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Posted: 4/3/2020 10:45 PM

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p4wn3r

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As I understand it, extra dimensions in physics are simply a mathematical construct to solve problems. This is a very powerful way to solve complicated systems, and also to come up with new theories. One example I like to give: the Kepler problem. Newton became famous when he showed that, assuming that the gravitational force is of the form 1/r^2, the orbits of the planets around the sun form ellipses with the sun at one of the centers. A proof of this is given in most introductory physics courses. Another proof that can be written goes like this: for the 1/r^2 force you can write a vector that's conserved by the motion. From this conservation property, it's not very hard to show that the planet's orbit is an ellipsis. Also, in the (non-relativistic) hydrogen atom, where the force is also 1/r^2, the s, p and d orbitals all have the same energy, and this property is only true if the force goes with the inverse of the square radius, for all others it does not work. When confronted with this observation, there are two possible attitudes: A) You don't care, you already have a set of equations that describe the physics of the phenomenon you want. If all you want is to compute predictions from the theory, the problem is already solved, and you can use the formalism you want. B) You want to know *why* the 1/r^2 force is so special, why it conserves such an exotic vector and makes the energy levels for different states in hydrogen agree. If your attitude is in line with (B), then going to extra dimensions gives you an explanation. It turns out that the motion of a particle subject to a force that goes like 1/r^2 is completely equivalent to the motion of a free particle on the "surface" of a "sphere" on four-dimensional space. That is, if you have four space coordinates x,y,z,w, and study the movement on a space that satisfies x²+y²+z²+w²=1, what we call a 3-sphere, you can transform the solution to a movement in 3-dimensional space under the force 1/r^2! Although it's a bit abstract, you can see where the strange properties come from. The ordinary sphere has some symmetries, you can roll it in different ways so that it stays the same. The 3-sphere has more symmetries than the ordinary sphere does, so the conservation of the LRL vector in classical mechanics, and the s and p orbitals having the same energy in quantum mechanics are simply artifacts of this extra symmetries, which are very hard to see in the ordinary space, but obvious in the 3-sphere. So, one exotic way of solving this problem would be: take four very weird variables from the problem, construct a four-dimensional Euclidean space from it, and check that the problem is just a free particle, then transform the solution back to ordinary space and find the answer. If your attitude is (A), when presented with this solution, you'll think it's a huge waste of time. Why construct weird geometric spaces, study their symmetries, just to solve a problem that can be destroyed with ordinary calculus? You gained no new insight into it. However, it's often the case that physicists need to generalize their theories because the current ones they have are not good enough, and this kind of understanding can lead to new insights. For example, you could try different geometries and see what kind of forces you can solve. The solutions you will find would be very difficult to find otherwise. Using time as a fourth dimension is a similar story. You can write relativity considering space and time as separate entities, no problem. Once you have all the equations, if you can solve them, no errors will happen. However, it just happens that if you construct a four-dimensional spacetime, the formulas can be derived elegantly from geometric properties. About the question whether these "dimensions" are real, I leave it for the reader. Many people would say that spacetime is "real", because the equations of relativity are more elegant on it. By the same argument, the 4-dimensional space in the hydrogen atom would also be "real" because the equations are also more elegant. For this reason, I prefer to treat it as a convenient mathematical abstraction. P.S.: This trick of working on higher dimensions to search for theories in lower ones is a common pattern in modern physics. For anyone who's interested, a classic case is general relativity on five-dimensions with the fifth curled up on a cylinder, which is equivalent to gravity+eletromagnetism. In the worldline formalism, all particle interactions can be derived by curling up extra dimensions to implement the forces of nature. Also, it's conjectured that string theory compactified on a Calabi-Yau manifold, a 6-dimensional space with some exotic properties, gives all forces of nature, including gravity.

Posted: 4/2/2020 8:02 PM

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p4wn3r

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Yup, very fun! The trick is to multiply by x+y (ax^2+by^2)(x+y) = ax^3 + by^3 + xy(ax+by) 7(x+y) = 16 + 3xy If you do the same to the cube sum: 16(x+y) = 42 + 7xy Solving the linear system: x+y=-14 xy=-38 Now do the same for the fourth power sum: 42(x+y) = ax^5 + by^5 +16xy Then: ax^5+by^5 = 16*38 - 42*14 = 4*(8*19-21*7)=4*(152-147)=20 I find it weird that it decreases, but it's possible depending on the sign of a,b,x,y

Posted: 3/20/2020 8:58 PM

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p4wn3r

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I work at the financial hub in São Paulo, lots of people traveling around the world all the time. The second confirmed case in Brazil was actually a guy who works on the building next to where I work. My company does not allow home office, but as the situation worsened, they rushed to prepare a VPN so that people could work from home. Some days ago, a guy two floors below tested positive, so they told everybody to do home office, and that they would clean the whole building thoroughly. They said we could return to it on Monday. About me, I'm feeling just as well as I always have. I wonder to what extent the quarantine measures work, if I don't stock lots of food, then I have to go to the supermarket regularly, but it's overloaded with people. I also saw many angry people yelling at others accusing them of spreading diseases. Because of that, I don't tell anyone I work at a building with a confirmed case, I'm afraid of them lynching me there!

