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# Common false beliefs in Physics [closed]

Well, in Mathematics there are somethings, which appear true but they aren't true. Naive students often get fooled by these results.

Let me consider a very simple example. As a child one learns this formula $$(a+b)^{2} =a^{2}+ 2 \cdot a \cdot b + b^{2}$$ But as one mature's he applies this same formula for Matrices. That is given any two $n \times n$ square matrices, one believes that this result is true: $$(A+B)^{2} = A^{2} + 2 \cdot A \cdot B +B^{2}$$ But eventually this is false as Matrices aren't necessarily commutative.

I would like to know whether there any such things happening with physics students as well. My motivation came from the following MO thread, which many of you might take a look into:

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## closed as not constructive by David Z♦Sep 30 '12 at 20:05

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Community wiki? –  Marek Nov 17 '10 at 23:10
@MArek: i didnt find the option. if anyone can do it they are welcome –  S.C. Nov 17 '10 at 23:15
@Chandru: AFAIK StackExchange recently changed its rules about this matter so that only moderators can make a question community wiki (the rationale being that the CW option is being misused at StackOverflow). –  Marek Nov 17 '10 at 23:26
While i perceive the question interesting, I think the example with matrices is rather 'a common silly mistake' than 'a common false belief'. –  Piotr Migdal Nov 17 '10 at 23:48
It seems a bit odd to be accepting a single answer to a soft question... –  David Z Dec 1 '10 at 0:21

Historically the concept of absolute velocity was commonly believed until the time of Galileo Galilei in the early 17th century. As a naive child, before studying basic physics, it is surprisingly easy to believe in this even these days!

The idea of absolute velocity states that all velocities are fixed with respect to an absolute frame of reference. Galileo showed that velocity is relative to your frame of reference, a principle known as Galilean relativity. This was later fully quantified by Isaac Newton, who also proposed that acceleration is invariant with respect to inertial frames.

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A mistake that I often come across and that is so easy to make: people somehow have a visceral belief that heavy objects fall faster than light ones. Setting aside of course problems of air resistance, this is obviously false, but it seems to be so counterintuitive and I think it is somehow tied to our intuitive understanding of mass as inertia. Since higher mass means higher inertia. People understand this intuitively as it takes more force to push a fat guy than a thin one. But they don't see that gravity is a force proportional to mass, so that more inertia is paralleled by more gravity as well. With as a result the same gravitational acceleration for light and heavy bodies.

I've even noticed the mistake being made by professional physicists in colloquial conversations.

EDIT: I just found this article today, shedding a new light on why misconceptions in physics or science in general are so common and so hard to get rid of.

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+1 for "setting aside air resistance"... it always bugs me when people point out how things fall at the same rate, while neglecting to mention that this assumes air resistance is negligible. Leaving out this clarification I think leads to more confusion, as all someone has to do is drop a rock next to a feather to (incorrectly) conclude that Galileo was wrong. –  Tim Goodman Nov 18 '10 at 3:22
If you are assuming a fixed earth, then you are correct. But if you let the earth move, then the Earth moves more rapidly when you drop a bigger mass than when you drop a smaller one. In that sense, bigger masses drop faster. This has nothing to do with the equivalence principle of course. –  Vagelford Nov 18 '10 at 17:49
Seems important to mention here Galileo's famous gedankenexperiment. If you follow the line of thought that heavy objects fall faster, and go split a heavy falling object in two, large one and big one, both should be falling slower than the original. The upper part would be pulling lower part up, wanting to go even slower, hence a contradiction. –  Pavel Radzivilovsky Dec 1 '10 at 9:44
There is also the complication of buoyancy - a dense object would be heavier at the same mass, and thus fall faster even if air was friction-less. –  romkyns Jan 12 '11 at 1:22
@Pavel You're ignoring tidal effects. Splitting something in two doesn't really change anything, because the two halves, being in close proximity, experience different tidal forces than those between two objects farther apart. –  ErikE Jan 21 '11 at 18:21

While the other answers are absolutely correct, they are very subtle in the way they appear. However one very large false-belief in electrical science is that current is taken to be "flowing" from the + terminal to the - terminal in DC current.

While it doesn't really matter which way you choose since you are dealing with close to light speed current, electrons are actually moving from - terminal to + terminal(do not get me wrong electrons are NOT moving anywhere near light speed while drifting. They are moving in the order of centimeters per second.). Hence the current actually moves from - to +.

And it would be interesting to note that not a soul in the professional area of electrical science considers the current from - to +, as it would cause inconsistency with his/her colleagues.

