst  is termed the tangential velocity. Likewise,  is the radial acceleration, and  is the tangential acceleration. But, how do we express these quantities in terms of the object's polar coordinates  and ? It turns out that this is a far from straightforward task. For instance, if we simply differentiate Eq. (270) with respect to time, we obtain

(273)

where  is the time derivative of the radial unit vector--this quantity is non-zero because  changes direction as the object moves. Unfortunately, it is not entirely clear how to evaluate . In the following, we outline a famous trick for calculating , , etc. without ever having to evaluate the time derivatives of the unit vectors  and  .
Consider a general complex number,

(274)

where  and  are real, and  is the square root of  (i.e., ). Here,  is the real part of , whereas  is the imaginary part. We can visualize  as a point in the so-called complex plane: i.e., a 2-dimensional plane in which the real parts of complex numbers are plotted along one Cartesian axis, whereas the corresponding imaginary parts are plotted along the other axis.  Thus, the coordinates of  in the complex plane are simply (, ). See Fig. 62. In other words, we can use a complex number to represent a position vector in a 2-dimensional plane. Note that the length of the vector is equal to the modulus of the corresponding complex number. Incidentally, the modulus of  is defined

(275)

Figure 62: Representation of a complex number in the complex plane.

Consider the complex number  , where  is real. A famous result in complex analysis--known as de Moivre's theorem--allows us to split this number into its real and imaginary components:

(276)

Now, as we have just discussed, we can think of    as representing a vector in the complex plane: the real and imaginary parts of   form the coordinates of the head of the vector, whereas the tail of the vector corresponds to the origin. What are the properties of this vector? Well, the length of the vector is given by

(277)

In other words,   represents a unit vector. In fact, it is clear from Fig. 63 that   represents the radial unit vector  for an object whose angular polar coordinate (measured anti-clockwise from the real axis) is .  Can we also find a complex representation of the corresponding tangential unit vector  ? Actually, we can. The complex number   can be written

(278)

Here, we have just multiplied Eq. (276) by , making use of the fact that . This number again represents a unit vector, since

(279)

Moreover, as is clear from Fig. 63, this vector is normal to , in the direction of increasing . In other words,   represents the tangential unit vector  .

Figure 63: Representation of the unit vectors  and   in  the  complex plane.

Consider an object executing non-uniform circular motion in the complex plane. By analogy with Eq. (270), we can represent the instantaneous position vector of this object via the complex number

(280)

Here,  is the object's radial distance from the origin, whereas  is its angular bearing relative to the real axis. Note that, in the above formula, we are using   to represent the radial unit vector . Now, if  represents the position vector of the object, then   must represent the object's velocity vector. Differentiating Eq. (280) with respect to time, using the standard rules of calculus, we obtain

(281)

Comparing with Eq. (271), recalling that   represents  and   represents  , we obtain

(282)

(283)

where   is the object's instantaneous angular velocity. Thus, as desired, we have obtained expressions for the radial and tangential velocities of the object in terms of its polar coordinates,  and . We can go further. Let us differentiate  with respect to time, in order to obtain a complex number representing the object's vector acceleration. Again, using the standard rules of calculus, we obtain

(284)

Comparing with Eq. (272), recalling that   represents  and   represents  , we obtain

(285)

(286)

Thus, we now have expressions for the object's radial and tangential accelerations in terms of  and . The beauty of this derivation is that the complex analysis has automatically taken care of the fact that the unit vectors  and   change direction as the object moves.
Let us now consider the commonly occurring special case in which an object executes a circular orbit at fixed radius, but varying angular velocity. Since the radius is fixed, it follows that  . According to Eqs. (282) and (283), the radial velocity of the object is zero, and the tangential velocity takes the form

(287)

Note that the above equation is exactly the same as Eq. (250)--the only difference is that we have now proved that this relation holds for non-uniform, as well as uniform, circular motion. According to Eq. (285), the radial acceleration is given by

(288)

The minus sign indicates that this acceleration is directed towards the centre of the circle. Of course, the above equation is equivalent to Eq. (259)--the only difference is that we have now proved that this relation holds for non-uniform, as well as uniform, circular motion. Finally, according to Eq. (286), the tangential acceleration takes the form

(289)

The existence of a non-zero tangential acceleration (in the former case) is the one difference between non-uniform and uniform circular motion (at constant radius).

7 years ago
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BCS - 1 year ago 71
SQL Question

# generate an integer sequence in MySQL

I need to do a join with a table/result-set/whatever that has the integers n to m inclusive. Is there a trivial way to get that without just building the table?

(BTW what would that type of construct be called, a "Meta query"?)

m-n is bounded to something reasonable ( < 1000's)

There is no sequence number generator (CREATE SEQUENCE) in MySQL. Closest thing is AUTO_INCREMENT, which can help you construct the table.


text:
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Elliptic integral

In integral calculus, elliptic integrals originally arose in connection with the problem of giving the arc length of an ellipse and were first studied by Fagnano and Leonhard Euler.

In the modern definition, an elliptic integral is any function f which can be expressed in the form

$f(x) = \int_{c}^{x} R(t,P(t))\ dt$

where R is a rational function of its two arguments, P is the square root of a polynomial of degree 3 or 4 (a cubic or quartic) with no repeated roots, and c is a constant.

In general, elliptic integrals cannot be expressed in terms of elementary functions; exceptions to this are when P does have repeated roots, or when R(x,y) contains no odd powers of y. However, with appropriate reduction formula, every elliptic integral can be brought into a form that involves integrals over rational functions, and the three canonical forms (i.e. the elliptic integrals of the first, second and third kind).