Posted: 3/16/2020 11:34 AM

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p4wn3r

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Here is a fun one I recently found out in a YouTube video, I haven't seen it published anywhere. Let A and B be vectors in R². An ellipse can be defined as the set of vectors P such that: P = A*cos(x) + B*sin(x) + C (*), where C is the vector that denotes the center of the ellipse. Notice that I do not require that A and B be perpendicular, as is commonly done. In fact, no requirement is necessary. Every pair of vectors A and B will define an ellipse, even linear dependent ones, if you consider a line segment as a degenerate ellipse. The consequence of this is that an ellipse does not have a unique representation by formula (*), there's an infinite choice of vectors A and B that will define the same curve. Besides that, we also have several formula for the ellipse in vector form: Area: S = pi*|A x B|, where A x B denotes the vector product Hypotenuse: D² = |A|² + |B|² Inside test: |(P-C)xA|² + |(P-C)xB|² < |A x B|², for a point P Tangent test: |R x A|² + |R x B|² = |Rx(P-C)|², for a point P with direction vector R The amazing thing is, all the previous formulas work for any parametrization of the ellipse with vectors A and B, they don't need to be perpendicular! Prove it.

Posted: 2/7/2020 1:30 PM

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p4wn3r

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Do you intend to release it for Android? I showed your game to several people at work. They liked it, but said they only play on mobile, so they are not interested in Steam or Windows games...

Posted: 12/1/2019 7:36 PM

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p4wn3r

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It's widely believed that the (asymptotically) fastest way to divide two large numbers is to reduce the problem to multiplication using Newton's method and doing enough multiplications to converge it. Multiplication of two large integers can be done in O(n log n) by reducing it to a Fast-Fourier-Transform. However, the details are extremely complex and the construction and proof of an actual algorithm that multiplies two integers in n*log n was done only recently. It's an academic algorithm, though, because it would take numbers much larger than those our computers can handle to make it faster than common fast multiplication ones. So, asymptotically, division is a bit slower than n log n, I don't remember the actual complexity. And this is a conjecture, because there's no proof that n log n is optimal for FFT. For practical purposes, algorithms like Barrett reduction should perform faster.

Posted: 9/7/2019 5:23 PM

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p4wn3r

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Yes, they are related. Take two vectors A and B and evaluate the dot product of A+B with itself. Expanding the product you get the square if magnitude of A+B in terms of the magnitude of A and B and the cosine if their relative angle. Geometrically, that's equivalent to a relation between three sides of a triangle and the cosine of one of the angles, the law of cosines. Also, using the fact that the cosine is always between 1 and -1 you can derive the Cauchy-Schwarz inequality, which holds in more general vector spaces.

Posted: 8/18/2019 3:29 PM

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p4wn3r

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Ah, the color of the oceans question. That's confusing, yes. Not many places give the correct explanation. There are many factors involved. The color of a body of water depends a lot on what is dissolved in it. Nevertheless, there are some general phenomena. The first one is reflection. Since water is a reflective surface, part of the light you see comes from what's reflected in the atmosphere. The second one is absorption. This one is a bit complicated to explain because it involves quantum mechanics. Water molecules can rotate or vibrate in a large number of ways, and using quantum mechanical considerations you expect some light frequencies enter in resonance with these modes, and that light gets absorbed. If you measure the things, you'll find that water absorbs frequencies towards the red more than ten times more than ones towards the blue. The third phenomenon is Rayleigh scattering. Basically, when water goes through a fluid, different wavelengths get scattered differently. The smaller the wavelength, the stronger the scattering. The blue color of the oceans comes mostly from the reflection from the atmosphere and from the scattered light rays that entered it and then left. These rays are mostly blue because it's the color that gets less absorbed, and also the one most likely to scatter. Now, why are the oceans (or the sky) not violet? That has to do with the composition of sunlight. Sunlight has a lot more blue than violet, and our eyes cannot see violet very well, so even when violet light is there, we cannot see it very well.

