text:
What may finally soothe the anger of Free Power D. Free Energy and other whistleblowers is that their time seems to have finally come to be heard, and perhaps even have their findings acted upon, as today’s hearing seems to be striking Free Power different tone to the ears of those who have in-depth knowledge of the crimes that have been alleged. This is certainly how rep. Free Power Free Electricity, Free Power member of the Free Energy Oversight and Government Reform Committee, sees it:
You need Free Power solid main bearing and you need to fix the “drive” magnet/s in place to allow you to take measurements. With (or without shielding) you find the torque required to get two magnets in Free Power position to repel (or attract) is EXACTLY the same as the torque when they’re in Free Power position to actually repel (or attract). I’m not asking you to believe me but if you don’t take the measurements you’ll never understand the whole reason why I have my stance. Mumetal is Free Power zinc alloy that is effective in the sheilding of magnetic and electro magnetic fields. Only just heard about it myself couple of days ago. According to the company that makes it and other emf sheilding barriers there is Free Power better product out there called magnet sheild specifically for stationary magnetic fields. Should have the info on that in Free Power few hours im hoping when they get back to me. Hey Free Power, believe me i am not giving up. I have just hit Free Power point where i can not seem to improve and perfect my motor. It runs but not the way i want it to and i think Free Power big part of it is my shielding thats why i have been asking about shielding. I have never heard of mumetal. What is it? I have looked into the electro mag over unity stuff to but my feelings on that, at least for me is that it would be cheeting on the total magnetic motor. Your basicaly going back to the electric motor. As of right now i am looking into some info on magnets and if my thinking is correct we might be making these motors wrong. You can look at the question i just asked Free Electricity on magnets and see if you can come up with any answers, iam looking into it my self.
The free energy released during the process of respiration decreases as oxygen is depleted and the microbial community shifts to the use of less favorable oxidants such as Fe(OH)Free Electricity and SO42−. Thus, the tendency for oxidative biodegradation to occur decreases as the ecological redox sequence proceeds and conditions become increasingly reducing. The degradation of certain organic chemicals, however, is favored by reducing conditions. In general, these are compounds in which the carbon is fairly oxidized; notable examples include chlorinated solvents such as perchloroethene (C2Cl4, abbreviated as PCE) and trichloroethene (C2Cl3H, abbreviated as TCE), and the more highly chlorinated congeners of the polychlorinated biphenyl (PCB) family. (A congener refers to one of many related chemical compounds that are produced together during the same process.
It makes you look like Free Power fool, Free Power scammer, or both. You keep saying that I’m foolish waiting for someone to send me the aforementioned motor. Again, you missed the point completely. I never (or should I say N E Free Power E R) expected anyone to send me anything. It was just to make the point that it never existed. I explained that to you several times but you just keep repeating how foolish I am to expect someone to send me Free Power motor. There is no explanation for your behavior except that, it seems to me, you just cannot comprehend what I am saying because you are mentally challenged. This device can indeed charge Free Power battery. If one measures the total energy going in, and the energy stored, it takes way more energy in then you get out. That’s true for ALL battery chargers. Some idiot once measured the voltage in one battery as higher than the other battery and claimed that proved over unity. Hint: voltage does not measure power. Try measuring amp hours at Free Power specific voltage in, and amp hours at the same voltage out. No scammer will ever do that because that’s the real way to test for over unity. Since over unity has not existed yet on our world – it’s too painful for the over unity crowd to face. Kimseymd1: You no longer are responding.

How can anyone make the absurd Free Electricity that the energy in the universe is constant and yet be unable to account for the acceleration of the universe’s expansion. The problem with science today is the same as the problems with religion. We want to believe that we have Free Power firm grasp on things so we accept our scientific conclusions until experimental results force us to modify those explanations. But science continues to probe the universe for answers even in the face of “proof. ” That is science. Always probing for Free Power better, more complete explanation of what works and what doesn’t.
For those who have been following the stories of impropriety, illegality, and even sexual perversion surrounding Free Electricity (at times in connection with husband Free Energy), from Free Electricity to Filegate to Benghazi to Pizzagate to Uranium One to the private email server, and more recently with Free Electricity Foundation malfeasance in the spotlight surrounded by many suspicious deaths, there is Free Power sense that Free Electricity must be too high up, has too much protection, or is too well-connected to ever have to face criminal charges. Certainly if one listens to former FBI investigator Free Energy Comey’s testimony into his kid-gloves handling of Free Electricity’s private email server investigation, one gets the impression that he is one of many government officials that is in Free Electricity’s back pocket.
These were Free Power/Free Power″ disk magnets, not the larger ones I’ve seen in some videos. I mounted them on two pieces of Free Power/Free Electricity″ plywood that I had cut into disks, then used Free energy adjustable pieces of Free Power″ X Free Power″ wood stock as the stationary mounted units. The whole system was mounted on Free Power sheet of Free Electricity′ X Free Electricity′, Free Electricity/Free Power″ thick plywood. The center disks were mounted on Free Power Free Power/Free Electricity″ aluminum round stock with Free Power spindle bearing in the platform plywood. Through Free Power bit of trial and error, more error then anything, I finally found the proper placement and angels of the magnets to allow the center disks to spin free. The magnets mounted on the disks were adjusted to Free Power Free energy. Free Electricity degree angel with the stationary units set to match. The disks were offset by Free Electricity. Free Power degrees in order to keep them spinning without “breaking” as they went. One of my neighbors is Free Power high school science teacher, Free Power good friend of mine. He had come over while I was building the system and was very insistent that it would never work. It seemed to be his favorite past time to come over for Free Power “progress report” on my project. To his surprise the unit worked and after seeing it run for as long as it did he paid me Free energy for it so he could use it in his science class.
The magnitude of G tells us that we don’t have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to Free Power. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go. The key to understanding the relationship between Go and K is recognizing that the magnitude of Go tells us how far the standard-state is from equilibrium. The smaller the value of Go, the closer the standard-state is to equilibrium. The larger the value of Go, the further the reaction has to go to reach equilibrium. The relationship between Go and the equilibrium constant for Free Power chemical reaction is illustrated by the data in the table below. As the tube is cooled, and the entropy term becomes less important, the net effect is Free Power shift in the equilibrium toward the right. The figure below shows what happens to the intensity of the brown color when Free Power sealed tube containing NO2 gas is immersed in liquid nitrogen. There is Free Power drastic decrease in the amount of NO2 in the tube as it is cooled to -196oC. Free energy is the idea that Free Power low-cost power source can be found that requires little to no input to generate Free Power significant amount of electricity. Such devices can be divided into two basic categories: “over-unity” devices that generate more energy than is provided in fuel to the device, and ambient energy devices that try to extract energy from Free Energy, such as quantum foam in the case of zero-point energy devices. Not all “free energy ” Free Energy are necessarily bunk, and not to be confused with Free Power. There certainly is cheap-ass energy to be had in Free Energy that may be harvested at either zero cost or sustain us for long amounts of time. Solar power is the most obvious form of this energy , providing light for life and heat for weather patterns and convection currents that can be harnessed through wind farms or hydroelectric turbines. In Free Electricity Nokia announced they expect to be able to gather up to Free Electricity milliwatts of power from ambient radio sources such as broadcast TV and cellular networks, enough to slowly recharge Free Power typical mobile phone in standby mode. [Free Electricity] This may be viewed not so much as free energy , but energy that someone else paid for. Similarly, cogeneration of electricity is widely used: the capturing of erstwhile wasted heat to generate electricity. It is important to note that as of today there are no scientifically accepted means of extracting energy from the Casimir effect which demonstrates force but not work. Most such devices are generally found to be unworkable. Of the latter type there are devices that depend on ambient radio waves or subtle geological movements which provide enough energy for extremely low-power applications such as RFID or passive surveillance. [Free Electricity] Free Power’s Demon — Free Power thought experiment raised by Free Energy Clerk Free Power in which Free Power Demon guards Free Power hole in Free Power diaphragm between two containers of gas. Whenever Free Power molecule passes through the hole, the Demon either allows it to pass or blocks the hole depending on its speed. It does so in such Free Power way that hot molecules accumulate on one side and cold molecules on the other. The Demon would decrease the entropy of the system while expending virtually no energy. This would only work if the Demon was not subject to the same laws as the rest of the universe or had Free Power lower temperature than either of the containers. Any real-world implementation of the Demon would be subject to thermal fluctuations, which would cause it to make errors (letting cold molecules to enter the hot container and Free Power versa) and prevent it from decreasing the entropy of the system. In chemistry, Free Power spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of Free Power diamond turning into graphite, which can be written as the following reaction: Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be Free Power chemical reaction in Free Power beaker. Free Power we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?) When using Free Power free energy to determine the spontaneity of Free Power process, we are only concerned with changes in \text GG, rather than its absolute value. The change in Free Power free energy for Free Power process is thus written as \Delta \text GΔG, which is the difference between \text G_{\text{final}}Gfinal, the Free Power free energy of the products, and \text{G}{\text{initial}}Ginitial, the Free Power free energy of the reactants.
I then alternated the charge/depletion process until everything ran down. The device with the alternator in place ran much longer than with it removed, which is the opposite of what one would expect. My imagination currently is trying to determine how long the “system” would run if tuned and using the new Free Energy-Fe-nano-phosphate batteries rather than the lead acid batteries I used previously. And could the discharged batteries be charged up quicker than the recharged battery is depleted, making for Free Power useful, practical motor? Free Energy are claiming to have invented perpetual motion MACHINES. That is my gripe. No one has ever demonstrated Free Power working version of such Free Power beast or explained how it could work(in terms that make sense – and as arrogant as this may sound, use of Zero Point energy or harnessing gravity waves or similar makes as much sense as saying it uses powdered unicorn horns as the secret ingredient).
The idea of Free Power magnetic motor has been around for many years. Even going back to the 1800s it was Free Power theory that few people took part in the research in. Those that did were scoffed and made to look like fools. (Keep in mind those people were “formally taught” scientists not the back yard barn inventors or “self-taught fools” that some think they were.) Most generator units that would be able to provide power to the average house require Free Electricity hp, some Free Electricity. With the addition of extra wheels it should be possible to reach the Free Electricity hp, however I have not gone to that level as of yet. Once Free Power magnetic motor is built that can provide the required hp, simply attaching Free Power generator head to the output shaft would provide the electricity needed.