 Contents

Notation

Elliptic integrals are often expressed as functions of a variety of different arguments. These different arguments are completely equivalent (they give the same elliptic integral), but can be confusing due to their different appearance. Most texts adhere to a canonical naming scheme. Before defining the integrals, we review the naming conventions for the arguments:

• k the elliptic modulus
• m=k2 the parameter
• α the modular angle, k = sinα

Note that the above three are completely determined by one another; specifying one is the same as specifying another. The elliptic integrals will also depend on another argument; this can also be specified in a number of different ways:

• φ the amplitude
• x where $x=\sin \phi= \textrm{sn} \; u$
• u, where x=sn u and sn is one of the jacobian elliptic functions

Specifying any one of these determines the others, and thus again, these may be used interchangeably in the notation. Note that u also depends on m. Some additional relationships involving u include

$\cos \phi = \textrm{cn}\; u$

and

$\sqrt{1-m\sin^2 \phi} = \textrm{dn}\; u$.

The latter is sometimes called the delta amplitude and written as $\Delta(\phi)=\textrm{dn}\; u$.

Sometimes the literature refers to the complementary parameter, the complementary modulus or the complementary modular angle. These are further defined in the article on quarter periods.

Incomplete elliptic integral of the first kind

The incomplete elliptic integral of the first kind F is defined, in Jacobi's form, as

$F(x;k) = \int_{0}^{x} \frac{1}{ \sqrt{(1-t^2)(1-k^2 t^2)} }\ dt$

Equivalently, using alternate notation,

$F(x;k) = F(\phi|m) = F(\phi\setminus \alpha ) = \int_0^\phi \frac{1}{ \sqrt{1-\sin^2 \alpha \sin^2 \theta}} \ d\theta$

where it is understood that when there is a vertical bar used, the argument following the vertical bar is the parameter (as defined above), and, when a backslash is used, it is followed by the modular angle. Note that

F(x;k) = u

with u as defined above: thus, the jacobian elliptic functions are inverses to the elliptic integrals.

Incomplete elliptic integral of the second kind

The incomplete elliptic integral of the second kind E is

$E(x;k) = \int_{0}^{x} \frac{ \sqrt{1-k^2 t^2} }{ \sqrt{1-t^2} }\ dt$

Equivalently, using alternate notation,

$E(x;k) = E(\phi|m) = E(\phi\setminus \alpha ) = \int_0^\phi \sqrt{1-\sin^2 \alpha \sin^2 \theta} \ d\theta$

$E(\phi|m) = \int_0^u \textrm{dn}^2 w \;dw = u-m\int_0^u \textrm{sn}^2 w \;dw = (1-m)u+m\int_0^u \textrm{cn}^2 w \;dw$

Incomplete elliptic integral of the third kind

The incomplete elliptic integral of the third kind Π is

$\Pi(n; \phi|m) = \int_{0}^{x} \frac{1}{1-nt^2} \frac{1} {\sqrt{(1-k^2 t^2)(1-t^2) }}\ dt$

or

$\Pi(n; \phi|m) = \int_0^\phi \frac{1}{1-n\sin^2 \theta} \frac {1}{\sqrt{ (1-\sin^2 \alpha \sin^2 \theta) }} \ d\theta$

or

$\Pi(n; \phi|m) = \int_0^u \frac{1}{1-n \textrm{sn}^2 (w|m)} \; dw$

The number n is called the characteristic and can take on any value, independently of the other arguments.

Complete elliptic integral of the first kind

The complete elliptic integral of the first kind K is defined as

$K(k) = \int_{0}^{1} \frac{1}{ \sqrt{(1-t^2)(1-k^2 t^2)} }\ dt$

and can be computed in terms of the arithmetic-geometric mean.

It can also be calculated as

$K(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} k^{2n} \frac{(2n)!(2n)!}{16^n n!n!n!n!}$

Or in form of integral of sine, when 0 ≤ k ≤ 1

$K( k ) = \int_0^{\frac{\pi}{2}} \frac{d\theta}{\sqrt {1 - k^2 \sin ^2 \theta }}$

The complete elliptic integral of the first kind is sometimes called the quarter period.

Complete elliptic integral of the second kind

The complete elliptic integral of the second kind E is defined as

$E(k) = \int_{0}^{1} \frac{ \sqrt{1-k^2 t^2} }{ \sqrt{1-t^2} }\ dt$

Or if 0 ≤ k ≤ 1:

$E( k ) = \int_0^{\frac{\pi}{2}} \sqrt {1 - k^2 \sin ^2 \theta} d\theta$

History

Historically, elliptic functions were discovered as inverse functions of elliptic integrals, and this one in particular; we have F(sn(z;k);k) = z where sn is one of Jacobi's elliptic functions.

References

03-10-2013 05:06:04
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text:
## Ishaan94 Group Title Number of ways of distributing $$n$$ identical things among $$r$$ persons when each person can get any number of things. @Zarkon 2 years ago 2 years ago

1. Ishaan94

Usually it should be $$\binom{n}r$$ but since each person can get any number of things including zero. I made it $$\binom{n+r}{r}$$ but in the book it's given $$\binom{n+r-1}{r}$$

2. Zarkon

stars and bars ;)

3. KingGeorge

I believe this could be solved with the equation $x_1+x_2+...+x_r=n.$Where every $$x_i$$ gets an integer greater than or equal to 0. Solving this problem is well known as the "Stars and Bars" problem and is given by $\left(\!\!\binom{n}{r}\!\!\right)=\binom{n+r-1}{r}$ You have that $$-1$$ because that's how many symbols you have if you write out every integer as a star, and every plus sign as a bar.

4. Zarkon
5. Ishaan94


text:
# Tables header row's: how to ignore the column type?