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If current always flows as electrons from - to +, then please explain how the Hall coefficient can be positive in many materials, such as p-type semiconductors. –  Keenan Pepper Nov 18 '10 at 0:02
The best explanation I heard for this was when they came up with the diagrams they had no idea what an electron was. So they just picked a direction. Turns out that they had it backward and what's actually flowing from + to - is the absence of electrons (holes). –  jcollum Nov 27 '10 at 18:25

One widespread belief (I think due to popular books such as Hawking's) is that GTR can never ever, ever be quantized and you always obtain infinities and blah blah. Well, it can, in many situations and in many theories. What is actually meant is that GTR is not a renormalizable quantum field theory in the naive way. But this specification is never explicitly pointed out so people get a false impression that quantum gravity is something completely out of the realms of current physics. Well, surprise, surprise, it's not. We can quantize many gravitational effects (such as waves), we understand that black holes have entropy, we understand they produce Hawking radiation and eventually disappear, etc. And assuming that string theory is correct we can predict whole deal more about it.

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Quantum mechanics is way too strange so it can't possibly be a correct description of the real world. Right? I think nothing else needs to be said about this.

Or maybe on a second thought, some more concrete beliefs and their solution are in order:

1. The physical world has to be deterministic (it doesn't).
2. Every possible question that can occur to you must have a precise answer by measurement (we observe only what we can, not what we want to).
3. The collapse of the wavefunction is in contradiction with finite speed of light (no information is being transmitted).
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@Noldorin: Bohr? Really? You wouldn't find a more stringent advocate of quantum mechanics than Bohr (who I think can rightly be called its father). Einstein on the other is a different story. He wasn't able to let go of his belief that physics must be complete and answer everything we want to know. But still, this led to nice results like EPR paradox. So his inquisitive mind arrived at interesting physics even though his prejudices didn't let him accept it :-) –  Marek Nov 17 '10 at 23:33
@Marek: That's why I put the question mark, silly. :P Who can blame Einstein in any case? I certainly don't. To call him prejudiced is not only arrogant but hugely ironic! I do not wish to continue with this debate, thank you. –  Noldorin Nov 17 '10 at 23:39
@Noldorin: why would it be arrogant? Just consider that Einstein himself called the inclusion of cosmological constant his biggest mistake, so he himself admitted that he was too prejudiced about stationary universe and unwilling to admit its expansion (but he eventually let go because of Hubble's experimental evidence). Even physicists (especially older ones) can be prejudiced. I am not saying this with any contempt, Einstein was one of the best minds of the human kind. Just that everyone has some prejudice or other. Although in this case it's quite ironic because he helped to create QM :-) –  Marek Nov 17 '10 at 23:47
@Marek: I think it is not only about lack of experience, but also about teaching QM in wrong way (how often do you hear "No-one knows if electric field really exists or only is our tool to describe how electron moves"? or "If someone says s(he) understand classical probability, s(he) must be lying!") –  Piotr Migdal Nov 18 '10 at 0:23
@Marek: Yes, it is ironic, as he was one of the great figures in very early quantum mechanics. Still, I don't think that even this day we can say he's wrong! There's nothing to stop some other more fundamental theory superseding quantum mechanics and proving Einstein right. In any case, fair enough, I just suggest you keep a slightly more open mind. :) (I know too many close-minded physicists.) –  Noldorin Nov 19 '10 at 21:39

A common misconception is that in a double slit experiment, the electrons or the photons go through both slits at the same time and interfere with themselves.

Nothing of the sort ever happens and that idea doesn’t come from quantum mechanics. It is just meaningless to ask the question (from which slit did the particle pass?) in the context of the particular experiment and even more meaningless to say that it went through both. The particle always passes from the one or the other slit if you setup an experiment that asks the question and in the case that you shoot the particles one at a time you always get one hit on the screen and not an interference of portions of the particle.

I think that the misconception has its roots in the wave analogy of the wave function description of quantum mechanics, where you must have a wave passing from both slits in order to have interference on the other side. Of course that picture doesn’t translate to having a particle passing from both slits, but it only states that there exists interference between the wave functions of the two independent events that translates to a distribution of probability for finding a particle on a particular point on the screen. That is a statement of Born’s interpretation that has been recently tested with a triple slit experiment.

One could ask at this point, “Ok, I get it for the electrons, but light behaves like a wave in everyday experience. What happens in that case?”, and the answer is that light behaves like a wave only when you have a large flux of photons and you can go to the continuous field approximation. It is then that the wave-like properties become apparent both for photons and electrons.

Update2: QM in your face

Update3: The Feynman Lectures on Physics vol3: Scattering from a crystal (neutrons).