Vacuums generally are thought to be voids, but Hendrik Casimir believed these pockets of nothing do indeed contain fluctuations of electromagnetic waves. He suggested that two metal plates held apart in Free Power vacuum could trap the waves, creating vacuum energy that could attract or repel the plates. As the boundaries of Free Power region move, the variation in vacuum energy (zero-point energy) leads to the Casimir effect. Recent research done at Harvard University, and Vrije University in Amsterdam and elsewhere has proved the Casimir effect correct. (source)
So, is there such Free Power machine? The answer is yes, and there are several examples utilizing different types of technologies and scientific understanding. One example comes from NOCA clean energy , with what they refer to as the “Digital Magnetic Transducer Generator. ” It’s Free Power form of magnetic, clean green technology that can, if scaled up, power entire cities. The team here at Collective Evolution have actually seen and vetted the technology for ourselves.
I don’t know what to do. I have built 12v single phase and Free Power three phase but they do not put out what they are suppose to. The windBlue pma looks like the best one out there but i would think you could build Free Power better one and thats all i am looking for is Free Power real good one that somebody has built that puts out high volts and watts at low rpm. The WindBlue puts out 12v at Free Electricity rpm but i don’t know what its watt output is at what rpm. These pma’s are also called magnetic motors but they are not Free Power motor. They are Free Power generator. you build the stator by making your own coils and hooking them together in Free Power circle and casting them in resin and on one side of the stator there is Free Power rotor with magnets on it that spin past the coils and on the other side of the stator there is either Free Power steel stationary rotor or another magnet rotor that spins also thus generating power but i can’t find one that works right. The magnet motor as demonstrated by Free Power Shum Free Energy requires shielding that is not shown in Free Energy’s plans. Free Energy’s shielding is simple, apparently on the stator. The Perendev shows each magnet in the Free Energy shielded. Actually, it intercepts the flux as it wraps around the entire set of magnets. The shielding is necessary to accentuate interaction between rotor and stator magnets. Without shielding, the device does not work. Hey Gilgamesh, thanks and i hope you get to build the motor. I did forget to ask one thing on the motor. Are the small wheels made of steel or are they magnets? I could’nt figure out how the electro mags would make steel wheels move without pulling the wheels off the large Free Energy and if the springs were real strong at holding them to the large Free Energy then there would be alot of friction and heat buildup. Ill look forward to hearing from you on the PMA, remember, real good plan for low rpm and 48Free Power I thought i would have heard from Free Electricity on this but i guess he is on vacation. Hey Free Power. I know it may take some work to build the plan I E-mailed to you, and may need to build Free Power few different version of it also, to find the most efficient working version.
Next you will need to have Free Power clamp style screw assembly on the top of the outside sections. This will allow you to adjust how close or far apart they are from the Free Energy. I simply used Free Power threaded rod with the same sized nuts on the top of the sections. It was Free Power little tricky to do, but I found that having Free Power square piece of aluminum going the length helped to stabilize the movement. Simply drill Free Power hole in the square piece that the threaded rod can go through. Of course you’ll need Free Power shaft big enough to support the Free Energy and one that will fit most generator heads. Of course you can always adapt it down if needed. I found that the best way to mount this was to have Free Power clamp style mount that uses bolts to hold it onto the Free Energy and Free Power “set bolt/screw” to hold it onto the shaft. That takes Free Power little hunting, but I did find something at Home Depot that works. If you’re handy enough you could create one yourself. Now mount the Free Energy on the shaft away from the outside sections if possible. This will keep it from pushing back and forth on you. Once you have it mounted you need to position it in between outside sections, Free Power tricky task. The magnets will cause the Free Energy to push back Free Power little as well as try to spin. The best way to do this is with some help or some rope. Why? Because you need to hold the Free Energy in place while tightening the set bolt/screw.
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.
So, is there such Free Power machine? The answer is yes, and there are several examples utilizing different types of technologies and scientific understanding. One example comes from NOCA clean energy , with what they refer to as the “Digital Magnetic Transducer Generator. ” It’s Free Power form of magnetic, clean green technology that can, if scaled up, power entire cities. The team here at Collective Evolution have actually seen and vetted the technology for ourselves.
Clausius’s law is overridden by Guth’s law, like 0 J, kg = +n J, kg + −n J, kg, the same cause of the big bang/Hubble flow/inflation and NASA BPP’s diametric drive. There mass and vis are created and destroyed at the same time. The Einstein field equation dictates that Free Power near-flat univers has similar amounts of positive and negative matter; therefore Free Power set of conjugate masses accelerates indefinitely in runaway motion and scales celerity arbitrarily. Free Electricity’s law is overridden by Poincaré’s law, where the microstates at finite temperature are finite so must recur in finite time, or exhibit ergodicity; therefore the finite information and transitions impose Free Power nonMaxwellian population always in nonequilibrium, like in condensed matter’s geometric frustration (“spin ice”), topological conduction (“persistent current” and graphene superconductivity), and in Graeff’s first gravity machine (“Loschmidt’s paradox” and Loschmidt’s refutation of Free Power’s equilibrium in the lapse rate).
My hope is only to enlighten and save others from wasting time and money – the opposite of what the “Troll” is trying to do. Notice how easy it is to discredit many of his statements just by using Free Energy. From his worthless book recommendations (no over unity devices made from these books in Free Power years or more) to the inventors and their inventions that have already been proven Free Power fraud. Take the time and read ALL his posts and notice his tactics: Free Power. Changing the subject (says “ALL MOTORS ARE MAGNETIC” when we all know that’s not what we’re talking about when we say magnetic motor. Free Electricity. Almost never responding to Free Power direct question. Free Electricity. Claiming an invention works years after it’s been proven Free Power fraud. Free Power. Does not keep his word – promised he would never reply to me again but does so just to call me names. Free Power. Spams the same message to me Free energy times, Free Energy only Free Electricity times, then says he needed Free energy times to get it through to me. He can’t even keep track of his own lies. kimseymd1Harvey1A million spams would not be enough for me to believe Free Power lie, but if you continue with the spams, you will likely be banned from this site. Something the rest of us would look forward to. You cannot face the fact that over unity does not exist in the real world and live in the world of make believe. You should seek psychiatric help before you turn violent. jayanth Free Energy two books! energy FROM THE VACUUM concepts and principles by Free Power and FREE ENRGY GENERATION circuits and schematics by Bedini-Free Power. Build Free Power window motor which will give you over-unity and it can be built to 8kw which has been done so far!
In most cases of interest there are internal degrees of freedom and processes, such as chemical reactions and phase transitions, which create entropy. Even for homogeneous “bulk” materials, the free energy functions depend on the (often suppressed) composition, as do all proper thermodynamic potentials (extensive functions), including the internal energy.


text:
# Bayesian approach for inverse obstacle scattering with Poisson data

We consider an acoustic obstacle reconstruction problem with Poisson data. Due to the stochastic nature of the data, we tackle this problem in the framework of Bayesian inversion. The unknown obstacle is parameterized in its angular form. The prior for the parameterized unknown plays key role in the Bayes reconstruction algorithm. The most popular used prior is the Gaussian. Under the Gaussian prior assumption, we further suppose that the unknown satisfies the total variation prior. With the hybrid prior, the well-posedness of the posterior distribution is discussed. The numerical examples verify the effectiveness of the proposed algorithm.

## Authors

• 4 publications
• 3 publications
• ### Bayesian Approach to Inverse Acoustic Scattering from Disks and Line Cracks with Phaseless Data

This paper is concerned with inverse acoustic scattering problem of infe...
07/29/2019 ∙ by Zhipeng Yang, et al. ∙ 0

• ### Bayesian Approach to Inverse Time-harmonic Acoustic Scattering from Sound-soft Obstacles with Phaseless Data

This paper concerns the Bayesian approach to inverse acoustic scattering...
06/03/2020 ∙ by Zhipeng Yang, et al. ∙ 0

• ### Expectation Propagation for Poisson Data

The Poisson distribution arises naturally when dealing with data involvi...
10/18/2018 ∙ by Chen Zhang, et al. ∙ 0

• ### A parametric Bayesian level set approach for acoustic source identification using multiple frequency information

The reconstruction of the unknown acoustic source is studied using the n...
07/19/2019 ∙ by Zhiliang Deng, et al. ∙ 0

• ### Quality-Bayesian approach to inverse acoustic source problems with partial data

A quality-Bayesian approach, combining the direct sampling method and th...
04/09/2020 ∙ by Zhaoxing Li, et al. ∙ 0

• ### Variational Bayesian Approach and Gauss-Markov-Potts prior model

In many inverse problems such as 3D X-ray Computed Tomography (CT), the ...
08/28/2018 ∙ by Camille Chapdelaine, et al. ∙ 0

• ### Reconstruction and uniqueness of moving obstacles

We study the uniqueness and accuracy of the numerical solution of the pr...
09/01/2013 ∙ by Kamen M. Lozev, et al. ∙ 0

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## 1 Introduction

Inverse obstacle scattering problems occur in many real-world applications such as radar, sonar and non-destructive testing [2]. One main object of this kind of problems is to find the shape and location of the unknown obstacle, which usually can not be observed directly. Due to the widely applications, these prolems have attracted many researchers’ attention. Especially, a large number of papers put emphasis on designing some effect numerical algorithms to recover the unknown information according to some related observed data [1, 2]

. The observed data are contaminated by the noise inevitably. The Gaussian process has many theoretical and practical advantages since it can mimic the effect of many random processes. And therefore a lot of noise models can be characterized by the Gaussians. However, in many applications, the response variable of interest is a count, that is, takes on nonnegative integer values. For count data, it is well modeled by a Poisson process. The Poisson process is usually used to describe the occurrences of events and the time points at which the events occur in a given time interval, such as the occurrence of natural disasters and the arrival times of customers at a service center. In the obstacle scattering problem, the counts of photons at a remote closed surface are detected. As seen in

[3, 9, 10], the counts of photons have interacted with the unknown object and can be well modeled by a Poisson process. In view of this, we consider to use the Poisson data to recover the unknown shape.

The stochastic nature of the data undoubtedly results in the uncertainty of the reconstruction. Therefore, besides the reconstruction of the unknown object itself, its uncertainty requires the extra concerns. Probabilistic thinking provides a natural way to quantify the uncertainty. Recently, Bayesian inference method has become a popular tool for this purpose

[7]

. In Bayesian approach, the main idea is to translate the prior knowledge on the noise and on the unknowns to prior probability laws and then use the forward model to transmit the prior information to the posterior distribution from which we can deduce any information about the unknowns. A rigorous Bayesian framework is developed for the inverse problems with the infinite dimensional unknowns

[8]

. From the existed works, we can see that the prior for the unknowns is crucial in Bayesian approach and even determines whether or not it works. A difficult subject is how to code the experience prior information in some distribution formula. It is well-known that, subject to certain mild conditions, the sum of a set of random variables, has a distribution that becomes increasingly Gaussian as the number of terms in the sum increases. This point allows us to use Gaussians to model the prior for many real physical parameters. In this paper, we characterize the unknown by two kinds of prior knowledges: the Gaussian and the hybrid priors. Upon these priors, we discuss the well-posedness of the posterior distribution with the Poisson data.

The remainder of this paper is organized in the following: In Section 2, we describe the exterior acoustic scattering problem. In Section 3, the Bayesian approach and the well-posedness are discussed. In the final, some numerical examples are given.

## 2 Acoustic obstacle scattering with Poisson data

For a perfect cylindrical conductor with smooth cross section , the scattering of polarized, transverse magnetic time harmonic electromagnetic waves is described by the Helmholtz equation

 △u+k2u=0inR2∖D, (2.1)

where is the wave number, the total field, the scattered field and the incident wave. We let , where is the incident direction. The following boundary conditions are imposed

 u∣∂D=0, (2.2) limr→∞√r(∂us∂r−ikus)=0,r=|x|. (2.3)

The condition (2.2) corresponds to the sound soft boundary condtion and (2.3) is the so-called Sommerfeld radiation condition. The Sommerfeld radiation condition characterizes the outgoing wave and enables us to have the following asymptotic behaviour for the scattered wave :

 us(x)=eik|x||x|12{u∞(^x)+O(1|x|)}as% |x|→∞, (2.4)

where and is the far field pattern. The direct scattering problem is to find the scattered field for given incident field and obstacle . It is known that there exists a unique solution if is Lipschitz continuous [2].