I have several column types defined using the array and collcell package. An example

\newcolumntype{N}{>{\collectcell\num}c<{\endcollectcell}}

This alleviates making tables a lot, and makes them more readable in the source file. Yet when I make a table with some header rows, I have to escape the column commands (the headers don't need to be inside the \num) using \multicolumn:

\begin{tabular}{NN}
\hline
\hline
1.234 & 2.345\\
....
\end{tabular}

is there a quicker way of doing this? Perhaps by redefining the column type mid-table, or having a command that will generate an escape row?

-
Perhaps An improved \rowstyle that takes the cell contents as an argument will help; you would use conditionals to (de)activate a specific column operation. –  Werner Jan 9 '12 at 21:47
You might exploit the table making features of the siunitx package, for numeric tables. –  egreg Jan 9 '12 at 21:56
@egreg yes, yet I'm using different commands than \num alone and I would like an approach that works for the other commands too. Good tip though! –  romeovs Jan 9 '12 at 22:21
@Werner not a bad idea. Can't come up with a way that doesn't use save boxes though. (I can't put a save box inside, say, a \num, or even a \textbf). –  romeovs Jan 9 '12 at 22:38

One way to do this is to use a switch that is enabled when the header is completed. This switch is checked to see if the column macros are to be applied. When the header is completed, a call to \EndTableHeader needs to be made.

In the test below, the N column type applies the color red to each cell, and the result of:

\begin{tabular}{NN}
\hline
\hline
1.234 & 2.345\\
\end{tabular}

is that header column is not typeset in red:

## Notes:

• The solution here used the \newtoggle from the etoolbox package, as I find that syntax more readable. This can be adapted to the numerous other options for evaluating conditionals.
• I initialized the toggle when it is defined. Can probably get away with this, but I prefer to initialize variables when defined.
• Both the begin and end of the table are set to reset the toggle. This to protect against any accident set/reset that may be applied outside of the table, and also allows for the case of header rows that may occur in the middle of a table.
• I am sure there unnecessary % at the end of the lines in the preamble, but I find it safer to just always add them in my preamble, rather than figure out if they are necessary

## Code:

\documentclass{article}
\usepackage{collcell}
\usepackage{xcolor}% only needed for testing
\usepackage{etoolbox}

% Redefine tabular to initialize \StartTableHeader at start and end
\let\OldTabular\tabular%
\let\OldEndTabular\endtabular%

% Define the column type:
\newcolumntype{N}{>{\collectcell\myColorCell}c<{\endcollectcell}}%

\begin{document}
\noindent
\begin{tabular}{NN}
\hline
\hline
1.234 & 2.345\\
\end{tabular}

\bigskip\noindent
Check again to make sure that settings are correct at start of table:

\noindent
\begin{tabular}{NN}
\hline


text:
Home > Trapezoidal Rule > Trapezoidal Method Error

# Trapezoidal Method Error

## Contents

It is simply not correct to say that all measurements are so unreliable as to rule out any such estimate. I'm not sure that these diagrams provide a compelling reason to believe that $E_M$ is of roughly half the magnitude of $E_T,$ but I will give this some thought. Solution First, for reference purposes, Maple gives the following value for this integral.                                                      In each case the width of the subintervals will be,                                                              and so the You should always take care, however, especially if you plan to use the method on functions that aren't so smooth. –Will Orrick Feb 13 '14 at 12:12 @will-orick but

## Trapezoidal Rule Error Calculator

Again, I apologize for the down time! The question says How large should $n$ be to guarantee the Trapezoidal Rule approximation for $\int_{0}^{\pi}x\cos x\,dx$ be accurate to within 0.0001 ? Learn MATLAB today!

1. Why can't the second fundamental theorem of calculus be proved in just two lines?
2. Also most classes have assignment problems for instructors to assign for homework (answers/solutions to the assignment problems are not given or available on the site).
3. In that case it would be necessary to use appropriate filters covering a larger span of points to get the necessary accuracy.
5. Privacy Statement - Privacy statement for the site.
6. asked 2 years ago viewed 9373 times active 4 days ago Related 3The error of the midpoint rule for quadrature1Midpoint Rule, Trapezoidal Rule, etc.: When the number of intervals increases by

Matt J Matt J (view profile) 94 questions 3,683 answers 1,447 accepted answers Reputation: 7,730 on 4 Jan 2013 Direct link to this comment: https://www.mathworks.com/matlabcentral/answers/57737#comment_120706 If your data is very accurate As an example I computed the integral of sin(x) from 0 to pi where the exact answer would be 2. You should see a gear icon (it should be right below the "x" icon for closing Internet Explorer). Trapezoidal Rule Formula Terms of Use - Terms of Use for the site.

Badbox when using package todonotes and command missingfigure Why are only passwords hashed? Trapezoidal Rule Error Proof We can do better than that by looking at the second derivative in more detail, say between $0$ and $\pi/4$, and between $\pi/4$ and $\pi/2$. Here are the results: 6 intervals actual error by trapz - 0.04590276668629 est. my response It is especially true for some exponents and occasionally a "double prime" 2nd derivative notation will look like a "single prime".

Here are the bounds for each rule.                                                                                In each case we can see that the errors are significantly smaller than the actual bounds. Error In Trapezoidal Rule Is Of Order Show Answer Answer/solutions to the assignment problems do not exist. Is it dangerous to use default router admin passwords if only trusted users are allowed on the network? I am hoping they update the program in the future to address this.