Posted: 7/31/2019 11:53 AM

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p4wn3r

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Here's an example to see how acceleration due to gravity is different than acceleration due to everything else, if the Einstein equivalence principle holds. If you have a point charge moving, and suddenly impart it with some acceleration, it will start radiating, in a phenomenon we call bremmstrahlung. This is measured everyday in particle accelerators. Now, if you are in a frame together with the electron, can you measure that it's accelerating? Well, yes. If you see radiation coming from it, the electron must be accelerating due to something , if there's no radiation, no acceleration. Consider now doing the same thing with gravity. Should an electron orbiting a gravity well should also radiate? Well, according to Newton it should, because there isn't much difference between the electric and gravitational force. However, if you assume the Einstein equivalence principle, you must conclude that it does not radiate. Because if it did, you would be able to measure gravitational acceleration. So, the gravitational force in GR must have a very special form so that electric charges subject to it do not radiate. (Incidentally, that's because astronomical bodies do a much better job than artificial accelerators and fusion reactors. In human made accelerators, we use the electric force to accelerate particles, this process is inefficient because of the energy lost due to radiation, while black hole accretion disks can subject charged particles to huge velocities without causing them to radiate. Same thing with fusion reactors, because we use magnetic fields to confine the hot plasma, there's energy loss due to radiation. The sun uses gravity to confine the plasma, which is much more efficient). So, as you can see, accelerating something with gravity is different from accelerating it with something else. When you read that, for someone standing on Earth, an accelerometer shows non-zero acceleration, you can think of it in two ways: (1) Gravity is exerting a force and the Earth is also exerting a force on the other direction. That applies some stress in your body, and the accelerometer measures this stress. (2) Gravity is not a force at all, and the only measurable acceleration you have is the one coming from the Earth. Since you are accelerating, the electrical properties in your body should change, and it's that change the accelerometer measures. It turns out both of these explanations are correct, because in GR what's called "force" and what's called "inertia" is essentially a matter of convention. Explanation (2) is in a language more in line with the philosophy of GRm but (1) is correct, too.

Posted: 7/25/2019 11:59 PM

Post subject: Would a cloud emulator running a copyrighted ROM be legal?

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p4wn3r

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This is a thought that occurred to me recently. Computing is very rapidly transitioning to a state where we don't install software in our hardware, and instead interact with software running server-side in the cloud. This transition is already starting to happen in games, with Google announcing their Stadia platform where all the hard processing is done in their servers and the client device only has to send the player input to the cloud and play the streaming video. Given that many of the games that we are interested in this site were released during the 1980s, 1990s, and early 2000s, it's unlikely that their developers considered the possibility that their games could be played over the cloud. So, in theory, say, a NES emulator on the cloud, where the user just picks a game code, and runs a copy of the ROM server side and only sends to a browser the information needed for the PPU to draw could arguably be legal, since the copyright would only be violated if the ROM itself would be transmitted over the network, which never happens in this case. This interpretation does have its precedent in the open source community, for example. If an open source software is licensed under the GPL, then any modifications to it must also be released under the GPL. However, if you modify the software and only expose an API of it that runs inside your server you don't have to disclose the modifications under the GPL, because as long as the GPL is concerned, running something in your server and letting people access it does not count as distributing the software, while putting the binary for download does. If the developer intends to block people from running their modified code on the cloud, they should choose the AGPL, a more restrictive license. Of course, copyright holders with lots of cash could find other reasons to take down something like this, and doing what I am proposing would surely net you a lawsuit from some developer, but what does everyone think? Of course, the infrastructure costs of doing something like this would be orders of magnitude larger than a forum+wiki website, but is a project like this worth the risk to free the legacy games?

Posted: 7/10/2019 2:33 PM

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p4wn3r

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Sorry for taking too long to answer, I could not check the forum for a while.

Warp wrote:

I have hard time believing that you'll get a megaton-range explosion by sublimating ice. It's the kinetic energy of the meteor (which is absolutely ginormous due to the speed it's traveling) that gets converted into heat due to colliding with air.

I don't think we can rule out one scenario or the other just analyzing the kinetic energy. Sure, since the meteor slows down, its energy has to go somewhere. To understand where it goes, we have to model it somehow. If you think it's improbable that a good chunk of the kinetic energy can be transferred to steam, please share with us why. I am not a specialist in this topic and am just guessing, it would be great to hear.

OmnipotentEntity wrote:

I think you're misunderstanding his point Warp. The explosion is caused by rapid heating of the air in your scenario. In p4wn3r's scenario it's that and the sublimation of ice, both contribute to the effect.

My point is that, for a complicated system like a meteor, I find it unlikely that you can point out the explosion mechanism to a single cause, and it's worth to throw around ideas to brainstorm it. Most problems in science today are not those that you can solve with absolute principles like "you cannot travel faster than the speed of light", or "the gravitational force falls with the inverse of the squared radius". The systems of interest today are pretty complicated and you have to model them to understand the mechanism. It's not very difficult to come up with a model that favors your favorite explanation, if you are willing to do it. Popular science expositions usually give the impression that a phenomenon is very deterministic and has a unique cause, but usually things are more blurry. No asteroid is the same as another one, so it's perfectly possible that the explosion in one is very different from the explosion from another one. In fact, if it's really true that every asteroid that crashes into the atmosphere explodes in the same way, then that should tell us a lot about the internal structure of asteroid, which would be a conclusion far more remarkable than the explosion mechanism itself!