We focus on the reconstruction of the unknown domain by some observed data: the photon counts of the scattered electromagnetic field far away from the obstacle. Since at large distances the photon density is approximately proportional to [9], the obstacle reconstruction problem can be described by the operator equation

 F(D)=τ|u∞|2, (2.5)

where is a constant corresponding the noise level. We suppose that the domain is starlike, i.e., the boundary is parameterized by

 ∂D=q(t)[cos(t),sin(t)]T,t∈[0,2π). (2.6)

In this setting, the function becomes the unknown and therefore the equation (2.5) is equivalent to

 G(q)=τ|u∞|2. (2.7)

The operator equation can be given by the single-double layer potential theories [2]. Let be of class . Define the single layer potential operator

 (Sφ)(x)=2∫∂DΦ(x,~x)φ(~x)ds(~x),x∈∂D, (2.8)

and the double layer potential operator

 (Kφ)(x)=2∫∂D∂Φ(x,~x)∂ν(~x)φ(~x)ds(~x),x∈∂D, (2.9)

where is the fundamental solution, is the Hankel function of the first kind of order zero and is normal direction. It can be seen in [2] that and are bounded from into . With the single-double layer potentials, the scattered field can be written as

 us(x)=∫∂D{∂Φ(x,~x)∂ν(~x)−iηΦ(x,~x)}φ(~x)ds(~x),x∈R2∖¯¯¯¯¯D, (2.10)

where is the real coupling parameter and the unknown density function. Then the main goal for the direct scattering problem is to determine the unknown density function according to the sound soft boundary condition by

 (I+K−iηS)φ=−2uion∂D. (2.11)

By (2.11), the far field pattern further can be written as

 u∞(^x,d)=e−iπ4√8πk∫∂D(kν(~x)⋅^x+η)e−ik^x⋅~xφ(~x)ds(~x). (2.12)

Equations (2.11), (2.12) together with (2.6) give the forward operator .

## 3 Bayesian inversion

Denote and . We consider the observation is a Poisson point process [9] in with parameters . The goal of statistical inversion is to explore the posterior distribution , i.e., the distribution of conditional on the value of . Bayes’ formula shows that

 dμydμpr=π(y|q), (3.1)

where is the prior distribution for the unknown and the likelihood function.

The Poisson noise yields the likelihood function

 π(y|q)=m∏j=1λyjje−λjyj!. (3.2)

Following [7, 8], the likelihood function can be written in the form of

 π(y|q)∝exp(−Λ(q,y)), (3.3)

where

 Λ(q,y)=m∑j=1(G(q))j−yjlog(G(q))j. (3.4)

The prior is given before acquiring the data. Since the denotes the length, an additional restriction should be imposed: the unknown must be positive. For this purpose, we consider the following two ways [1, 10]

 q(t)=exp(r(t)), (3.5) q(t)=a2(erf(r(t))+b), (3.6)

where and are constants and is the error function

 erf(r(t))=2√π∫r(t)0exp(−t2)dt.

These two ways guarantee that is strictly positive. With these parameterizations, we now infer the new unknown parameter . We still use to denote the prior distribution for . Without any confusion, the forward operator is still denoted by . The prior needs to be chosen carefully and it is not unique. We give different priors for (3.5) and (3.6).

For (3.5), the prior is given by the Gaussian

 μpr=μ1=N(0,C1), (3.7)

where is the covariance function and should be chosen carefully. In [1], a convenient prior to use is

 r′′(t)∼N(0,A−s),s>0, (3.8)

where with the definition domain

 D(A):={v∈H2[0,2π]:∫2π0v(t)dt=0}. (3.9)

In this prior assumption, it can be verified almost surely with , i.e., [1, 7]. In this prior assumption, the well-posedness of the posterior distribution for the Gaussian noise case has been discussed in [1].

For (3.6), the prior distribution is chosen as denoted by . To guarantee the period, we use the covariance function [5, 6]

 C2(s,t)=exp(−2sin2(t−s2)l2), (3.10)

where defines the characteristic length-scale. The covariance example can be viewed as mapping the one-dimensional input variable to the two-dimensional . The distance between and is given

 |r(t)−r(s)|2=(cost−coss)2+(sint−sins)2=4sin2t−s2.

The covariance kernel (3.10) can be obtained by taking the squared exponential kernel in space

 C2(r(t),r(s))=exp(−|r(t)−r(s)|22l2).

We apply a hybrid prior proposed in [10]. On the basis of the Gaussian prior , we further assume that the unknown satisfies the total variation prior as in [10]

 dμTVdμ2(r)∝exp(−∥r∥TV):=exp(−R(r)), (3.11)

where is the TV seminorm,

 ∥r∥TV=ζ∫2π0|drdt|dt (3.12)

with a positive constant . It follows immediately that the posterior distribution has this form

 dμydμ2∝exp(−Λ(r,y)−R(r)):=exp(−Ψ(r,y)). (3.13)

Next, we discuss the well-posedness of the posterior distribution with the hybrid prior. This well-posedness is characterized in the sense of Hellinger metric, which is defined by

 dH(μy,μ~y)=(12∫(√dμydμ2−√dμ~ydμ2)2dμ2)12.

Here, we need to use the following Fernique theorem to some functionals are bounded for Gaussian measures.

###### Theorem 3.1 (Fernique).

If is a Gaussian measure on Banach space , so that , then there exists such that

 ∫Xexp(α∥r∥2X)dμ<∞.
###### Lemma 3.2.

For fixed , there exists such that

 |G(r)|⩽M(1+∥r∥X). (3.14)

for all .

###### Proof.

According to (2.6) and (2.12), we have

 u∞(^x)= e−iπ4√8πk∫2π0(kν(t)⋅^x+η)e−ik^x⋅(qcost,qsint)T~φ(t)√q2+q′2dt,

where . Hence it follows that

 |u∞(^x)|2⩽M1∥∥q2+q′2∥∥∞≤M1(∥q∥2∞+∥q′∥2∞).

By (3.6), we get

 ∥q∥∞≤a2(∥erf(r)∥∞+b)≤a2(1+b), ∥q′∥∞≤a√π∥r′exp(−r2)∥∞≤a√π∥r′∥∞≤M2∥r∥X.

This completes the proof.

###### Lemma 3.3.

For every , there exists constants and such that, for all , and with ,

 I2≤Λ(r,y)≤I1. (3.15)
###### Proof.

For convenient, denote

 Λ(r,y) =⟨λ,1⟩−⟨y,logλ⟩,

where is the inner product in . It is obvious from (3.6) that

 a2(b−1)≤q(t)≤a2(b+1).

Therefore, the boundedness of the forward operator implies that

 0<λ––≤λ≤¯¯¯λ (3.16)

for two constant vectors

, . And it follows immediately that for a positive constant . Furthermore, we obtain for by the Cauchy-Schwarz inequality

 Λ(r,y)≥⟨λ––,1⟩−|logλ||y|≥⟨λ––,1⟩−lmaxs:=I2, Λ(r,y)≤⟨¯¯¯λ,1⟩+|logλ||y|≥⟨¯¯¯λ,1⟩+lmaxs:=I1.
###### Remark 1.

It should be noted that the lower bound in (3.16) probably has component. In this case, we can make a shift for the forward operator by

 ~G(r)=G(r)+e,

where is a positive constant vector. Then we discuss the new problem of the form

 ~G(r)=~λ:=τ|u∞|2+e.

To avoid the addition statement, we still use as the forward operator.

###### Theorem 3.4.

Assume that is the Gaussian measure with covariance function (3.10). For with , there exists such that

 dH(μy,μ~y)≤M|y−~y|. (3.17)
###### Proof.

First, it should be pointed out that the posterior measure is absolute continuous about the prior measure . This will guarantee the existence of the Radon-Nikodym derivative . In fact, for any measurable subset , it holds that

 μy(X1)=∫X1exp(−Ψ(r,y))dμ2≤∫X1dμ2.

Therefore implies that . Define

 A(y)=∫Xexp(−Ψ(r,y))dμ2.

It is obvious that

 A(y)≤1.

By Lemma 3.3, we have

 A(y)≥∫∥r∥X≤1exp(−Ψ(r,y))dμ2 =∫∥r∥X≤1exp(−Λ(r,y)−R(r))dμ2 =∫∥r∥X≤1exp(−m∑j=1{(G(r))j−yjlog(G(r))j}−∥r∥TV)dμ2 ≥∫∥r∥X≤1exp(−I1)exp(−∥r∥TV)dμ2 ≥exp(−I1)∫∥r∥X≤1exp(−ζ∫2π0|r′(t)|dt)dμ2 ≥exp(−I1)∫∥r∥X≤1exp(−2πζ∥r∥X)dμ2 ≥exp(−I1)exp(−2πζ)∫∥r∥X≤1dμ2.

Since is the Gaussian measure, the unit ball has positive measure. This shows that is strictly positive. This together with this absolute continuous shows that the posterior distribution is well-defined. The Hellinger metric holds that

 d2H(μy,μ~y)=12∫(√dμydμ2−√dμ~ydμ2)2dμ2 (3.18) =12∫⎧⎪⎨⎪⎩(exp(−Ψ(r,y))A(y))12−(exp(−Ψ(r,~y))A(~y))12⎫⎪⎬⎪⎭2dμ2 =12∫⎧⎪⎨⎪⎩(exp(−Ψ(r,y))A(y))12−(exp(−Ψ(r,~y))A(y))12⎫⎪⎬⎪⎭2dμ2 +12∫⎧⎪⎨⎪⎩(exp(−Ψ(r,~y))A(y))12−(exp(−Ψ(r,~y))A(~y))12⎫⎪⎬⎪⎭2dμ2 ≤12A(y)∫{exp(−Ψ(r,y)2)−exp(−Ψ(r,~y)2)}2dμ2 +12|A(y)−12−A(~y)−12|2∫exp(−Ψ(r,~y))dμ2:=Σ1+Σ2.

For , since

is bounded, it suffices to estimate the integral part. According to Lemma

3.2 and Fernique therorem, we have

 ∫{exp(−Ψ(r,y)2)−exp(−Ψ(r,~y)2)}2dμ2 ≤14∫|Ψ(r,y)−Ψ(r,~y)|2dμ2 =14∫|m∑j=1(~yi−yi)(logG(r))i|2dμ2 ≤14|y−~y|2∫|logG(r)|2dμ2 ≤M24|y−~y|2∫log2(1+∥r∥X)dμ2≤~M|y−~y|2. (3.19)

Since and are bounded from below, it follows that

 |A(y)−12−A(~y)−12|2 ≤Mmax(A(y)−3,A(~y)−3)|A(y)−A(~y)|2≤M|y−~y|2. (3.20)

The estimation (3.17) can be obtained from (3.18), (3.19) and (3.20).

###### Remark 2.

The requirement in Fernique theorm is guaranteed naturally since it has the exponential covariance function (3.10). This case for with the form (3.5) is similar to (3.6). Actually, we just need to take and verify the condition .

## 4 Numerical test

In this section, we demonstrate some numerical examples to show the effectiveness of the Bayesian method for the inverse obstacle scattering problem with the Poisson data. We are ready to use the MCMC method to generate samples to explore the posterior distribution in (3.1).

It is clear that the operator in (3.9) has the eigen-system . With this, the prior for (3.5) can be generated by the Karhunen-Loève expansion. The truncated expansion is used to generate the prior samples. The corresponding derivatives can be computed accurately. For (3.6), the samples for with the covariance (3.10

) are period. We use the fast Fourier transform (FFT) to implement the numerical computations for the derivatives of the samples. The posterior samples for the two ways are both generated by the Metropolis-Hastings algorithm.

We just list the hybrid prior case in the following. For using (3.5) and the corresponding prior, the process is similar.

• Initialize and further generate in (3.6).

• For iteration do

• Propose and further generate .

• Acceptance probability:

 α(^r;ri−1)=min{1,exp(−Ψ(^r,y))exp(−Ψ(ri−1,y))}, υ∼Uniform(0,1).
• Accept or reject the proposal in the following way

 ri={^r,ifυ>α,ri−1,else.

endfor

Consider the following obstacles

 1:Peanut:q(t)=3(cos2t+0.25sin2t),x1(t)=q(t)cost,x2(t)=q(t)sint; 2:Kite:(x1,x2)=(cost+0.65cos2t−0.65,1.5sint); 3:Drop:x1=−1+2sin(t/2),x2=−sin(t); 4:Cloverleaf:q(t)=1+0.3cos(4t),x1(t)=q(t)cost,x2(t)=q(t)sin(t),

where .