## Trapezoidal Rule Error Proof

Midpoint Trapezoid Simpson’s n Approx. What exactly do you mean by "typical second finite differences in the data"? Trapezoidal Rule Error Calculator How to apply for UK visit visa after four refusal Why does Deep Space Nine spin? Simpson's Rule Error Formula error, 2nd diff. - 0.04363323129986 100 intervals actual error by trapz - 0.00016449611255687 est.

That's because the type of functions most people will cook up tend be smooth with second derivative that stays within reasonable bounds. http://degital.net/trapezoidal-rule/trapezoidal-error-bound.html If there's nothing stopping you from assuming your discrete samples came from this piece-wise linear function, then voila, you're done, and your area calculation was perfect! Or if you did want accurate estimates of errors that small, mightn't the additional errors introduced by the finite differencing make that difficult? I only have 350 data points (x,y) with intervals of dx = 0.5, so I can't change my samplings. Trapezoidal Rule Error Analysis

up vote 1 down vote favorite 1 I stack about Error Bounds of Trapezoidal Rule. Let me know what page you are on and just what you feel the typo/mistake is. That is, you know how to convert a pair of X and Y values into an estimate of the area for a single segment, and you can use error propagation http://en.wikipedia.org/wiki/Propagation_of_uncertainty this contact form Usually then, $f''$ will be more unpleasant still, and finding the maximum of its absolute value could be very difficult.

Those are intended for use by instructors to assign for homework problems if they want to. Error Formula For Trapezoidal Rule Calculator Sign in to add this to Watch Later Add to Loading playlists... Not the answer you're looking for?

## Let’s get first develop the methods and then we’ll try to estimate the integral shown above.

How could a language that uses a single word extremely often sustain itself? Then \begin{aligned} A[x^2]&=\int_a^bx^2\,dx=\frac{b^3-a^3}{3},\\ T[x^2]&=\frac{b-a}{2}(b^2+a^2)=\frac{b^3-ab^2+a^2b-a^3}{2}\\ M[x^2]&=(b-a)\left(\frac{b+a}{2}\right)^2=\frac{b^3+ab^2-a^2b-b^3}{4}. \end{aligned} So \begin{aligned} E_T[x^2]&=T[x^2]-A[x^2]=\frac{b^3-a^3}{6}-ab\frac{b-a}{2},\\ E_M[x^2]&=M[x^2]-A[x^2]=-\frac{b^3-a^3}{12}+ab\frac{b-a}{4}=-\frac{1}{2}E_T[x^2].\\ \end{aligned} Likewise \begin{aligned} A[x^3]&=\int_a^bx^3\,dx=\frac{b^4-a^4}{4},\\ T[x^3]&=\frac{b-a}{2}(b^3+a^3)=\frac{b^4-ab^3+a^3b-a^4}{2}\\ M[x^3]&=(b-a)\left(\frac{b+a}{2}\right)^3=(b-a)\frac{b^3+3ab^2+3a^2b+a^3}{8}\\ &=\frac{b^4+2ab^3-2a^3b-a^4}{8}. \end{aligned} So \begin{aligned} E_T[x^3]&=T[x^3]-A[x^3]=\frac{b^4-a^4}{4}-\frac{ab}{2}(b^2-a^2),\\ E_M[x^3]&=M[x^3]-A[x^3]=-\frac{b^4-a^4}{8}+\frac{ab}{4}(b^2-a^2)=-\frac{1}{2}E_T[x^3].\\ \end{aligned} Show Answer There are a variety of ways to download pdf versions of the material on the site. Trapezoidal Rule Example In addition, using the maximum of $|f''(x)|$ usually gives a needlessly pessimistic error estimate.

A series of pairs of plots is shown below. More detailed analysis can be found in.[3][4] "Rough" functions This section needs expansion. The system returned: (22) Invalid argument The remote host or network may be down. navigate here Sign in Don't like this video?

Working... Here is a graph of the fourth derivative. Show Answer Short Answer : No. And I don't think it would be correct to take the fitted curve as the 'underlying function' and then base errors on this unless you could be confident that the correct

However in the second case the data has to be very accurate to achieve this with second differencing.The remedy when data is not sufficiently accurate is to widen the span of How do I download pdf versions of the pages? Science. Long Answer with Explanation : I'm not trying to be a jerk with the previous two answers but the answer really is "No".

Then we know that the error has absolute value which is less than or equal to $$\frac{3.6\pi^3}{12n^2}.$$ We want to make sure that the above quantity is $\le 0.0001$. You will be presented with a variety of links for pdf files associated with the page you are on. error, 2nd diff. - 0.00016446634993921In each case the estimated error is fairly accurate percentage-wise. When to use conjunction and when not?

Browse other questions tagged calculus sequences-and-series or ask your own question. Up next Error Estimates (Midpoint Rule, Trapezoid Rule, Simpson's Rule) - Duration: 9:37. Once on the Download Page simply select the topic you wish to download pdfs from. When working with experimental data, there is no known underlying function.

The system returned: (22) Invalid argument The remote host or network may be down. Why cast an A-lister for Groot?


text:
Optimising PNG Generation in Python

From: andrew cooke <andrew@...>

Date: Sat, 16 Jul 2011 01:38:21 -0400

I'm generating images in Python using pypng http://code.google.com/p/pypng/

It's a great library (and pure Python, so it can be used in AppEngine), but I
needed the code to run as fast as possible.  Anyway, after some profiling, I
can say that for my images (which are small, stylised graphics with a
restricted number of colours), using a palette is a big win - it reduces time
spent creating (in my code) and writing (in pypng) the image.  It also reduces
the amount of memory used (which means I can cache more!).

And as far as I can tell, the "compression" parameter to Write should be 9
(the maximumum for zlib).  This does have an effect on size, and even for my
simple images it was significant and progressive.  But it doesn't have much
effect on speed, so turning to max (9) makes sense.