... Let’s review the physics of this experiment. If we could, in principle, distinguish the alternative final states (even though you do not bother to do so), the total, final probability is obtained by calculating the probability for each state (not the amplitude) and then adding them together. If you cannot distinguish the final states even in principle, then the probability amplitudes must be summed before taking the absolute square to find the actual probability. The thing you should notice particularly is that if you were to try to represent the neutron by a wave alone, you would get the same kind of distribution for the scattering of a down spinning neutron as for an up spinning neutron. You would have to say that the “wave” would come from all the different atoms and interfere just as for the up spinning one with the same amplitude. But we know that this is not the way it works. So as we stated earlier, we must be careful not to attribute too much reality to the waves in space. They are useful for certain problems but not for all.

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Actually, what you said is not true. First, electron go through both slits at once (if you want to say something contrary, it's much more about taste or interpretation, that 'fighting with a false belief'). Second, while properties of a single photon are interesting, I would not say it does not behave like a wave (it certainly have many wave-like properties). –  Piotr Migdal Nov 20 '10 at 9:56
Very strange answer. Actually, it only depends on interpretation whether you think that electron passes through both slits. So this means that it is not a misconception at all. Actually, the most attractive interpretation to me (Feynman's path integral) tells you that the electron travels through all paths and all of them are equally important in determining the final amplitude. In particular, the paths through left and right slits interfere. –  Marek Dec 1 '10 at 0:04
It seems to me like you are trying to propose ancient preconceptions (like the ones Einstein had) that quantum effects appear just as statistics after letting many electrons through the slits. Well, this is obviously not the case. Every single electron behaves quantum mechanically and travels through all the paths (in path integral view) or is a wave that passes through both slits (in Schroedingerian view). –  Marek Dec 1 '10 at 0:08
Sorry to step in abruptly, but QM does not tell us ANYTHING about the path of the electron. The ONLY objective, measurable and sustainable thing that QM tells us is the OUTCOME of the MEASUREMENTS. I.e. that electrons will hit the screen in some pattern. All the rest is (non-physical) interpretation and subjective. –  Sklivvz Dec 3 '10 at 21:16
@Sklivvz: but you are implicitly using one such interpretation yourself. Namely, one that tells you that it's unphysical to ask anything except measurements. You have arbitrarily chosen to make measurement the basis of what is physical (and seems you are not even aware of it). The point is, I can take another interpretation, one that tells me where the electrons went and it also perfectly reproduces all the measurements, so there is no problem with it. It's as valid (and physical) an interpretation as yours is ;-) –  Marek Dec 4 '10 at 16:12

As a tutor, I frequently have conversations like this:

"So we worked out that if when I toss my pen up at 2m/s, it will go 20cm high. How high will it go if I toss it up at 4m/s?"

"40 cm."

"Well, okay, let's check by working through that equation again..."

[We find out that the answer is 80 cm]

"So when I throw it up twice as fast, it goes four times as high, because it takes twice as long to get to the top, but is also going twice as fast."

"Okay."

"Now what if I throw it up three times as fast? How many times as high will it go?"

"Six."

This isn't a misconception about kinetic energy, so much as a lack of comprehension about what scaling is. When students are missing this concept, almost all of physics is more difficult to discuss.

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@Mark No, not really. They "know" the term is squared. They can instantly recite the formula KE = 1/2mv^2. But they don't really get what that means. They think that the only way to answer questions about what would happen if we throw the thing twice as fast is to plug in some numbers to the formula. The idea of looking at the exponents in formulas and gaining physical insight from that is alien to them. It's natural to you because you know basic physics and math quite well, but to students it is a weird and unusual trick. It's easy to forget how little you knew a long time ago. –  Mark Eichenlaub Nov 19 '10 at 22:25
Yes - exactly. Building intuition about math is a slow process, but it's interesting to watch a student progress throughout the course of a year. –  Mark Eichenlaub Nov 19 '10 at 22:46

I would say that for most people, the quadratic scaling of kinetic energy with speed is somewhat of a mystery.

People don't understand how if you go twice as fast, a car accident actually four times as energetic, hence the high number of reckless drivers and deadly accidents.

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A couple of space and sci-fi derived misconceptions:

• An orbiting satellite needs propulsion and that orbiting is different from free fall.
• You can actually see a laser beam in free space! I've seen this in my experimental physics class a few years back, before keychain lasers were common.
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And the misconception that orbiting satellites and other space vessels make cool whooshing sounds as they go by at 1/1000th of their actual speed by the pretend-stationary camera. –  romkyns Jan 12 '11 at 1:22
Seeing a laser beam in free space? Aren't all photons supposed to go straight in free space instead of some of them changing path to land in my eyes? –  Kim Kim Mar 17 '11 at 16:21
@KimKim I didn't express myself properly: I've seen people believing that. –  Sklivvz Sep 23 '11 at 12:40

Amazingly, Wikipedia has an article titled "List of common misconceptions". There is a (short) section dedicated to Physics, which mentions:

• The role of the Coriolis effect in bathtubs and sink drains
• The role of angular momentum in bicycle stability
• The "equal time" fallacy in explaining the lift developed by an airfoil
• Glass isn't actually a high viscosity fluid
• Composition of air
• "Lightning never strikes twice"

The Astronomy section has some good ones too:

• When a star collapses into a black hole, its gravitational pull does not actually increase.
• Meteorites are not actually hot when they land; usually they are cold. (I would add: the heating of meteors is more due to the compression of the air in front of them than to 'friction with the air' as commonly believed.)