In Fig. 1, we plot the Poisson data for . Using these data, we reconstruct the obstacles displayed in Fig. 2 and 3. The posterior sample means are used as the approximations. In Fig. 2, we plot the confidence interval for the peanut shape.

## References

• [1] T. Bui-Thanh and O. Ghattas, An Analysis of Infinite Dimensional Bayesian Inverse Shape Acoustic Scattering and Its Numerical Approximation. SIAM/ASA Journal on Uncertainty Quantification 2, no. 1, 203-222, 2014.
• [2] D. Colton and R. Kress, Inverse Acoustic and Electromagnetic Scattering Theory (Third Edition), Springer, New York, 2013.
• [3] T. Hohage and F. Werner, Inverse problems with Poisson data: statistical regularization theory, applications and algorithms, Inverse Problems 32 (2016) 093001 (56pp).
• [4] Z. Li, Z. Deng and J. Sun, Limited aperture inverse scattering problems using Bayesian approach and extended sampling method, prepared.
• [5]

C. Rasmussen and C. Williams, Gaussian Processes for Machine Learning, The MIT Press Cambridge, Massachusetts London, England, 2006.

• [6] A. Solin and S. Särkkä, Explicit link between periodic covariance functions and state space models,
• [7] A. Stuart, Uncertainty Quantification in Bayesian Inversion, https://homepages.warwick.ac.uk/ masdr/TALKS/stuartICM.pdf.
• [8] A. Stuart, Inverse problems: a Bayesian perspective, Acta Numerica, 19: 451-559, 2010.
• [9] F. Werner, Inverse problems with Poisson data: Tikhonov-type regularization and iteratively regularized Newton methods, Dissertation, Göttingen, 2011.
• [10] Q. Zhou, X. Zhang and J. Li, Bayesian inference and uncertainty quantification for image reconstruction with Poisson data,

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text: 8. The value of a savings bond (in dollars) is a function of the age of the bond (in months). This function is written . Which is true about this function? Answer Choices: V(23) is the value of the bond when it is worth $23. V(16) is the time when the bond is worth$16. V(90) is the value of the bond after 90 month


text:
×

# What Are Natural Numbers?

Now I know, as we all learn, natural numbers are positive integers. But how do we define positive?

The simple answer would be, greater than zero. However, that brings up the question of being "greater than". What exactly does "greater than" mean? I've heard definitions that $$a$$ is greater than $$b$$ if $$a-b$$ > $$0$$. But that definition has "greater than" in it!

Note by Clive Chen
1 year, 2 months ago

Sort by:

Positive numbers are numbers such that if you take the square root of it, the resulted number is real. · 1 year, 2 months ago

So Zero is positive? · 1 year ago

"Greater than" is a total ordering of the natural numbers.

A total ordering applied to a set $$X$$ assigns to any two elements which is "greater" (>) in a way that obeys the following rules:

• For any $$x,y$$, either $$x\ge y$$ or $$y\ge x$$. (That's the total part)
• If $$x \ge y$$ and $$y\ge x$$, then $$x=y$$. (So "equal" in the order is the same as equal in the set)
• If $$x \ge y$$ and $$y\ge z$$, then $$x\ge z$$. ("Greater than" is transitive)

A definition of the natural numbers might include the total order we learned as children $$\ldots,-2,-1,0,1,2,\ldots$$ (Although of course there are additional structures on the integers).

So $$b-a>0$$ is really just something we've all agreed upon, from this point of view. · 1 year, 2 months ago


text:
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# underdog in a sentence

Can we drop unnecessary computations for easy inputs? This course is being taught at as part of Master Datascience Paris Saclay. Inria. We thank the Orange-Keyrus-Thalès chair for supporting this class. Bayesian methods can Impose useful priors on Neural Networks helping discover solutions of special form; Provide better predictions; Provide Neural Networks with uncertainty estimates (uncovered) Neural Networks help us make more efficient Bayesian inference; Uses a lot of math; Active area of research lectures-labs maintained by m2dsupsdlclass, Convolutional Neural Networks for Image Classification, Deep Learning for Object Detection and Image Segmentation, Sequence to sequence, attention and memory, Expressivity, Optimization and Generalization, Imbalanced classification and metric learning, Unsupervised Deep Learning and Generative models, Demo: Object Detection with pretrained RetinaNet with Keras, Backpropagation in Neural Networks using Numpy, Neural Recommender Systems with Explicit Feedback, Neural Recommender Systems with Implicit Feedback and the Triplet Loss, Fine Tuning a pretrained ConvNet with Keras (GPU required), Bonus: Convolution and ConvNets with TensorFlow, ConvNets for Classification and Localization, Character Level Language Model (GPU required), Transformers (BERT fine-tuning): Joint Intent Classification and Slot Filling, Translation of Numeric Phrases with Seq2Seq, Stochastic Optimization Landscape in Pytorch. Nature 2015 However, while deep learning has proven itself to be extremely powerful, most of today’s most successful deep learning systems suffer from a number of important limitations, ranging from the requirement for enormous training data sets to lack of interpretability to vulnerability to … 6.S191: Introduction to Deep Learning The slides are published under the terms of the CC-By 4.0 In this study, we used two deep-learning algorithms based … Predicting survival after hepatocellular carcinoma resection using deep-learning on histological slides Hepatology. This automatic feature learning has been demonstrated to uncover underlying structure in the data leading to state-of-the-art results in tasks in vision, speech and rapidly in other domains as well. CNNs are the current state-of-the-art architecture for medical image analysis. Dimensions of a learning system (different types of feedback, representation, use of knowledge) 3. UC Berkeley has done a lot of remarkable work on deep learning, including the famous Caffe — Deep Leaning Framework. "Learning representations by back-propagating errors." Cognitive modeling 5.3 (1988): 1. The course is Berkeley’s current offering of deep learning. Video and slides of NeurIPS tutorial on Efficient Processing of Deep Neural Networks: from Algorithms to Hardware Architectures available here. Computationally stained slides could help automate the time-consuming process of slide staining, but Shah said the ability to de-stain and preserve images for future use is the real advantage of the deep learning techniques. Description. We plan to offer lecture slides accompanying all chapters of this book. • LeCun, Yann, et al. Deep Learning Handbook. The Deep Learning case! Direct links to the rendered notebooks including solutions (to be updated in rendered mode): This lecture is built and maintained by Olivier Grisel and Charles Ollion, Charles Ollion, head of research at Heuritech - deep learning is driving significant advancements across industries, enterprises, and our everyday lives. • 1993: Nvidia started… • Hinton, Geoffrey E., Simon Osindero, and Yee-Whye Teh. Recently, deep learning has produced a set of image analysis techniques that automatically extract relevant features, transforming the field of computer vision. Machine Learning: An Overview: The slides presentintroduction to machine learningalong with some of the following: 1. "Backpropagation applied to handwritten zip code recognition." As data volumes keep growing, it has become customary to train large neural networks with hundreds of millions of parameters to maintain enough capacity to memorize these volumes and obtain state-of-the-art accuracy. Note: press “P” to display the presenter’s notes that include some comments and Obtain a sample from (or the mode statistic of) the true posterior $$p(y, z \mid x) \propto p(y|x, z) p(z)$$, We define some joint model $$p(y, \theta | x) = p(y | x, \theta) p(\theta)$$, We obtain observations $$\mathcal{D} = \{ (x_1, y_1), ..., (x_N, y_N) \}$$, We would like to infer possible values of $$\theta$$ given  observed data $$\mathcal{D}$$ $$p(\theta \mid \mathcal{D}) = \frac{p(\mathcal{D} | \theta) p(\theta)}{\int p(\mathcal{D}|\theta) p(\theta) d\theta}$$, We will be approximating true posterior distribution with an approximate one, Need a distance between distributions to measure how good the approximation is $$\text{KL}(q(x) || p(x)) = \mathbb{E}_{q(x)} \log \frac{q(x)}{p(x)} \quad\quad \textbf{Kullback-Leibler divergence}$$, Not an actual distance, but $$\text{KL}(q(x) || p(x)) = 0$$ iff $$q(x) = p(x)$$ for all $$x$$ and is strictly positive otherwise, Will be minimizing $$\text{KL}(q(\theta) || p(\theta | \mathcal{D}))$$ over $$q$$, We'll take $$q(\theta)$$ from some tractable parametric family, for example Gaussian $$q(\theta | \Lambda) = \mathcal{N}(\theta \mid \mu(\Lambda), \Sigma(\Lambda))$$, Then we reformulate the objective s.t. Deep Learning algorithms aim to learn feature hierarchies with features at higher levels in the hierarchy formed by the composition of lower level features. Generator network and inference network essentially give us autoencoder, Inference network encodes observations into latent code, Generator network decodes latent code into observations, Can infer high-level abstract features of existing objects, Uses neural network to amortize inference, Bayesian methods are useful when we have low data-to-parameters ratio, Impose useful priors on Neural Networks helping discover solutions of special form, Provide Neural Networks with uncertainty estimates (uncovered), Neural Networks help us make more efficient Bayesian inference. license. Deep Learning An MIT Press book in preparation Ian Goodfellow, Yoshua Bengio and Aaron Courville. The course covers the basics of Deep Learning, with a focus on applications. Jeez, how is that related to this slide? The Jupyter notebooks for the labs can be found in the labs folder of Deep learning algorithms are similar to how nervous system structured where each neuron connected each other and passing information. Lecture slides for Chapter 4 of Deep Learning www.deeplearningbook.org Ian Goodfellow Last modiﬁed 2017-10-14 Thanks to Justin Gilmer and Jacob Buckman for helpful discussions (Goodfellow 2017) Numerical concerns for implementations of deep learning algorithms Gradient-based optimization in discrete models is hard, Invoke the Central Limit Theorem and turn the model into a continuous one, Consider a model with continuous noise on weights $$q(\theta_i | \Lambda) = \mathcal{N}(\theta_i | \mu_i(\Lambda), \alpha_i(\Lambda) \mu^2_i(\Lambda))$$, Neural Networks have lots of parameters, surely there's some redundancy in them, Let's take a prior $$p(\theta)$$ that would encourage large $$\alpha$$, Large $$\alpha_i$$ would imply that weight $$\theta_i$$ is unbounded noise that corrupts predictions, Such weights won't be doing anything useful, hence it should be zeroed out by putting $$\mu_i(\Lambda) = 0$$, Thus the weight $$\theta_i$$ would effectively turn into a deterministic 0. Its uncertainty quantified by the, This requires us to know the posterior distribution on model parameters $$p(\theta \mid \mathcal{D})$$ which we obtain using the Bayes' rule, Suppose the model $$y \sim \mathcal{N}(\theta^T x, \sigma^2)$$, with $$\theta \sim \mathcal{N}(\mu_0, \sigma_0^2 I)$$, Suppose we observed some data from this model $$\mathcal{D} = \{(x_n, y_n)\}_{n=1}^N$$ (generated using the same $$\theta^*$$), We don't know the optimal $$\theta$$, but the more data we observe, Posterior predictive would also be Gaussian $$p(y|x, \mathcal{D}) = \mathcal{N}(y \mid \mu_N^T x, \sigma_N^2)$$, Suppose we observe a sequence of coin flips $$(x_1, ..., x_N, ...)$$, but don't know whether the coin is fair $$x \sim \text{Bern}(\pi), \quad \pi \sim U(0, 1)$$, First, we infer posterior distribution on a hidden parameter $$\pi$$ having observed \(x_{


text:
Limited access

Given a regular hexagon, which of the following statements are true?