After a couple of hours of work, both total running time and image size are
about half what they were (I did some other code optimisatiosn too -
particularly keeping the image in HSV format until the last moment, rather
than swapping multiple times between RGB and HSV :o)

Andrew

Bit Depth?

From: andrew cooke <andrew@...>

Date: Sat, 16 Jul 2011 01:40:19 -0400

PS Curiously, changing bit depth seems to have no effect at all for paletted
images.  Too tire to work out why now, though....


text:
## Rate-independent systems. Theory and application.(English)Zbl 1339.35006

Applied Mathematical Sciences 193. New York, NY: Springer (ISBN 978-1-4939-2705-0/hbk; 978-1-4939-2706-7/ebook). xxii, 660 p. (2015).
The book focuses on the study of quasi-static evolutionary systems which are rate-independent and gives examples of such situations, mainly in continuum mechanics. It is divided in five chapters completed with a short introduction which gives a an historical presentation of the material and with three appendices.
In the short Chapter 1, the authors present the basic notions of evolutionary systems. They start with the ode $$M\overset{..}{q}+F(\overset{. }{q})+Kq=\widehat{l}(t)$$, which is a balance equation which describes the evolutions of the state $$q$$ with respect to the time parameter, under the action of the external force $$\widehat{l}$$. $$M$$ and $$K$$ are symmetric and positive-definite matrices and $$F$$ is the possibly nonlinear damping. The authors exhibit three time scales respectively associated to $$M^{-1}K$$, $$V^{-1}K$$ (assuming $$F(q)=Vq$$ for a symmetric and positive-definite matrix $$V$$ ) and $$\inf \left\| \frac{d}{dt}\widehat{l}(t)\right\|$$. They define the notion of rate-independent system when the two first time scales are much faster than the third one. They give an example considering $$\widehat{l} (t)=l(\varepsilon t)$$ with $$\varepsilon >0$$ and $$F(\overset{.}{q})=\nu \left| \overset{.}{q}\right| ^{\alpha -1}\overset{.}{q}$$. They indeed obtain the final ode $$\varepsilon ^{2}M\overset{..}{q}+\varepsilon ^{\alpha }\nu \left| \overset{.}{q}\right| ^{\alpha -1}\overset{.}{q} +Kq=l(t)$$, leading to the inclusion problem $$\partial \mathcal{R}(\overset{.} {q})+Kq\ni l(t)$$, where $$\mathcal{R}(\overset{.}{q})=\nu \left| \overset{ .}{q}\right| ^{\alpha }$$ is the dissipation potential, when taking the limit when $$\varepsilon$$ tends to 0. This last problem may be generalized as Biot’s equation $$\partial \mathcal{R}(\overset{.}{q})+D_{q}\mathcal{E} (t,q)\ni 0$$, introducing $$\mathcal{E}(t,q)=\frac{1}{2}\left\langle q,Kq\right\rangle -\left\langle l(t),q\right\rangle$$, $$D_{q}$$ being the partial differential operator with respect to $$q$$. The authors then give the definition of a rate-independent system, among other definitions: admissible, concatenation property, restriction property and causal systems. They give examples of such systems. They also define the notion of general evolutionary system as $$\partial _{\overset{.}{q}}\mathcal{R}(q,\overset{.}{q })+D_{q}\mathcal{E}(t,q)\ni 0$$, starting from an initial state $$q_{1}$$ at time $$t_{1}$$. Here $$\mathcal{R}:\mathbf{Q}\rightarrow [ 0,\infty ]$$ is convex and lower semicontinuous satisfying $$\mathcal{R}(0)=0$$ and $$\partial _{\overset{.}{q}}$$ denotes the subdifferential operator with respect to $$\overset{.}{q}$$ and $$D_{q}\mathcal{E}$$ means the Gateaux derivative of $$\mathcal{E}$$ with respect to $$q\in \mathbf{Q}$$, the Banach state space. In order to obtain a rate-independent system, the authors have to impose a $$p$$-homogeneous property on the dissipation potential $$\mathcal{R}$$: $$\mathcal{R }(q,\cdot)$$ is $$p$$-homogeneous for every $$q\in Q$$. This may be generalized splitting the state variable into its nondissipative and dissipative components and assuming that the dissipation potential only depends on the dissipative component. The authors then derive some general properties of such problems among which are a priori estimates on the state $$q$$. They define the notion of energetic solution, they prove the existence of an energetic solution under hypotheses on the initial condition and they analyze some examples. They also build an approximation scheme for such an energetic solution. They also introduce other notions of solutions for special problems and they compare these notions in the 1D case.
In Chapter 2, the authors analyze energetic rate-independent systems. They here consider a state space $$\mathcal{Q}$$ which splits as $$\mathcal{Q}= \mathcal{Y}\times \mathcal{Z}$$. After the general definition of such systems and of dissipative distance, from which they deduce that of energetic solution, they prove several existence results in different realistic situations. The chapter ends with an application of $$\Gamma$$-convergence tools in the present context in order to obtain relaxation results.
In Chapter 3, the authors assume that the state space $$\mathbf{Q}$$ splits as $$\mathbf{Q}=\mathbf{Y}\times \mathbf{Z}$$, where $$\mathbf{Y}$$ and $$\mathbf{Z}$$ are separate and reflexive Banach spaces. The purpose of this chapter is to define the appropriate notions of rate-independent system and of solution within this context. The authors start with the notion of energetic solution and they prove an existence result under hypotheses on the initial condition. In order to define more general solutions, they recall the notions of monotone, maximal monotone, responsive and maximal responsive operators $$A:\mathbf{Z}\rightrightarrows \mathbf{Z}^{\ast }$$ and they establish properties of such operators. After the presentation of the notions of convex, Fréchet, limiting, Clarke and Dini subdifferentials, the authors introduce the notions of differential, semi-differential, continuous dissipative (CD), local and weak solution for evolutionary systems. They establish some links between these notions and an existence result for an energetic solution which is also a CD solution. The case of convex systems, that is such that $$\mathcal{R}$$ is translation-invariant and $$\mathcal{E}(t,.)$$ is convex, is considered. The authors assume that $$\mathcal{Q}$$ is a convex subset of the Banach state space $$\mathbf{Q}$$. They prove that energetic solutions may be generated by a.e. local solutions and every energetic solution is proved to be continuous with respect to the time parameter, under some hypotheses. Then every energetic solution may be a differential or a semi-differential solution under appropriate hypotheses. The authors also prove some uniqueness results for energetic solutions of rate-independent and convex systems. The next section of this chapter focuses on the special case where $$\mathcal{E}(t,q)=\frac{1}{2}\left\langle q,Kq\right\rangle -\left\langle l(t),q\right\rangle$$, with $$l\in W^{1,1}(0,T;\mathbf{Q}^{\ast })$$ and $$A:\mathbf{Q}\rightarrow \mathbf{Q} ^{\ast }$$ is a symmetric, positive semi-definite, bounded and linear operator. When $$A$$ is positive definite, the authors prove an existence and uniqueness result for an energetic solution $$q\in W^{1,1}(0,T;\mathbf{Q})$$ starting from an initial condition $$q_{0}\in A^{-1}(l(0)-\Sigma )$$. When $$A$$ is only positive semi-definite, they prove an existence result under further hypotheses on $$\mathbf{Q}$$. This section ends with applications of $$\Gamma$$ -convergence tools and homogenization results. The next section describes some numerical tools for the computation of an energetic solution and the authors prove stability and convergence results. The chapter ends with two further considerations on rate-independent systems in this context.
The very long Chapter 4 moves to applications of the notions and tools which have been presented in the previous chapters in continuum mechanics and physics of solids. The chapter starts with a detailed presentation of continuum mechanics. They recall the notions of stress, strain, energy in the case of small or large strains, then that of elasticity and plasticity. They prove an existence result for an energetic solution in the polyconvexity case under hypotheses on the energy and they propose a numerical scheme for some examples. A long section is then devoted to the study of inelastic processes at small strains. The authors again prove an existence results for a unique solution which belongs to $$W^{1,\infty }(0,T;H^{1}(\Omega ,\mathbb{R}^{d}))$$, and they prove the strong convergence of the associated numerical scheme. They present several examples. The chapter ends with a section dedicated to the study of ferromagnetic materials. Once again, the authors prove an existence result and the convergence of the numerical scheme.
In the final Chapter 5, the authors analyze different realistic examples involving evolutionary systems. They start with the case of a hyperviscoelastic material with the Kelvin-Voigt rheology leading to the momentum equilibrium $$\rho \overset{..}{u}-\text{div}\sigma =f$$, with $$\sigma (e,z)=W_{e}^{\prime }(e,z)+R_{\overset{.}{e}}^{\prime }(e,z,\overset{. }{e})$$, $$e$$ being the linearized strain tensor. Here $$R$$ is a pseudopotential of viscous like dissipative forces. The authors put this problem in an abstract way $$M\overset{..}{u}+V(u,z)\overset{.}{u}+\partial _{u}\mathcal{E}(t,u,z)=f(t,u,z)$$, and they add initial conditions for $$u$$, $$\overset{.}{u}$$ and $$z$$. They prove an existence result for local and energetic solutions and an approximation result by time discretization. They then prove an existence result for a problem written in an abstract setting and the chapter ends with applications in plasticity, damages in a viscoelastic material, delaminations of viscoelastic bodies...
Two appendices complete this nice and detailed presentation. The first one recalls useful notions of topology and of functional analysis, while the second one is dedicated to elements of measure theory and of function spaces. The third appendix presents the basis of Young measures and their properties.
The book is written by two famous specialists of evolutionary systems as they worked for a long time on this topic. Throughout the whole book, the authors give lots of precise details concerning the particularities of the rate-independent systems in different and realistic settings. They take care to illustrate their theoretical notions with many examples and they complete their computations with many figures.