Posted: 7/5/2019 1:19 PM

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p4wn3r

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It could also be due to the presence of ice in the meteor. If heated extremely fast, ice will quickly sublimate into steam, and the resulting steam explosion is strong enough to destroy it. An example of this was the explosion from the accident at Chernobyl, reactor fuel cannot explode like a nuclear weapon does, because its uranium is not enriched enough. However, it can quickly boil the water coolant off and cause a steam explosion. In Chernobyl's case, it was strong enough to destroy the containment building and release radioactive materials in the air.

Posted: 6/24/2019 1:05 AM

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p4wn3r

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Warp wrote:

The product of square roots rule for real numbers assumes that the numbers are non-negative, as for example stated here. For complex numbers it becomes more confusing, with some questions at Stack Exchange asking why doesn't multiplying square roots of imaginary numbers follow the rule, and another trying to prove that it does, with confusing answers.

Well, what I'll write is similar to one of the SE answers, but I hope it will help. The confusion is mostly because the square root function has properties in the complex numbers that are a bit different from those in the real numbers. To begin with, the real numbers are totally ordered, while the complex numbers are not. That means you cannot choose a subset of complex numbers and call them "non-negative". Because of that, the definition of the real square root of x, the unique real number whose square gives x and is non-negative, cannot be used for complex numbers, because it assumes the existence of a total order, which does not exist in the complex case. That does not mean you cannot define the complex square root, but it will have some weird properties. If you write the complex number in polar form, you get an amplitude and a phase. The square root of the complex number is given by the square root of the amplitude and half the phase. The big issue is that there are two ways to halve the phase. Think about the angle of 30 degrees. Because two angles that differ by 360 are identical, there are two ways to halve 30 degrees, you can use 15 degrees or 195 degrees, because 2*195 = 390 = 360 + 30. So, if the complex number is not zero, you always have two choices for the square root. We usually call this general phenomenon "ramification". If you look at the problem purely algebraically, then the product formula is in fact true. That's because you can read it as "if x is the square root of a, and y is the square root of b, then xy is the square root of ab". This proposition is true, and in fact very easy to prove, you just need to use the fact that multiplication of complex numbers is multiplicative. Now, one property that the real square root has is continuity. The reason that you read that this formula is not true is because if you try to define the complex square root continuously, you get a lot of problems. One way to try to do it is choose the phase of a complex number z to be between 0 and 360 degrees, and set sqrt(z) to have half that, between 0 and 180. Now, look at the square root of 1, it's 1. But if you look at a number 1-ie, where e is very small, by the rules you should choose the phase as something close to 360, so sqrt(1-e) is in fact something close to -1. Because of this, the definition of the square root is not continuous. You can try to fix the problem by using phases from -180 to 180 degrees, but that only transfers the discontinuous point (actually, the entire half-line) from 1 to -1. It can be proved that you can never define the square root to be continuous in the whole complex plane, it will always have a branch cut. If you understand sqrt(a) as the value of the function discussed previously, because of the branch you can always find two values a and b for which sqrt(ab)=sqrt(a)sqrt(b) fails. There is a way to define the square root while keeping continuity though, you can define it as a Riemann surface , which is a way to complete the surface "going around the branch". It has some weird properties, for example if you take a closed path around the origin, the phase you choose for the square root is changed (this is what we call [url=https://en.wikipedia.org/wiki/Monodromy]monodromy), but depending on what you understand for "sqrt(a)", the product rule can be true or false, depending on your definition. An even weirder way to define it is to take the square root as a multivalued function, (or, equivalently, by a reduction of the Riemann surface at the points whose square gives the same complex number), it in fact is continuous, and gives a manifold that's isomorphic to the Möbius strip. In this multivalued function case, the product formula is true. TL;DR: For many cases, mathematical notation is not precise and you should not assume that by simply writing sqrt(z) for complex z there is a unique definition. For this case, there are many possible and reasonable definitions. Because of this, "Is sqrt(ab)=sqrt(a)sqrt(b) for complex a and b?" is not a good question. To understand this, it's important to study the mathematical concepts and understand the relations between them.

Posted: 6/1/2019 1:32 PM

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p4wn3r

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FractalFusion wrote:

p4wn3r wrote:
A category consists of a collection of objects, and for each pair of objects, a set of morphisms between them.
I never really understood what category theory is supposed to be. Isn't a morphism just a "function", and an object just a "set"? Now I suppose the difference is that a function is a map from a set to another, whereas object and morphism are mathematical things that no one can make any assumptions about; the morphisms from A to B are literally defined by the elements comprising Mor(A,B) and nothing else.