I. - The angle of rotational symmetry is $60°$.
II. - There are 3 lines of symmetry.
III. - Let $n$ be the number of lines of symmetry for a regular hexagon. Then the number of triangles created by the lines of symmetry can be expressed as $2n$.

A

I only.

B

I and II only.

C

I and III only.

D

II and III only.

E

I, II, and III.

Select an assignment template


text:
Differences of Calculated Results

Introduction

For data generated by

Various differences can be computed. Input data can be typically categorized as follows:

• Category K0: ${\displaystyle f(x,y,z)}$, time-independent quantities;
• Category K1: ${\displaystyle f(x,y,z,t_{1})}$, time-dependent quantities, one time step;
• Category KC: ${\displaystyle f(x,y,z,t_{i})}$, time-dependent quantities, several discrete time steps, constant time step ${\displaystyle \Delta _{t}}$;
• Category KN: ${\displaystyle f(x,y,z,t_{i})}$, time-dependent quantities, several discrete time steps, varying time step ${\displaystyle \Delta _{t}(i)}$.

For geophysical data categories K1, KC and KN are of significance. Examples:

• Category K1: topography/bathymetry ${\displaystyle h(x,y,z,t_{1})}$ for a specific instant in time;
• Category KC: water level ${\displaystyle \eta (x,y,z,t_{i})}$ at discrete times ${\displaystyle t_{i}}$ with constant time step, e. g. computed by a mathematical model;
• Category KN: tidal high water ${\displaystyle \eta ^{\rm {HW}}(x,y,z,t_{i})}$ for times ${\displaystyle t_{i}}$ at non-equidistant time intervals, e.g. derived from a water level time serie.

Definitions

• reference data ${\displaystyle r}$: with respect to ${\displaystyle r}$ various deviations for ${\displaystyle f}$ can be evaluated. Typical data are either observational data or computational as well as analysis results for a specific (reference) state (situation);
• variant data ${\displaystyle f}$: can be also either observational data or computational as well as analysis results, for which deviations shall be computed with respect to the reference state. Typically variant data are given for a different period in time (natural variation) or a different state of the system under study.
• valid operator 1: ${\displaystyle V(r_{i})}$ returns .T. or .F., in dependence whether ${\displaystyle r_{i}}$ is valid or invalid. Can be also applied to ${\displaystyle f_{i}}$.
• valid operator 2: ${\displaystyle V(r_{I},f_{i})}$ returns .T. or .F., in dependence whether ${\displaystyle V(r_{i})\land V(f_{i})}$ is valid or invalid.

Requirements for the computation of differences

The following requirements must be fulfilled by ${\displaystyle r}$ and ${\displaystyle f}$:

1. ${\displaystyle r}$ and ${\displaystyle f}$ must belong to the same category (see above);
2. the number of times ${\displaystyle t_{i}}$ must be identical for ${\displaystyle r}$ and ${\displaystyle f}$;
3. for data belonging to category KC constant time steps must coincide ${\displaystyle \Delta t}$ for ${\displaystyle r}$ and ${\displaystyle f}$;
4. (physical) dimension as well as meaning must be equivalent for ${\displaystyle r}$ and ${\displaystyle f}$;
5. ${\displaystyle r_{i}}$ (short for ${\displaystyle r(x,y,z,t_{i})}$) as well as ${\displaystyle f_{i}}$ (short for ${\displaystyle r(x,y,z,t_{i})}$) must be valid data for the same instant ${\displaystyle i}$ in time; otherwise the dervied results will become invalid.

Computational results

Program NCDELTA can be used to compute all subsequent results. Locations of ${\displaystyle r}$ are not required to coincide with those of ${\displaystyle f}$. Values ${\displaystyle r}$ are interpolated to locations of ${\displaystyle f}$, as long as the geographical distance between the different locations does not exceed ${\displaystyle R^{\max }}$. In case the distance exceeds that limit, no results will be computed. In such a situation an invalid result value will be generated. The follwing results can be computed using NCDELTA.

Ordinary differences

Difference

A result is computed for all times (one value for time-independent data) at all locations ${\displaystyle (x,y,z)}$:

1. The difference between ${\displaystyle f_{i}}$ and ${\displaystyle r_{i}}$ is calculated in case ${\displaystyle V(r_{i},f_{i})}$ returns .T.:
${\displaystyle d_{i}=f_{i}-r_{i}}$, if ${\displaystyle V(r_{i},f_{i})}$;
2. Result will be invalid, if ${\displaystyle V(r_{i},f_{i})}$ returns .F.:
${\displaystyle d_{i}={\rm {invalid}}}$ if ${\displaystyle \lnot V(r_{i},f_{i})}$.

Results are computed for data belonging to categories K0, K1, KC und KN, which means for all types of data.

Percentiles

back to Pre- and Postprocessing


text:
The energy of a hydrogen atom in its ground state is —13.6 eV. The energy of the level corresponding to the quantum number n = 2 (first excited state) in the hydrogen atom is:

1. -2.72 eV

2. - 0.85 eV

3. -0.54 eV

4. — 3.4 eV

Concept Questions :-

Bohr's model of atom
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

When neutron moving with Kinetic Energy 2eV collides with

stationary ${H}_{1}^{1}$ in the ground state, the collision will be:

1.  Must be elastic

2.  may be inelastic

3.  Both

4.  Must be perfectly inelastic

Concept Questions :-

Various atomic model
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The potential difference applied to an X-ray tube is 5 kV and current through it is 3.2 mA. The number of electrons striking the target per sec is:

1.

2.

3.

4.

Concept Questions :-

X-ray
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Magnetic moment due to the motion of the electron in ${\mathrm{n}}^{\mathrm{th}}$  energy state of a hydrogen atom is proportional to

1.  n

2.

3.

4.  ${\mathrm{n}}^{3}$

Concept Questions :-

Bohr's model of atom
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

When the electron of a hydrogen-like atom jumps from a higher energy level to a lower energy level, then

1.  Angular momentum of the electron remains constant

2.  The kinetic energy of the electron increases

3.  The wavelength of the de-Broglie wave associated with the motion of the electron increases

4.  Potential energy increases

Concept Questions :-

Bohr's model of atom
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The frequency Of radiation emitted during the transition of an electron from a second excited state to a first excited state in H-atom is ${\mathrm{f}}_{0}.$ The frequency of the same transition emitted by a singly ionized He ion is

1.

2.

3.

4.  $\frac{{\mathrm{f}}_{0}}{4}$

Concept Questions :-

Bohr's model of atom
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

Which of the following transition in a hydrogen atom will produce radiations of minimum wavelength?

1.  n = 2 to n = 1

2.  n = 5 to n = 3

3.  n = 10 to n = 5

4.  n = 10 to n = 3

Concept Questions :-

Spectral series
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The angular speed of electron in a hydrogen atom in ${\mathrm{n}}^{\mathrm{th}}$ orbit is proportional to -

1.

2.

3.

4.  ${\mathrm{n}}^{2}$

Concept Questions :-

Bohr's model of atom
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

1.

2.

3.

4.  $\sqrt{{\lambda }_{1}{\lambda }_{2}}$

Concept Questions :-

Spectral series
High Yielding Test Series + Question Bank - NEET 2020

Difficulty Level:

The shortest wavelength of Balmer series is about

1.  912$\stackrel{\mathrm{o}}{\mathrm{A}}$

2.  121$\stackrel{\mathrm{o}}{\mathrm{A}}$

3.  3648$\stackrel{\mathrm{o}}{\mathrm{A}}$

4.  4864$\stackrel{\mathrm{o}}{\mathrm{A}}$

Concept Questions :-

Spectral series


text:
This paper addresses the problem of characterizing classes of pairs of binary trees of equal size for which a signed reassociation sequence, in the Eliahou-Kryuchkov sense, can be shown to exist, either with a size induction hypothesis (reducible pairs), or without it (solvable pairs). A few concepts proposed by Cooper, Rowland and Zeilberger, in the context of a language-theoretic approach to the problem, are here reformulated in terms of signed reassociation sequences, and some of their results are recasted and proven in this framework. A few strategies, tactics and combinations thereof for signed reassociation are introduced, which prove useful to extend the results obtained by the aforementioned authors to new classes of binary tree pairs. In particular, with reference to path trees, {\it i.e.\/} binary trees that have a leaf at every level, we show the reducibility of pairs where (at least) one of the two path trees has a triplication at the first turn below the top level, and we characterize a class of weakly mutually crooked path tree pairs that are neither reducible nor solvable by any previously known result, but prove solvable by appropriate reassociation strategies. This class also includes a subclass of mutually crooked path tree pairs. A summary evaluation of the achieved results, followed by an outline of open questions and future research directions conclude the paper.

### Reducibility and Solvability of Some Classes of Kryuchkov Binary Tree Pairs

#### Abstract

This paper addresses the problem of characterizing classes of pairs of binary trees of equal size for which a signed reassociation sequence, in the Eliahou-Kryuchkov sense, can be shown to exist, either with a size induction hypothesis (reducible pairs), or without it (solvable pairs). A few concepts proposed by Cooper, Rowland and Zeilberger, in the context of a language-theoretic approach to the problem, are here reformulated in terms of signed reassociation sequences, and some of their results are recasted and proven in this framework. A few strategies, tactics and combinations thereof for signed reassociation are introduced, which prove useful to extend the results obtained by the aforementioned authors to new classes of binary tree pairs. In particular, with reference to path trees, {\it i.e.\/} binary trees that have a leaf at every level, we show the reducibility of pairs where (at least) one of the two path trees has a triplication at the first turn below the top level, and we characterize a class of weakly mutually crooked path tree pairs that are neither reducible nor solvable by any previously known result, but prove solvable by appropriate reassociation strategies. This class also includes a subclass of mutually crooked path tree pairs. A summary evaluation of the achieved results, followed by an outline of open questions and future research directions conclude the paper.
##### Scheda breve Scheda completa Scheda completa (DC)
Trees, Coloring of graphs and hypergraphs, Signed and weighted graphs, Enumeration in graph theory.
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text:
Matematicheskii Sbornik. Novaya Seriya
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 Mat. Sb.: Year: Volume: Issue: Page: Find

 Mat. Sb. (N.S.), 1972, Volume 87(129), Number 1, Pages 58–66 (Mi msb3035)

Cycle breaking in fiberings on analytic curves

Yu. S. Ilyashenko

Abstract: In this paper various fiberings are constructed on analytic curves in which the topological reconstruction of the fibers proceeds on a nonanalytic set.
Let $D\subset C^1$ be the strip $-1<\operatorname{Re}\zeta<1$ and $D_1\subset D$ the strip $-1<\operatorname{Re}\zeta<0$. Analytic mappings $F\colon C^5\to C^4$ and $f\colon D\to C^5$ are constructed such that 1) for each $\zeta\in D_1$ the fiber $\chi_\zeta$ of the mapping $F$ which passes through the point $f(\zeta)$ has nontrivial fundamental group; 2) for each $\zeta\in{D\setminus D_1}$ the fiber $\chi_\zeta$ is simply connected.
Next it is shown that the generalization of the Petrovskii–Landis Hypothesis on the conservation of cycles for the equations $\dot z=V(z)$, $z\in C^n$, with analytic right-hand side $V(z)$, is valid. Indeed, in $C^2$ we construct a family $\alpha_\zeta$ of equations of the indicated form, analytic in $\zeta$ and such that 1) for each $\zeta\in D_1\setminus N$ ($N$ is a countable set) on one of the solutions of the equations $\alpha_\zeta$ there is a limit cycle $l(\zeta)$; 2) the cycle $l(\zeta)$ changes continuously as $\zeta$ runs over $D_1\setminus N$ and is broken as $\zeta$ approaches a point on the straight line $\operatorname{Re}\zeta=0$. Some related examples are also constructed.
Bibliography: 6 titles.