### MSC:

 35-02 Research exposition (monographs, survey articles) pertaining to partial differential equations 35K85 Unilateral problems for linear parabolic equations and variational inequalities with linear parabolic operators 35Q74 PDEs in connection with mechanics of deformable solids 47J35 Nonlinear evolution equations 49J40 Variational inequalities 49J45 Methods involving semicontinuity and convergence; relaxation 49S05 Variational principles of physics 74-02 Research exposition (monographs, survey articles) pertaining to mechanics of deformable solids 74C15 Large-strain, rate-independent theories of plasticity (including nonlinear plasticity) 35B27 Homogenization in context of PDEs; PDEs in media with periodic structure
Full Text:


text:
# Every nonsingular $m\times m$ matrix is row equivalent to identity matrix $I_m$

Could anyone give an easy to understand proof of this fact or at least tell if the proof below is correct?

We know that every matrix is row equivalent to a reduced row echelon form. An $m\times m$ square matrix $A$ of rank $m$ is row equivalent to an identity matrix from the definition of reduced row echelon form:

1) all non-zero rows are above rows of all zeros

2) the leading coefficient is strictly to the right of the leading coefficient of the row above

3) the leading coefficient is $1$ and is the only non-zero number in its column

Plus, there's a theorem saying that the rank of any matrix equals the number of non-zero rows in its reduced row echelon form.