Thanks for pointing out about associativity. I somehow forgot to copy that part. The category of sets has more structure than a general category. The category of sets somehow has to encode the recursive enumerability of sets, so that you can have peano arithmetic and the cantor diagonal argument on it. There are multiple ways to define the category of sets, though, each with its own characteristics.

FractalFusion wrote:

OK, that did feel like a lot of "abstract nonsense". Yet it feels like something I went through over and over during my university studies.

The thing is, there's nothing in category theory that is "new" in category theory, because the whole point of it is to abstract things from other disciplines. Category theory is just language, but it's language with logic that you already know. If you look at the development of mathematics, you can see that the greatest progress has been made when people realized that two distinct areas were actually "the same thing", and then started to apply techniques from one area to the other. For example, the series 1 - 1/3 + 1/5 - 1/7 + ... = pi/4 The proof of this involves some seemingly uncorrelated areas of mathematics. It arises by looking at the 1/1+x² function and integrating it. From it, you have algebra (expanding it into a geometric series), analysis (integrating it term by term and showing convergence), and geometry (the result is the arctan function, which has a geometric interpretation). From this point of view, you can try to come up with more general objects, and if they describe simultaneously algebra, analysis and geometry, you expect to find more general identities. Category theory allows you to do this systematically, it's in practice very hard to do this in an ad hoc way if the mathematical objects are too complicated. Historically, the first problem that gave widespread adoption to category theory was that of elliptic curves over finite fields. If elliptic curves are defined over infinite fields (or, more precisely, those of zero characteristic), it was well known that you could interpret elliptic curves by looking at Riemann surfaces, lattices, modular groups of complex functions, and binary quadratic forms in number theory. Yet, no one could think of a similar analogy for those over finite fields. It took nearly two decades to understand that they encoded the topological properties of some geometric spaces, because the construction was so difficult. In the Grothendieck philosophy, if you can "look" at distinct objects at different ways, then they should be defined in the same way. That culminated in the development of the notion of a scheme. Its definition is obviously highly abstract, because it actually tries to abstract four or five areas of mathematics at the same time! But once you do understand the definitions, the proofs are kinda natural. It's very difficult to explain the huge amount of abstraction in modern mathematics. Like Fractal said, it's necessary to go through many different proofs and actually see that many are the same concept repeated over and over again. If someone has never done this, it does look like "abstract nonsense". I like to make an analogy with software engineering. When you have a very complex problem to solve, you can actually just write a bunch of code, and solve it. Most likely, though, you'll end up with something that's very difficult to understand separately, and people will take a lot of time to analyze it. Another way is to write an API to solve some very general abstract problems, and go slowly making it more concrete until you solve the problem you have. That second approach usually works much better. That has something to do with the inventor's paradox. In the words of Grothendieck:

Grothendieck wrote:

I can illustrate the ... approach with the ... image of a nut to be opened. The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months — when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado! A different image came to me a few weeks ago. The unknown thing to be known appeared to me as some stretch of earth or hard marl, resisting penetration ... the sea advances insensibly in silence, nothing seems to happen, nothing moves, the water is so far off you hardly hear it ... yet finally it surrounds the resistant substance.

It's usually said in popular circles that modern mathematics is too useless, but in my view, it has advanced a lot more recently than some hyped up fields, like materials science, and biotech. It seems that every year they promise some miraculous thing that after ten years turns out to not work. In contrast, I see that people highly trained in mathematics in enterprise environments are increasingly able to take more tasks, and more efficiently, at the same time. There's an anecdote that physicists that joined the development of the radar in WW2 were outperforming people that were electrical engineers their whole life, just like today you see startups of 15 people scaring huge corporations. It all has to do because these smaller companies could understand some abstract parts of their business that their competitors still refuse to see.

Posted: 5/30/2019 9:02 PM
(Edited: 6/1/2019 12:56 PM)

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p4wn3r

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I think it's time for an "abstract nonsense" exercise. A category consists of a collection of objects, and for each pair of objects, a set of morphisms between them. The collection of objects of a category C is often denoted obj(C), but the collection is usually also denoted by C. If A,B∈C, then the set of morphisms from A to B is denoted Mor(A,B). A morphism is often written f:A→B, and A is said to be the source of f, and B the target of f. Morphisms in a category have the notion of composition. There is a composition ◦: Mor(B,C)×Mor(A,B)→Mor(A,C), and if f ∈ Mor(A,B) and g∈Mor(B,C), then their composition is denoted g◦f. Morphisms are associative: f◦(g◦h) = (f◦g)◦h For each object A∈C, there is always an identity morphism id_A:A→A, such that when you (left- or right-)compose a morphism with the identity, you get the same morphism. We also have a notion of isomorphism between two objects of a category (a morphism f:A→B such that there exists some - necessarily unique - morphism g:B→A, where f◦g and g◦f are the identity on B and A respectively), and a notion of automorphism of an object (an isomorphism of the object with itself). Now, the problem: If A is an object in a category C, show that the invertible elements of Mor(A,A) form a group, the so-called automorphism group of A, denoted by Aut(A). Show that two isomorphic objects have isomorphic automorphism groups.