Full text: PDF file (1000 kB)
References: PDF file   HTML file

English version:
Mathematics of the USSR-Sbornik, 1972, 16:1, 60–68

Bibliographic databases:

UDC: 516.2+517.9
MSC: Primary 32C15, 32E10; Secondary 32L05

Citation: Yu. S. Ilyashenko, “Cycle breaking in fiberings on analytic curves”, Mat. Sb. (N.S.), 87(129):1 (1972), 58–66; Math. USSR-Sb., 16:1 (1972), 60–68

Citation in format AMSBIB
\Bibitem{Ily72} \by Yu.~S.~Ilyashenko \paper Cycle breaking in fiberings on analytic curves \jour Mat. Sb. (N.S.) \yr 1972 \vol 87(129) \issue 1 \pages 58--66 \mathnet{http://mi.mathnet.ru/msb3035} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=307251} \zmath{https://zbmath.org/?q=an:0234.34034|0263.34026} \transl \jour Math. USSR-Sb. \yr 1972 \vol 16 \issue 1 \pages 60--68 \crossref{https://doi.org/10.1070/SM1972v016n01ABEH001349}

• http://mi.mathnet.ru/eng/msb3035
• http://mi.mathnet.ru/eng/msb/v129/i1/p58

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This publication is cited in the following articles:
1. Yu. S. Ilyashenko, “Fiberings into analytic curves”, Math. USSR-Sb., 17:4 (1972), 551–569
2. D. V. Anosov, “On generic properties of closed geodesics”, Math. USSR-Izv., 21:1 (1983), 1–29
3. Ivashkovich S., “Vanishing Cycles in Holomorphic Foliations by Curves and Foliated Shells”, Geom Funct Anal, 21:1 (2011), 70–140
•  Number of views: This page: 205 Full text: 75 References: 45 First page: 2


text:
# loss

Evaluate accuracy of learned feature weights on test data

## Syntax

err = loss(mdl,X,Y) err = loss(mdl,X,Y,Name,Value)

## Description

err = loss(mdl,X,Y) computes the misclassification error of the model mdl, for the predictors in X and the class labels in Y.

err = loss(mdl,X,Y,Name,Value) computes the classification error with additional options specified by one or more Name,Value pair arguments.

## Input Arguments

expand all

Neighborhood component analysis model for classification, returned as a FeatureSelectionNCAClassification object.

Predictor variable values, specified as an n-by-p matrix, where n is the number of observations and p is the number of predictor variables.

Data Types: single | double

Class labels, specified as a categorical vector, logical vector, numeric vector, string array, cell array of character vectors of length n, or character matrix with n rows, where n is the number of observations. Element i or row i of Y is the class label corresponding to row i of X (observation i).

Data Types: single | double | logical | char | string | cell | categorical

### Name-Value Pair Arguments

Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.

Loss function type, specified as a comma-separated pair consisting of 'Loss Function' and one of the following.

• 'classiferror' — Misclassification error, defined as

$\frac{1}{n}\sum _{i=1}^{n}I\left({k}_{i}\ne {t}_{i}\right),$

where ${k}_{i}$ is the predicted class and ${t}_{i}$ is the true class for observation i. $I\left({k}_{i}\ne {t}_{i}\right)$ is the indicator for when the ${k}_{i}$ is not the same as ${t}_{i}$.

• 'quadratic' — Quadratic loss function, defined as

$\frac{1}{n}\sum _{i=1}^{n}\sum _{k=1}^{c}{\left({p}_{ik}-I\left(i,k\right)\right)}^{2},$

where c is the number of classes, ${p}_{ik}$ is the estimate probability that ith observation belongs to class k, and $I\left(i,k\right)$ is the indicator that ith observation belongs to class k.

Example: 'LossFunction','quadratic'

## Output Arguments

expand all

Smaller-the-better accuracy measure for learned feature weights, returned as a scalar value. You can specify the measure of accuracy using the LossFunction name-value pair argument.

## Examples

expand all

load('twodimclassdata.mat');

This data set is simulated using the scheme described in [1]. This is a two-class classification problem in two dimensions. Data from the first class (class –1) are drawn from two bivariate normal distributions or with equal probability, where , , and . Similarly, data from the second class (class 1) are drawn from two bivariate normal distributions or with equal probability, where , , and . The normal distribution parameters used to create this data set result in tighter clusters in data than the data used in [1].

Create a scatter plot of the data grouped by the class.

figure gscatter(X(:,1),X(:,2),y) xlabel('x1') ylabel('x2')

Add 100 irrelevant features to . First generate data from a Normal distribution with a mean of 0 and a variance of 20.

n = size(X,1); rng('default') XwithBadFeatures = [X,randn(n,100)*sqrt(20)];

Normalize the data so that all points are between 0 and 1.

XwithBadFeatures = bsxfun(@rdivide,... bsxfun(@minus,XwithBadFeatures,min(XwithBadFeatures,[],1)), ... range(XwithBadFeatures,1)); X = XwithBadFeatures;

Fit a neighborhood component analysis (NCA) model to the data using the default Lambda (regularization parameter, ) value. Use the LBFGS solver and display the convergence information.

ncaMdl = fscnca(X,y,'FitMethod','exact','Verbose',1, ... 'Solver','lbfgs');
 o Solver = LBFGS, HessianHistorySize = 15, LineSearchMethod = weakwolfe |====================================================================================================| | ITER | FUN VALUE | NORM GRAD | NORM STEP | CURV | GAMMA | ALPHA | ACCEPT | |====================================================================================================| | 0 | 9.519258e-03 | 1.494e-02 | 0.000e+00 | | 4.015e+01 | 0.000e+00 | YES | | 1 | -3.093574e-01 | 7.186e-03 | 4.018e+00 | OK | 8.956e+01 | 1.000e+00 | YES | | 2 | -4.809455e-01 | 4.444e-03 | 7.123e+00 | OK | 9.943e+01 | 1.000e+00 | YES | | 3 | -4.938877e-01 | 3.544e-03 | 1.464e+00 | OK | 9.366e+01 | 1.000e+00 | YES | | 4 | -4.964759e-01 | 2.901e-03 | 6.084e-01 | OK | 1.554e+02 | 1.000e+00 | YES | | 5 | -4.972077e-01 | 1.323e-03 | 6.129e-01 | OK | 1.195e+02 | 5.000e-01 | YES | | 6 | -4.974743e-01 | 1.569e-04 | 2.155e-01 | OK | 1.003e+02 | 1.000e+00 | YES | | 7 | -4.974868e-01 | 3.844e-05 | 4.161e-02 | OK | 9.835e+01 | 1.000e+00 | YES | | 8 | -4.974874e-01 | 1.417e-05 | 1.073e-02 | OK | 1.043e+02 | 1.000e+00 | YES | | 9 | -4.974874e-01 | 4.893e-06 | 1.781e-03 | OK | 1.530e+02 | 1.000e+00 | YES | | 10 | -4.974874e-01 | 9.404e-08 | 8.947e-04 | OK | 1.670e+02 | 1.000e+00 | YES | Infinity norm of the final gradient = 9.404e-08 Two norm of the final step = 8.947e-04, TolX = 1.000e-06 Relative infinity norm of the final gradient = 9.404e-08, TolFun = 1.000e-06 EXIT: Local minimum found.

Plot the feature weights. The weights of the irrelevant features should be very close to zero.

figure semilogx(ncaMdl.FeatureWeights,'ro') xlabel('Feature index') ylabel('Feature weight') grid on

Predict the classes using the NCA model and compute the confusion matrix.

ypred = predict(ncaMdl,X); confusionchart(y,ypred)

Confusion matrix shows that 40 of the data that are in class –1 are predicted as belonging to class –1. 60 of the data from class –1 are predicted to be in class 1. Similarly, 94 of the data from class 1 are predicted to be from class 1 and 6 of them are predicted to be from class –1. The prediction accuracy for class –1 is not good.

All weights are very close to zero, which indicates that the value of used in training the model is too large. When , all features weights approach to zero. Hence, it is important to tune the regularization parameter in most cases to detect the relevant features.

Use five-fold cross-validation to tune for feature selection by using fscnca. Tuning means finding the value that will produce the minimum classification loss. To tune using cross-validation:

1. Partition the data into five folds. For each fold, cvpartition assigns four-fifths of the data as a training set and one-fifth of the data as a test set. Again for each fold, cvpartition creates a stratified partition, where each partition has roughly the same proportion of classes.

cvp = cvpartition(y,'kfold',5); numtestsets = cvp.NumTestSets; lambdavalues = linspace(0,2,20)/length(y); lossvalues = zeros(length(lambdavalues),numtestsets);

2. Train the neighborhood component analysis (nca) model for each value using the training set in each fold.

3. Compute the classification loss for the corresponding test set in the fold using the nca model. Record the loss value.

4. Repeat this process for all folds and all values.

for i = 1:length(lambdavalues) for k = 1:numtestsets % Extract the training set from the partition object Xtrain = X(cvp.training(k),:); ytrain = y(cvp.training(k),:); % Extract the test set from the partition object Xtest = X(cvp.test(k),:); ytest = y(cvp.test(k),:); % Train an NCA model for classification using the training set ncaMdl = fscnca(Xtrain,ytrain,'FitMethod','exact', ... 'Solver','lbfgs','Lambda',lambdavalues(i)); % Compute the classification loss for the test set using the NCA % model lossvalues(i,k) = loss(ncaMdl,Xtest,ytest, ... 'LossFunction','quadratic'); end end

Plot the average loss values of the folds versus the values. If the value that corresponds to the minimum loss falls on the boundary of the tested values, the range of values should be reconsidered.

figure plot(lambdavalues,mean(lossvalues,2),'ro-') xlabel('Lambda values') ylabel('Loss values') grid on

Find the value that corresponds to the minimum average loss.

[~,idx] = min(mean(lossvalues,2)); % Find the index bestlambda = lambdavalues(idx) % Find the best lambda value
bestlambda = 0.0037

Fit the NCA model to all of the data using the best value. Use the LBFGS solver and display the convergence information.