Now we know that $A$ has $m$ non-zero rows, then every row contains a leading coefficient $1$ that is shifted to the right with respect to the leading coefficient in the row above. The only way we can arrange $m$ 'ones' in this manner is by placing them on diagonal (we can prove it by noting the the required number of 'shifts to the right' of the leading coefficients when going one row down is $m$ - actually - is it clear enough for a proof?). And we know there is nothing except $1$ in any of those columns - we have a diagonal matrix.

Can you make it shorter without referring to some more complicated terminology?

• I think it should be "Every invertible (or regular, or non-singular, etc.) matrix is ...." etc. – Timbuc Apr 23 '15 at 15:27
• this wont work with the zero matrix. – abel Apr 23 '15 at 15:34
• I've corrected the title. Do you have any objections? – user4205580 Apr 23 '15 at 15:47

Matrix equivalence can be completely characterized in terms of a factorization: $A \sim B$ if and only if there exist nonsingular $P,Q$ of appropriate order so that $A=PBQ$. The matrices $P$ and $Q$ are the products of elementary row and column operation matrices respectively.
Now if $A$ is a nonsingular $m \times m$ matrix, we have $I=A^{-1}A=A^{-1}AI$, with $A^{-1}$ and $I$ nonsingular...so $A \sim I$.
Edit: Sorry I see you refer to row equivalence specifically. There is a similar characterization: $A \stackrel{R}{\sim} B$ iff there exists nonsingular $P$ so that $A=PB$ where $P$ is a product of elementary matrices representing elementary row operations. You can fill in the details....


text:
When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:02:23Z http://mathoverflow.net/feeds/question/108643 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108643/when-a-semigroup-extension-s-times-a-of-a-semigroup-s-with-a-commutative-gro When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$? Marius Buliga 2012-10-02T15:13:26Z 2012-10-02T18:10:04Z
Maybe is a trivial question, but I don't know how to handle it.

Setting: Let $S$ be a semigroup (i.e. has an associative operation with neutral element $e$) and let $(A,+)$ be a commutative group (with neutral element $0$).

A semigroup extension of $S$ with $A$ is any operation on $S \times A$, of the form

$(s,a)(s',a') = (s s' , a+a'+ \lambda(s,s')$

where $\lambda: S \times S \rightarrow A$ is a function such that $S \times A$ with the mentioned operation is a semigroup with neutral element $(e,0)$. Obviously, the operation is encoded by the function $\lambda$, which satisfies certain equations.

Problem: Describe the class of functions $\lambda$ with the property: there is an injective morphism of semigroups from $S$ to $S \times A$ (with the semigroup extension operation induced by $\lambda$). Is there an elegant way to describe this class?

Motivation: I am puzzled by the following two examples, I cannot put the finger on the essential difference between those.

Let $X$ and $Y$ be topological, locally convex, real vector spaces of dual variables $x \in X$ and $y \in Y$, with the duality product $\langle \cdot , \cdot \rangle : X \times Y \rightarrow \mathbb{R}$.

The spaces $X, Y$ have topologies compatible with the duality product, in the sense that for any continuous linear functional on $X$ there is an $y \in Y$ which puts the functional into the form $x \mapsto \langle x,y\rangle$ (respectively any continuous linear functional on $Y$ has the form $y \mapsto \langle x,y\rangle$, for a $x \in X$).

Example 1: (Heisenberg group) Let $S = X \times Y$ with the operation of addition of pairs of vectors and let $A = \mathbb{R}$ with addition. We may define a Heisenberg group over the pair $(X,Y)$ as $H(X,Y) = S \times A$ with the operation

$(x,y,a)(x',y',b) = (x+x', y+y', a+a'+ \langle x, y'\rangle - \langle x', y \rangle)$

There is no injective morphism from $(X\times Y, +)$ to $H(X,Y)$.

Example 2: Let this time $S = (X \times Y)^{*}$, the free semigroup generated by $X \times Y$, i.e. the collection of all finite words with letters from $X \times Y$, together with the empty word $e$, with the operation of concatenation of words.

Let $A$ be the set of bi-affine real functions on $X \times Y$, i.e. the collection of all functions $a: X \times Y \rightarrow \mathbb{R}$ which are affine and continuous in each argument. $A$ is a commutative group with the addition of real valued functions operation.

We define the function $\lambda: S \times S \rightarrow A$ by:

$\lambda(e, c)(x,y) = \lambda(c,e)(x,y)=0$ for any $c \in S$ and any $(x,y) \in X \times Y$.

if $c, h \in S$ are words $c = (x_{1},y_{1})...(x_{n}, y_{n})$ and $h = (u_{1},v_{1})...(u_{m}, v_{m})$, with $m,n \geq 1$, then

$\lambda(c,h)(x,y) = \langle u_{1} - x , y_{n} - y \rangle$ for any $(x,y) \in X \times Y$.

This $\lambda$ induces a semigroup extension operation on $S \times A$ and there is an injective morphism $F: S \rightarrow S \times A$, with the expression $F(c) = (c, E(c))$. Here, for any $c = (x_{1},y_{1})...(x_{n}, y_{n})$ the expression $E(c)(x,y)$ is the well known circular sum associated to the "discrete cycle" $(x_{1},y_{1})...(x_{n}, y_{n})(x,y)$, namely:

$E(c)(x,y) = \langle x_{1},y\rangle + \langle x,y_{n}\rangle - \langle x,y \rangle + \sum_{k=1}^{n-1}\langle x_{k+1}, y_{k}\rangle - \sum_{k=1}^{n} \langle x_{k},y_{k} \rangle$

dear to convex analysis.
 http://mathoverflow.net/questions/108643/when-a-semigroup-extension-s-times-a-of-a-semigroup-s-with-a-commutative-gro/108644#108644 Answer by Benjamin Steinberg for When a semigroup extension $S \times A$ of a semigroup $S$ with a commutative group $A$ contains a copy of $S$? Benjamin Steinberg 2012-10-02T15:25:03Z 2012-10-02T15:25:03Z
First of all I would not call what you write a semidirect product. It will be such in your setting precisely when the natural maps to $S$ splits, in which case it will actually be isomorphic to a direct product.