Posted: 5/29/2019 10:44 PM

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p4wn3r

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BrunoVisnadi wrote:

p4wn3r wrote:
Now, for general N, it's easy to see if you solve for N prime, you solved for general N.
I don't find it that obvious. It's easy to see that by solving for all primes you solve for powers of primes, but how about products of different primes? Not that this maters if for all N there is a Pascal Triangle row with the property you conjectured. Looking at small values of m, though, I couldn't find a row with all elements multiples of 4.

Well, if you have N = pq, then you can make p groups of people, each having q. Then you perform the process for p, to choose the group which survives. Then, you change the process to select the q people in the group. But now, that I think of it, this new process might require you to change p, so this would not work. About Pascal's triangle, it seems that you're right. I made a program to compute it mod 4 and did not find a row for the first 100 iterations. Maybe that property only works for prime n, I don't remember. Anyway, I still think you can adapt this method to work. The equation p^m + (1-p)^m = 1/N can always be solved for large enough m. That's because the maximum is at p={0,1}, which is simply 1, and the minimum is at p=1/2, which is a very small number if m is very large. So, even if the rows in Pascal's triangle are not all divisible, you can pick a huge m, and set person A at the extremes, and for every row you put A on m mod (N-1) combinations. If m is large enough, these new combinations should be very small, and the equation still solvable. For example, if N=5, looking at Pascal's triangle mod 4, at row 8: 8: 1 0 0 0 2 0 0 0 1 So, the equation now is this, which also works to get a fair probability for everyone.

Posted: 5/29/2019 8:36 PM

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p4wn3r

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Here's what I came up with. Toss the unfair coin 4 times. If it lands all heads or all tails, choose person A. Now, for the other possibilities, notice that there are 4 ways of getting 1 H and 3 T's, 6 ways of getting 2 H's and 2 T's and 4 ways of getting 3 H's and 1 T. For each of these, choose half and assign to person B, and the other half assign to person C. With this, you guarantee that the probability of B being selected is the same as C. So, if you make the probability of choosing A 1/3, the probability of choosing the other two will be 1/3 for each one also. It turns out that the probability of choosing A is simply p^4 + (1-p)^4. Therefore, solving p^4+(1-p)^4 = 1/3, which does give a number between 0 and 1, you can make your method work. Now, for general N, it's easy to see if you solve for N prime, you solved for general N. For prime N, you should flip the coin a number of times m such that the m-th row in Pascal's triangle (except the extremes) are divisible by N-1. For that number of flips, you can put N-1 people in the combinations of these rows, and assign the last one to the extremes. Divisibility patterns in Pascal's triangle are complicated, but iirc you do get an infinite number of m for which all the non-extreme values in the row are multiples of a given number. From that you have to solve p^m + (1-p)^m = 1/N. Since m can get arbitrarily large, you should always be able to find a p between 0 and 1 satisfying that, so that construction should always work.

Posted: 5/28/2019 12:27 AM

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p4wn3r

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That's a complicated question. To answer the simple part: No, it might fail to converge. First notice that the absolute values of the terms is 1/n. Therefore the series of the absolute values is the harmonic series, which diverges. From this, we conclude that the series, if it does converge, it does so conditionally. To make it diverge, simply set a = 1, b = 2pi/3. Now look at the remainder of n when divided by 3: * If it's 1, then the term is 1/n * If it's 2, then the term is -1/n * If it1's 0, then the term is 1/n From this, you can rearrange it to a series with term: 1/(3n+1) - 1/(3n+2) + 1/(3n+3) = [(3n+2)(3n+3)-(3n+1)(3n+3)+(3n+1)(3n+2)]/(3n+1)(3n+2)(3n+3) Now you have some messy algebra, but you get 9n² + something in the numerator and 27n³ + something in the denominator. From this, you can set up some inequalities like: 9n² + something > 8n² after some large n (that happens because a quadratic term grows faster than the rest). 27n³ + something <28n> 8/28n, so your term is larger than an harmonic one. Since the harmonic series diverges, so does your series. Since it already diverges for n in the denominator, changing it to n^c, for 0<c<1 only gives you a larger term, so this series can also diverge. ------------ Now, for the general question of when it diverges, and when it converges. If b/2pi is a rational number, then the exponents are eventually periodic, you can grab all terms in the period, and if you have an equal amount of terms with + and - signs, you can prove it's smaller than something times 1/n², which converges. Therefore, the series converges. If there are different quantities of signs, it will diverge. Now, if b/2pi is irrational, the sequence of signs is not periodic. However, we have the equidistribution theorem to help. Basically, as n goes to infinity, the probability of hitting a point in the interval [0,2pi[ is uniform. From there, you expect a*sin(bn) to behave like an arcsine distribution. From this, you can find using integrals the probability of the floor values. If it's 1/2 for even/odd values, it should converge. The problem is to prove this rigorously, the switch from a series to a continuous probability distribution is very complicated, because you have to make the error terms explicit and be creative with the bounds to show that the error terms do not spoil the asymptotic behavior. This reminds me a lot of proofs in analytic number theory, I wonder if this exercise was inspired by that. About the series with n^c, when the signs alternate (+-+-+-...), it's the same as the Dirichlet eta function, which does converge for all c>0. The proof is more elaborate, because it involves some complex bounds. See here.