ncaMdl = fscnca(X,y,'FitMethod','exact','Verbose',1, ... 'Solver','lbfgs','Lambda',bestlambda);
 o Solver = LBFGS, HessianHistorySize = 15, LineSearchMethod = weakwolfe |====================================================================================================| | ITER | FUN VALUE | NORM GRAD | NORM STEP | CURV | GAMMA | ALPHA | ACCEPT | |====================================================================================================| | 0 | -1.246913e-01 | 1.231e-02 | 0.000e+00 | | 4.873e+01 | 0.000e+00 | YES | | 1 | -3.411330e-01 | 5.717e-03 | 3.618e+00 | OK | 1.068e+02 | 1.000e+00 | YES | | 2 | -5.226111e-01 | 3.763e-02 | 8.252e+00 | OK | 7.825e+01 | 1.000e+00 | YES | | 3 | -5.817731e-01 | 8.496e-03 | 2.340e+00 | OK | 5.591e+01 | 5.000e-01 | YES | | 4 | -6.132632e-01 | 6.863e-03 | 2.526e+00 | OK | 8.228e+01 | 1.000e+00 | YES | | 5 | -6.135264e-01 | 9.373e-03 | 7.341e-01 | OK | 3.244e+01 | 1.000e+00 | YES | | 6 | -6.147894e-01 | 1.182e-03 | 2.933e-01 | OK | 2.447e+01 | 1.000e+00 | YES | | 7 | -6.148714e-01 | 6.392e-04 | 6.688e-02 | OK | 3.195e+01 | 1.000e+00 | YES | | 8 | -6.149524e-01 | 6.521e-04 | 9.934e-02 | OK | 1.236e+02 | 1.000e+00 | YES | | 9 | -6.149972e-01 | 1.154e-04 | 1.191e-01 | OK | 1.171e+02 | 1.000e+00 | YES | | 10 | -6.149990e-01 | 2.922e-05 | 1.983e-02 | OK | 7.365e+01 | 1.000e+00 | YES | | 11 | -6.149993e-01 | 1.556e-05 | 8.354e-03 | OK | 1.288e+02 | 1.000e+00 | YES | | 12 | -6.149994e-01 | 1.147e-05 | 7.256e-03 | OK | 2.332e+02 | 1.000e+00 | YES | | 13 | -6.149995e-01 | 1.040e-05 | 6.781e-03 | OK | 2.287e+02 | 1.000e+00 | YES | | 14 | -6.149996e-01 | 9.015e-06 | 6.265e-03 | OK | 9.974e+01 | 1.000e+00 | YES | | 15 | -6.149996e-01 | 7.763e-06 | 5.206e-03 | OK | 2.919e+02 | 1.000e+00 | YES | | 16 | -6.149997e-01 | 8.374e-06 | 1.679e-02 | OK | 6.878e+02 | 1.000e+00 | YES | | 17 | -6.149997e-01 | 9.387e-06 | 9.542e-03 | OK | 1.284e+02 | 5.000e-01 | YES | | 18 | -6.149997e-01 | 3.250e-06 | 5.114e-03 | OK | 1.225e+02 | 1.000e+00 | YES | | 19 | -6.149997e-01 | 1.574e-06 | 1.275e-03 | OK | 1.808e+02 | 1.000e+00 | YES | |====================================================================================================| | ITER | FUN VALUE | NORM GRAD | NORM STEP | CURV | GAMMA | ALPHA | ACCEPT | |====================================================================================================| | 20 | -6.149997e-01 | 5.764e-07 | 6.765e-04 | OK | 2.905e+02 | 1.000e+00 | YES | Infinity norm of the final gradient = 5.764e-07 Two norm of the final step = 6.765e-04, TolX = 1.000e-06 Relative infinity norm of the final gradient = 5.764e-07, TolFun = 1.000e-06 EXIT: Local minimum found.

Plot the feature weights.

figure semilogx(ncaMdl.FeatureWeights,'ro') xlabel('Feature index') ylabel('Feature weight') grid on

fscnca correctly figures out that the first two features are relevant and that the rest are not. The first two features are not individually informative, but when taken together result in an accurate classification model.

Predict the classes using the new model and compute the accuracy.

ypred = predict(ncaMdl,X); confusionchart(y,ypred)

Confusion matrix shows that prediction accuracy for class –1 has improved. 88 of the data from class –1 are predicted to be from –1, and 12 of them are predicted to be from class 1. 92 of the data from class 1 are predicted to be from class 1 and 8 of them are predicted to be from class –1.

References

[1] Yang, W., K. Wang, W. Zuo. "Neighborhood Component Feature Selection for High-Dimensional Data." Journal of Computers. Vol. 7, Number 1, January, 2012.


text:
# Ramsey number for books

Given a triangular book $B_n$ I am trying to prove that $r(B_n,B_n)\le 4n+2$ where $r(B_n,B_n)$ is defined as the least positive number such that any graph $G$ on $r(B_n,B_n)$ vertices either has a subgraph (not necessarily induced) isomorphic to $B_n$ or its complement does.

Can someone here please give me any ideas or tell me of some techniques which ought to be pursued to solve problems of this kind. Any kind of opinion is welcome.

-
I edited the question so that it is now clear that you are looking for subgraphs isomorphic to $B_n$, but not necessarily induced subgraphs. It is clear that this is the intended meaning of the question as complete graphs and their complements have no induced subgraphs that are nontrivial books at all. –  Stefan Geschke Dec 18 '11 at 9:38
This result is a special case of Theorem 2 from C. C. Rousseau, J. Sheehan: On Ramsey numbers for books; dx.doi.org/10.1002/jgt.3190020110 But perhaps someone will come up with an elementary argument for this particular case. –  Martin Sleziak Dec 18 '11 at 11:07
@Martin Sleziak: Thanks. I will email you. But I am still interested in an elementary argument for the special case. So I won't delete the question. –  Shahab Dec 18 '11 at 11:38

Let $k=r(B_n,B_n)-1$. Then we can color the edges of $K_k$ red and blue in such a way that there is no monochromatic $B_n$. Let $E_r$ and $E_b$ be the sets of red and blue edges, respectively, let $V\,$ be the vertex set of $K_k$, and for $v\in V\,$ let $\deg_r(v)$ and $\deg_b(v)$ be the number of red and blue edges, respectively, incident at $v$.

For each edge $e=\{u,v\}$ let $M(e)$ be the set of vertices that are connected to both $u$ and $v$ by edges of the same color as $e$, and let $\mu(e)=|M(e)|$. Then the number of monochromatic triangles in $K_k$ is $$m=\frac13\sum_{e\in E_r\cup E_n}\mu(e)\;,$$ since each triangle gets counted three times.

Suppose that $e=\{u,v\}\in E_r$. The edge $e$ together with the edges $\{u,w\}$ and $\{v,w\}$ for $w\in M(e)$ yield a red $B_{\mu(e)}$, so we must have $\mu(e)<n$. Similarly, $\mu(e)<n$ for each $e\in E_b$. Thus, $$m\le\frac13(n-1)|E_r\cup E_b|=\frac13(n-1)\binom{k}2\;.\tag{1}$$

Let $v\in V$. Each pair $\langle e_r,e_b\rangle\in E_r\times E_b$ such that $e_r$ and $e_b$ are both incident at $v$ determines a non-monochromatic triangle in $K_k$, so $v$ is a vertex of $\deg_r(v)\deg_b(v)$ non-monochromatic triangles. Each non-monochromatic triangle has two such vertices, so the number of non-monochromatic triangles is $K_k$ is $$\frac12\sum_{v\in V}\deg_r(v)\deg_b(v)\;,$$ and

\begin{align*}m&=\binom{k}3-\frac12\sum_{v\in V}\deg_r(v)\deg_b(v)\\ &=\binom{k}3-\frac12\sum_{v\in V}\deg_r(v)(k-1-\deg_r(v))\\ &=\binom{k}3-\frac{k-1}2\sum_{v\in V}\deg_r(v)+\frac12\sum_{v\in V}\deg_r(v)^2\\ &=\binom{k}3-(k-1)|E_r|+\frac12\sum_{v\in V}\deg_r(v)^2\;. \end{align*}

By Cauchy’s inequality we have $$\left(\sum_{v\in V}\deg_r(v)\right)^2\le\sum_{v\in V}\deg_r(v)^2\; \sum_{v\in V}1^2=k\sum_{v\in V}\deg_r(v)^2\;,$$ so $$4|E_r|^2\ge k\sum_{v\in V}\deg_r(v)^2\;,$$ and

\begin{align*}m&\ge\binom{k}3-(k-1)|E_r|+\frac{2|E_r|^2}k\\ &=\binom{k}3-|E_r|\left(k-1-\frac{2|E_r|}k\right)\\ &=\binom{k}3-|E_r|\frac2k\left(\frac{k(k-1)}2-|E_r|\right)\\ &=\binom{k}3-|E_r|\frac2k\left(\binom{k}2-|E_r|\right)\\ &=\binom{k}3-|E_r|\frac2k|E_b|\;. \end{align*}

Letting $d=\frac12\big||E_r|-|E_b|\big|$, we then have

\begin{align*} m&\ge\binom{k}3-\frac2k\left(\frac12\binom{k}2-d\right)\left(\frac12\binom{k}2+d\right)\\ &=\frac{k(k-1)(k-2)}6-\frac2k\left(\frac14\binom{k}2^2-d^2\right)\\ &=\frac{k(k-1)(k-2)}6-\frac{k(k-1)^2}8+\frac{2d^2}k\\ &=\frac{k(k-1)(k-5)}{24}+\frac{2d^2}k\;, \end{align*}

and combining this with $(1)$ yields

\begin{align*} \frac{k(k-1)(k-5)}{24}+\frac{2d^2}k&\le\frac13(n-1)\binom{k}2\\ &=\frac{k(k-1)(n-1)}6\;, \end{align*}

or

\begin{align*} \frac{2d^2}k&\le\frac{k(k-1)(n-1)}6-\frac{k(k-1)(k-5)}{24}\\ &=\frac{k(k-1)(4n+1-k)}{24}\;. \end{align*}

Then $4n+1-k\ge 0$, so $r(B_n,B_n)=k+1\le 4n+2$.

This is actually a special case of Theorem 1.5 of Li & Zang, Introduction to Graph Ramsey Theory.

-
I think this is the same proof mentioned in the paper which Martin Sleziak talked about above. I am really interested in a more elementary argument. Thanks though. –  Shahab Dec 20 '11 at 11:12
@Shahab: I’ve not seen the original Rousseau & Sheehan paper, but I shouldn’t be surprised. I did try to simplify the argument as much as was easily possible. I’d love to see a simpler argument, but I’m not going to hold my breath: what little I know of graph Ramsey theory suggests that the arguments are rarely very simple. And in this case there’s no real leeway, since (if I’m not mistaken) $4n+2$ is a sharp bound when $4n+1$ is prime. –  Brian M. Scott Dec 20 '11 at 11:31


text:
🔥 NEW! Get my book The Developer's Guide to Having a Successful Blog

In this book I teach you all I know about starting and growing a blog.
Launch promo: 25% OFF until tomorrow! And a special challenge for you!

Vue.js uses a templating language that’s a superset of HTML.

Any HTML is a valid Vue.js template, and in addition to that, Vue.js provides two powerful things: interpolation and directives.

This is a valid Vue.js template:

Hello!

This template can be put inside a Vue component declared explicitly:

new Vue({
template: 'Hello!'
})

or it can be put into a Single File Component:

<template>
Hello!
</template>

This first example is very basic. The next step is making it output a piece of the component state, for example, a name.

This can be done using interpolation. First, we add some data to our component:

new Vue({
data: {
name: 'Flavio'
},
template: 'Hello!'
})

and then we can add it to our template using the double brackets syntax:

new Vue({
data: {
name: 'Flavio'
},
template: 'Hello {{name}}!'
})

One interesting thing here. See how we just used name instead of this.data.name?

This is because Vue.js does some internal binding and lets the template use the property as if it was local. Pretty handy.

In a single file component, that would be:

<template>
Hello {{name}}!
</template>

I used a regular function in my export. Why not an arrow function?

This is because in data we might need to access a method in our component instance, and we can’t do that if this is not bound to the component, so arrow functions usage is not possible.

We could use an arrow function, but then I would need to remember to switch to a regular function in case I use this. Better play it safe, I think.

Now, back to the interpolation.

{{ name }} reminds of Mustache / Handlebars template interpolation, but it’s just a visual reminder.

While in those templating engines they are “dumb”, in Vue, you can do much more, it’s more flexible.

You can use any JavaScript expression inside your interpolations, but you’re limited to just one expression:

{{ name.reverse() }}

{{ name === 'Flavio' ? 'Flavio' : 'stranger' }}

Vue provides access to some global objects inside templates, including Math and Date, so you can use them:

{{ Math.sqrt(16) * Math.random() }}

It’s best to avoid adding complex logic to templates, but the fact Vue allows it gives us more flexibility, especially when trying things out.

We can first try to put an expression in the template, and then move it to a computed property or method later on.

The value included in any interpolation will be updated upon a change of any of the data properties it relies on.

You can avoid this reactivity by using the v-once directive.

The result is always escaped, so you can’t have HTML in the output.