More precisely, you are viewing $A$ here as a trivial module over $\mathbb ZS$. Then $\lambda$ is a 2-cocycle and you want it to be equivalent in $H^2(S,A)$ with the trivial cocycle.


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Intuition for why the exact length of the object is unnecessary when calculating the forces using equilibrium of the torque

For example, for the following scenario:

A $$m \;\text{kg}$$ person is walking across a level bridge and stops three-fourths of the way from one end. The bridge is uniform and weighs $$M \;\text{kg}$$. What are the values of the vertical forces exerted on each end of the bridge by its supports?

We can solve it by considering the net torque on the bridge about the end closest to the person is $$\frac{mL}{4} + \frac{ML}{2}−FL= 0$$ which has a solution for the supporting force on the far end of F.

What's the intuition for why the exact length of the bridge is unnecessary when solving questions such as the above one?

Imagine stretching or contracting the bridge. This changes the total length $$L$$ by some factor $$L \to \lambda L$$. Intuitively the bridge will still be in equilibrium even after this rescaling.

This implies that the equilibrium condition cannot depend on the total length of the bridge, but only on the relative position of where the various forces are applied.

My reasoning works because the system is symmetric under scaling along the length of the bridge. This is the statement that you can only measure things that do not depend on that $$\lambda$$ parameter I said at the beginning. In your case this scaling also causes the position of where the force is applied $$x$$ to scale in the same way, thus everything only depends on the $$\lambda$$ independent ratio $$x/L$$.
This only works however given the assumptions that the bridge is rigid (can't be deformed) and so $$L$$ and $$x$$ are the only length scale of the problem. Introducing the deformations in the mix will introduce other parameters such as the stress-tensor of the bridge and the deformations that will transform with some powers of $$\lambda$$ and make the whole thing more complicated.
You could use a dimensional argument: the only lengths available in the problem are $$x$$ (position on the bridge) and $$L$$. The Pi theorem (Vaschy Buckingham) requires that only the ratio $$x/L$$ appears in the final result: $$T=mg\varphi(x/L)$$
This is not intuitive because it feels like a short bridge is stronger than a long bridge. But bond strength cannot be measured simply with the total force alone. If you had calculated the bending moments within the bridge, the length of the bridge would be involved. For example, imagine a rigid diving board attached to only one end: the global force at the attachment point is the opposite of the weight of the diving board $$T=mg$$ But the moment of actions at the point of fixation is $$-Mgl/2$$, the greater the longer the diving board. And intuitively, this moment plays a critical role in the risk of rupture.


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# Finite temperature effects on Majorana bound states in chiral $p$-wave superconductors

### Submission summary

 As Contributors: Henrik Røising Arxiv Link: https://arxiv.org/abs/1901.09933v1 Date submitted: 2019-02-06 Submitted by: Røising, Henrik Submitted to: SciPost Physics Domain(s): Theor. & Comp. Subject area: Quantum Physics

### Abstract

We study Majorana fermions bound to vortex cores in a chiral $p$-wave superconductor at temperatures non-negligible compared to the superconducting gap. Thermal occupation of Caroli de Gennes-Matricon states, below the full gap, causes the free energy difference between the two fermionic parity sectors to decay algebraically with increasing temperature. The power law acquires an additional factor of $T^{-1}$ for each bound state thermally excited. The zero-temperature result is exponentially recovered well below the minigap (lowest-lying CdGM level). Our results suggest that temperatures larger than the minigap may not be disastrous for topological quantum computation.

###### Current status:
Editor-in-charge assigned

### Submission & Refereeing History

Submission 1901.09933v1 (6 February 2019)


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# Search for new resonances decaying via WZ to leptons in proton-proton collisions at sqrt(s)=8 TeV

Abstract : A search is performed in proton-proton collisions at sqrt(s) = 8 TeV for exotic particles decaying via WZ to fully leptonic final states with electrons, muons, and neutrinos. The data set corresponds to an integrated luminosity of 19.5 inverse femtobarns. No significant excess is observed above the expected standard model background. Upper bounds at 95% confidence level are set on the production cross section of a W' boson as predicted by an extended gauge model, and on the W'WZ coupling. The expected and observed mass limits for a W' boson, as predicted by this model, are 1.55 and 1.47 TeV, respectively. Stringent limits are also set in the context of low-scale technicolor models under a range of assumptions for the model parameters.
Document type :
Journal articles

http://hal.in2p3.fr/in2p3-01023624
Contributor : Sylvie Flores <>
Submitted on : Tuesday, July 15, 2014 - 7:41:44 AM
Last modification on : Thursday, June 17, 2021 - 3:19:02 PM

### Citation

V. Khachatryan, S. Baffioni, F. Beaudette, P. Busson, C. Charlot, et al.. Search for new resonances decaying via WZ to leptons in proton-proton collisions at sqrt(s)=8 TeV. Physics Letters B, Elsevier, 2015, 740, pp.83-104. ⟨10.1016/j.physletb.2014.11.026⟩. ⟨in2p3-01023624⟩

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