Posted: 5/25/2019 8:47 PM

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p4wn3r

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Aroymart wrote:

Hello, I've been looking around for glitches here on and off for a while, but didn't think to look on this forum. Last year I discovered that you can get this glitch without zapdos, instead using Devolution Spray. This lets us start the glitch very early into the game (no need to entirely beat it first!), and it puts the glitch on our own board, meaning we can manipulate it a lot easier. I haven't found anything entirely useful (though a few leads I suppose). I also haven't looked into it since finding it last year. I definitely think this could be used to get ace, though something else interesting might have to be discovered first.

Amazing! Now start breaking the game! I'm looking forward to some developments, too. The memory corruption of this glitch is so huge that there's probably ACE waiting to be discovered. EDIT: This is all speculation, because I haven't looked at the code for this game, for many years. But here is how I think ACE could be obtained: * There are two basic ways to get ACE, the first and easy one is to find where the game is loading function pointers, and corrupt them. The best place to look for it would be PokéPowers. It's possible that the programmers implemented pokepowers by putting a specific address at each card and telling the code to jump to it and execute a function there. If they did that, you just have to write something to that pointer and force the code to go to a convenient location which is programmed to take over the game. * The other way, which is a lot harder, is to override the return values at the stack. Basically, the code stores lots of addresses in a stack that it has to return to every time you call a function. If you can write on that, you can get ACE. The reason this is harder is because.most of the time you cannot write just two bytes, and have to write over a ton of stuff. You have to be incredibly lucky that the game does not crash. * The natural place to store the payload would be in the deck arrangement. The game gives you a lot of freedom to change the order of the cards, and the card IDs come in many types of bytes. * The hardest part seems to be how to get some controlled corruption from the arena. I suspect it should be possible to get some misalignments and for example make the game believe you placed 200 energy cards in a pokemon, so that when you remove an energy card it starts to shift lots of bytes.

Posted: 5/24/2019 12:13 AM

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p4wn3r

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Wow, that was fast. Another solution is to multiply and divide by sin(20^o), use sin(10^o)=cos(80^o) and use the double angle formula: sin(2x)=2sin(x)cos(x) sin(20^o)cos(20^o)cos(40^o)cos(80^o)/sin(20^o) = 1/2*sin(40^o)cos(40^o)cos(80^o)/sin(20^o) = 1/4*sin(80^o)cos(80^o)/sin(20^o) = 1/8 * sin(160^o)/sin(20^o) And since sin(160^o)=sin(20^o), it equals 1/8. ----------------- Number theory: Find all pairs of positive integers n and m, such that n/m + m/n is also an integer.

Posted: 5/22/2019 1:53 AM

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p4wn3r

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Yup, that's the solution. Interestingly, even though it makes no sense for the sequence to be periodic, by reading Conway's book where he treats these kind of games, he proves that if you compute the sequence long enough, the computation itself is proof that it'll keep periodic forever! That's very interesting and comes from the definition of the mex function (which you use to calculate the nim numbers) when applied to a periodic array. ------------------- Now, some trigonometry: Prove that sin(10^o)cos(20^o)cos(40^o) = 1/8

Posted: 5/12/2019 5:05 PM

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p4wn3r

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Nice solution, Fractal. I also used the determinant. I'm still wondering if we can derive this formula geometrically, though. The big square root inside is proportional to the area of the triangle of sides a,b,c. I have no idea how to carry it out, though. I also have to confess something about this next one, I haven't proved it! But if you understand the combinatorics, and run some cases on a computer, an obvious pattern emerges. You are the owner of a casino and obviously want to rip people off using math. You have come up with a brand new game for your clients. The game works like this. The dealer deals an amount N>=2 of cards face down in a straight line (the content of the cards doesn't matter). The player and the dealer take alternating turns, with the player starting. At any point, the person who's playing can either: (a) Remove exactly two consecutive cards from any point in the interval. (b) Remove a single card, provided that it's isolated from the rest of the pack. The game goes on like this until it's someone's turn and there are no cards left. In that case, the person who has the turn loses. To add complexity to the game, the number N of cards initially dealt is random. Given an amount of cards, assuming the player and the dealer both play perfectly, the outcome depends only on the initial number N. For perfect skilled dealers and players, what is the fair payoff for this game? (i.e., if the player loses all the money when losing, how many dollars for every dollar bet do you return to the player when he wins to break even?)