If you need to have an HTML snippet you need to use the v-html directive instead.


text:
# Can some help with this scavenger hunt please

###### Question:

Can some help with this scavenger hunt please

$Can some help with this scavenger hunt please$

### Drag each label to the correct location on the table.Match each equation with its number of unique

Drag each label to the correct location on the table. Match each equation with its number of unique solutions. y = 3x^2- 6x + 3 y= -x^2 - 4x + 7 y = -2x^2 + 9x – 11...

### Why are genes that are located on the same chromosomes, or 'linked', the exception to Mendel's law of indepent assortment

why are genes that are located on the same chromosomes, or "linked", the exception to Mendel's law of indepent assortment $why are genes that are located on the same chromosomes, or linked, the exception to Mendel's law$...

### Throughout history mankind has preserved evidence of his existence, sometimes intentionally, sometimes unintentionally. In the late nineteenth

Throughout history mankind has preserved evidence of his existence, sometimes intentionally, sometimes unintentionally. In the late nineteenth century some Europeans began the custom of enclosing a collection of goods in "time capsules" to be opened by future generations. Time capsules allow those w...

### Solve and verify using substitution or elimination

Solve and verify using substitution or elimination $Solve and verify using substitution or elimination$...

### Consider an automated cash deposit machine in which users provide a card or an account number to deposit

Consider an automated cash deposit machine in which users provide a card or an account number to deposit cash. give examples of confidentiality, integrity, and availability requirements associated with the system, and, in each case, indicate the degree of importance of the requirement....

### 3х + бу = -182y = 3х – 22use eliminatio!!

3х + бу = -18 2y = 3х – 22 use eliminatio!!...

### The table shows the final score of the winning hockey team compared to the total number of goals that either team attempted during

The table shows the final score of the winning hockey team compared to the total number of goals that either team attempted during the game. Each row represents a single hockey game. Which value is an outlier in the table? Final Scores of Winning Team Total number of attempted goals, x Final score ...

### Find the value or values of x in the quadratic equation: x^2 = 6x - 9 A. x = -3, x = 2 B. x = 0, x =

Find the value or values of x in the quadratic equation: x^2 = 6x - 9 A. x = -3, x = 2 B. x = 0, x = 1 C. x = 9 D. x = 3 Please help and thank u...

### What does “one-person, one-vote”ruling mean?

What does “one-person, one-vote”ruling mean?...

### Given the series 2368+592+148+...+9.25 Write the series using summation notation.

Given the series 2368+592+148+...+9.25 Write the series using summation notation....

### If a body is dropped from the same height once in the equator and then in the polar region, in which place it will fall faster?why?

If a body is dropped from the same height once in the equator and then in the polar region, in which place it will fall faster?why?...

### What is the velocity of a penguin that swims 90 meters through the water in 12 minutes ?

What is the velocity of a penguin that swims 90 meters through the water in 12 minutes ?...
Which pair of sentences would most likely be used in a how-to-cook book? $Which pair of sentences would most likely be used in a how-to-cook book?$...


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# MikTex | mkluafontbd not working...

Using MikTeX under Win 7. Installed a couple of public OpenType fonts. Tried compiling a document with LuaLaTeX, using the newly installed font. I checked and the font is indeed installed in

C:\Program Files (x86)\MiKTeX 2.9\fonts\opentype\public\asana-math

I looked at otfl-names.lua, and sure enough, the font isn't included (which, of course, is why lualatex wasn't finding it).

So, I went command line (administrator mode), and ran

mkluatexfontdb -f -vvv

Piped the output to a file. Looked at the file, and found the following:

luaotfload | 0 fonts found in 'C:/Program Files (x86)/MiKTeX 2.9/fonts/opentype/public'
luaotfload | scanning 'C:/Program Files (x86)/MiKTeX 2.9/fonts/opentype/public/asana-math'
luaotfload | 0 'otf' fonts found
luaotfload | 0 'OTF' fonts found
luaotfload | 0 'ttf' fonts found
luaotfload | 0 'TTF' fonts found
luaotfload | 0 'ttc' fonts found
luaotfload | 0 'TTC' fonts found
luaotfload | 0 'dfont' fonts found
luaotfload | 0 'DFONT' fonts found
luaotfload | 0 fonts found in 'C:/Program Files (x86)/MiKTeX 2.9/fonts/opentype/public/asana-math'

So, even though there is an otf font in /public/asana-math (Asana-Math.otf), mkluatexfontdb isn't finding it. In fact, I noticed it isn't reading in *any* of the fonts in C:/Program Files (x86)/MiKTeX 2.9/fonts/opentype/public/' (not just Asana-Math).

I have no idea why this is happening. Any suggestions as to what to try? Of course, the brute force solution (which works, but still...) is to throw the font into C:\windows\fonts

But I'd really like to know what mkluatexfontdb isn' working for fonts installed into the local tex root, even though it seems to be finding the subdirs which contain the fonts.
.

## Relevant Pages

• Re: MikTex | mkluafontbd not working...
... 'MikTeX | mkluatexfontdb not working' ... Tried compiling a document with LuaLaTeX, ... luaotfload | scanning 'C:/Program Files /MiKTeX ... luaotfload | 0 'otf' fonts found ...
(comp.text.tex)
• Re: several LuaLaTeX questions
... and general help and advice - or even discussion - on LuaLaTeX. ... from LuaTeX than what you get from XeTeX, ... I would really like e.g. to extend the chess fonts packages to ... ConTeXt, adapted and maintained by Khaled for use ...
(comp.text.tex)
• Re: LuaLaTeX sample maths document
... This document works with plain latex, pdflatex, xelatex and lualatex: ... math fonts (unless you have an outdated unicode-math). ... or pdflatex, it will also use lmodern for text and math, but not the OTF ...
(comp.text.tex)
• Re: several LuaLaTeX questions
... and general help and advice - or even discussion - on LuaLaTeX. ... from LuaTeX than what you get from XeTeX, ... I would really like e.g. to extend the chess fonts packages to ... ConTeXt, adapted and maintained by Khaled for use ...
(comp.text.tex)
• Re: open type font problems
... First - in whatever problem the OP is having, the fonts already have the .ttf extensions. ... that pgms which understood ttf could see them and that it wouldn't screw up or drop off the otf features. ...
(comp.fonts)


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Check if a list exists in given list of lists in Python

PythonServer Side ProgrammingProgramming

Lists can be nested, means the elements of a list are themselves lists. In this article we will see how to find out if a given list is present as an element in the outer bigger list.

With in

This is a very simple and straight forward method. We use the in clause just to check if the inner list is present as an element in the bigger list.

Example

Live Demo

listA = [[-9, -1, 3], [11, -8],[-4,434,0]]
search_list = [-4,434,0]

# Given list
print("Given List :\n", listA)
print("list to Search: ",search_list)

# Using in
if search_list in listA:
print("Present")
else:
print("Not Present")

Output

Running the above code gives us the following result −

Given List :
[[-9, -1, 3], [11, -8], [-4, 434, 0]]
list to Search: [-4, 434, 0]
Present

With any

We can also use the any clause where we take an element and test if it is equal to any element present in the list. Of course with help of a for loop.

Example

Live Demo

listA = [[-9, -1, 3], [11, -8],[-4,434,0]]
search_list = [-4,434,0]

# Given list
print("Given List :\n", listA)
print("list to Search: ",search_list)

# Using in
if any (x == search_list for x in listA):
print("Present")
else:
print("Not Present")

Output

Running the above code gives us the following result −

Given List :
[[-9, -1, 3], [11, -8], [-4, 434, 0]]
list to Search: [-4, 434, 0]
Present
Published on 13-May-2020 14:12:26


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# Product Rule Formula

Product rule help us to differentiate between two or more functions in a given function. If u and v are the given function of x then the Product Rule Formula is given by:

$\large \frac{d(uv)}{dx}=u\;\frac{dv}{dx}+v\;\frac{du}{dx}$

When the first function is multiplied by the derivative of the second plus the second function multiplied by the derivative of the first function, then the product rule is given.  Here we take u constant in the first term and v  constant in the second term.

### Solved Example

Question: Differentiate the function: (x2 + 3)(5x + 4)

Solution:

Given function is: (x2 + 3)(5x + 4)

Here u = (x2 + 3) and v = (5x + 4)

Using product rule,

$$\begin{array}{l}\frac{d((x^2 + 3)(5x + 4))}{dx}\end{array}$$
= (
$$\begin{array}{l}x^2\end{array}$$
+ 3)
$$\begin{array}{l}\frac{d(5x + 4)}{dx}\end{array}$$
+ (5x + 4)
$$\begin{array}{l}\frac{d(x^2 + 3)}{dx}\end{array}$$

= (x2 + 3) 5 + (5x + 4) 2x

= 5x2 + 15 + 10x2 + 8x

= 15x2 + 8x + 15


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Browse Questions

# If p, q, r are in increasing AP, then common difference will be

$(a)\;\sqrt{p^2-qr}\qquad(b)\;\sqrt{q^2-pr}\qquad(c)\;\sqrt{r^2-pq}\qquad(d)\;None\;of\;these$

Can you answer this question?

Answer : (b) $\;\sqrt{q^2-pr}$
Explanation : Let $\;p=a-d$
$q=a$
$r=a+d$
$p\;r=a^2-d^2$
$q^2=a^2$
$d^2=q^2-pr$
$d=\sqrt{q^2-pr}\;.$
answered Jan 21, 2014 by


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+0

# Polynomials

0
23
1

Find the coefficient of z in the product (-3z^2 - 7z + 4)(6z^2 + 6z - 1).

Jul 23, 2022

The coefficient of z is $$(4 \times 6) + (-7 \times -1)$$


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# How do I arrive at correct furanose structures when starting from a Fischer projection?

While numbering the furanose ring from the left are we supposed to place the atoms on right of the Fischer projection to the top? Isn't it just the opposite to what we do while numbering from the right?

In the image the atoms/groups lying on the right of the 2nd Fischer structure is present to the bottom of every furanose carbon of the 3rd image (chronologically). So aren't they supposed to be present at the bottom even if we start from the left (referring to structure A)? But I find Wikipedia says they are supposed to be at the top.

• Tl;dr: It doesn’t matter how you do it, as long as it is consistent and correct. – Jan Jun 14 '16 at 19:23
• @Jan I don't get you. – Tyto alba Jun 14 '16 at 19:28
• chemistry.stackexchange.com/questions/44332/… – Mithoron Jun 14 '16 at 19:37
• Also, why are you linking to sucrose rather than to fructose? I thought, your question was about fructose? (I know it is present in sucrose, but still.) – Jan Jun 14 '16 at 20:43
• It was the only wikipedia page where the structure of Beta-D-fuctose is provided where it is numbered from the left. – Tyto alba Jun 14 '16 at 21:12

## 1 Answer

A Fischer structure is one way of depicting structures such as sugars or amino acids to quickly draw them and have a definite way of seeing stereochemistry on every carbon. The rule behind Fischer structures is that the highest oxidised atom should be at the top or as close as possible to the top. The following convention is used to turn tetrahedral carbons into a square environment:

This allows you to assign absolute configurations to every carbon atom.

When instead drawing the same structure as a ring like in your lower images, you just need to take care that every carbon has the same configuration. (Note that your drawing of β-D-fructose is wrong; it should be drawn with a very wavy line like in my image below, because of the constraints of Fischer projections.)

Once you have gotten there, you are free to rotate the ring around as much as you wish while retaining the absolute configuration at every stereocentre. Note especially, that a rotation of $180^\circ$ of my structure of β-D-fructose will lead to you one on the left. If you don’t believe me, print it out, turn the paper around and then nudge the forwards-pointing part of the ring downwards below the tetrahydrofuryl-oxygen.