Posted: 5/7/2019 6:38 PM

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p4wn3r

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Correct! Now, the next one: Prove or disprove: Let a, b, c be real numbers. Suppose that l is a real number satisfying l > max(|a-b|,|b-c|,|c-a|) and l < min(a+b,b+c,c+a), so that it's possible to form a triangle having one of the sides measuring l, and any two numbers from a, b and c. Then, in these conditions, it's always possible to construct a triangular pyramid having an equilateral triangle of side l as basis, and having a, b and c as the lengths of the remaining edges. If false, try to come up with a condition that allows the pyramid with equilateral basis to be constructed.

Posted: 4/25/2019 2:48 AM

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p4wn3r

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My hobby: coming up with useless series that converge to pi! :D I will name this one the trinity series because it has lots of 3's, and also the number 4 can be explained because according to alchemy the world has four elements! Prove that:

Posted: 4/19/2019 4:14 PM

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p4wn3r

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Yes, that's an extremely hard problem. It's not difficult to find the answer with a computer algebra system if you know how to look for it. However, proving that it's the smaller is essentially mechanical application of algorithms. This was extracted from this paper: http://ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf The main ideas are: 1) Notice that the equation x/(y+z) + y/(x+z) + z/(x+y) = 4 represents a space of dimension 2. This is seen by noticing that, if you have a solution (x,y,z), you automatically obtain another one by transforming to (ka,kb,kc). 2) So, it should be possible to reduce this to a projective curve. 3) Since you only care about integral solutions, you are actually looking for rational points in that projective curve. One thing about algebraic curves is that they sometimes turn out to be equivalent to each other (think about isomorphic graphs, for example). 4) In that case, the pretty name for what we are looking for is birational equivalence. It essentially means that you can find a transformation that maps rational numbers to rational numbers in another curve, so that any operation you do on one of them gets translated to another. 5) It turns out that this projective curve is isomorphic to an elliptic curve, the transformation can be found in equation (2.1) of the paper I linked. Keep in mind that although it's a complicated thing to find, today most mathematical software implements some good algorithms and can tell you if the curve is isomorphic to an elliptic one, and the rational transformation if it is. 6) Reducing the problem is a good thing, because it means it can be solved (but it's not easy). The nice thing about them is that their algebra has a simple structure, you can "add" two points in the curve to obtain a third one. The details can be seen here. 7) If you find a rational point at the curve, you can understand the algebra by adding the point to itself. Say you find a point A. Then you can compute a sequence: A -> A + A -> A + A + A -> A + A + A + A -> ... It might happen that you eventually reach a cycle. In that case, we have what we call a finite group. It also might happen that it goes on forever and you have an infinite chain. 8) These properties are encoded in the so-called rank of the elliptic curve. A rank of 0 means you have a finite number of rational points. A rank of 1 means you have besides those an infinite chain, a rank of 2 means you have two infinite chains and so on. 9) Now, for the particular elliptic curve we obtain from that equation, it turns out that its rank is 1. This is very complicated to prove, but computer programs today can instantly find rational points and compute the rank. 10) When you check that the rank is 1 it's simple to enumerate all possible solutions and check if they give you positive x, y and z. You first check the ones in the finite group . If none of them work, you go to the ones in the infinite chain. When you keep summing the point to itself you get bigger and bigger numbers, so at the smallest "multiple" of the point you get the smallest solution. In the paper, the authors prove that repeating this always works to get positive x, y and z, but the probability is very small, and the answer turns out to be very large. 11) If you look at table 2, you see that the smallest solution for the equation has 81 digits! Don't be disappointed if you don't understand all the steps in the solution. It's a very hard problem and most of the time the only way to work with this is to use computers. Elliptic curves are very complicated objects, they are used in cryptography because their algebra is simple to implement, but extremely tough to understand. To have a notion of the complexity, the solution of Fermat's Last Theorem can be summarized as saying that every elliptic curve can be understood as a modular curve, which is some sort of projection of a generalized complex plane. It turns out that proving this simple statement took a ridiculously long time and the proof is extremely technical. Today it's widely believed that every algebraic structure arising in problems like this are essentially the same mathematical object (automorphic forms) viewed in different ways (see here). No one has any good idea on how to prove this, but if it's done, Millennium problems like the Riemann hypothesis and the Birch and Swinnenton-Dyer conjecture will all be simple exercises in that theory! That's how complicated that subject is, if you understand it, you basically solved 21st century math!

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