Some that I would add:

• "Once something is in orbit it is free from Earth's gravity." Even educated people get tripped up on this one; the internet is rife with people suggesting we just "nudge" the International Space Station into lunar orbit. At a much more basic level of misunderstanding, there is the idea that astronauts are "weightless" because they are far away from the earth.

• "There is a high tide on the opposite side of the earth from the moon/sun because the earth 'shields' the ocean from the gravitational pull."

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Especially the equal time fallacy. Nobody believes you when you tell them it's not true. –  dan_waterworth Dec 15 '10 at 7:24
Just to clarify the 6 physics examples are things that aren't true (which people believe) - the 2 astronomy ones are things that are true (but people don't believe) –  Martin Beckett Mar 19 '11 at 5:51

"Long hair grows slower" is due to biological effects

In fact, this is a purely mathematical phenomenon. The bigger average length, the more decrease is caused by each hair fall. This leads to a differential equation $$\frac{dL}{dt} = K - \alpha L$$ that has a solutions which decay exponentially to an equilibrium at $$K/\alpha$$.

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@Pavel: sigh... are you intentionally misinterpreting my statements? I never said single hair is important. Just that your average doesn't account of single hairs (in particular, the long ones) and that the average is not important for this question. In reality what's important is that there is enough long hairs. The girl couldn't possibly distinguish whether she had 10000 30cm hairs or 5000 20cm hairs and 5000 40cm hairs. Your average would come out the same but obviously in the second case her hair would seem 10cm songer. Your model doesn't account for this at all and therefore it's useless. –  Marek Dec 3 '10 at 9:18
@Pavel: sure, that's what I am saying. But there are many such characteristics and you used just one: the average. And this is the most crudest one and most irrelevant too. –  Marek Dec 5 '10 at 12:38

"Burning coal for heating is more efficient than electrical due to thermodynamical losses at coal power plant"

This is false, even though it is true that converting electrical power to heat would be even worse. The correct way to heat with a given amount of heat source is this:

• burn it at high temperature
• make work with heat machine between Thigh and the environment
• use the work to power an air conditioner to bring heat from Tenvironment into Troom.

This gives efficiency more than 1 (more heat brought into room than heat produced by burning), and net cooling of the surrounding environment.

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I think one common false belief is that a light mill rotates because photons deposit more momentum on the shiny side (where they are reflected) than on the black side (where they are absorbed).

I find it quite astonishing to see that many people think so despite the fact that a light mill spins to the opposide direction than predicted by that explanation.

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It depends on the vacuum you have. When there isn't a high enough vacuum, the black side is heated more than the shiny side and because of convection of air and some edge effect there is motion from the black side to the shiny side. When there is a high enough vacuum, then the radiation pressure is what moves the mill. So it depends on the light mill. –  Vagelford Nov 30 '10 at 23:04
Someone should invent a light mill where the pressure inside can be changed enough to observe it working either way. It would be fun to show that to students and non-scientists and ask them to explain. –  DarenW Mar 5 '11 at 21:52

Something cannot come from nothing

Yes, it can: In quantum field theories, the vacuum is not empty.

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Some sport instructors would tell you:

Running on a treadmill is easier because you only have to jump, while on the street you also push forward

I suggested then that running in a train should be the easiest, by this line of thought. However, it is true that starting to run (or, accelerating) is indeed easier. Or the air resistance, unless there's a slowest forward wind. Or, the randomly changing slope. Also, lack of the air conditioner. That pretty much summarizes the difference.

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Some I have heard:

• Heavier objects fall faster (this is just plain wrong.) However, bigger and smaller objects in a typical Earth environment would fall at different rates due to air resistance, but actual mass has no influence.

• Two cars colliding at 60 mph is the same as one car colliding into a wall at 120 mph. I think MythBusters did something on this.

• Electrons travel really fast around a circuit: actually, they travel really slow but it is similar to Newton's cradle in that a small movement in one ball can transfer the energy to the last one almost instaneously.

• The laws of thermodynamics have been broken by some guy in a garage with some magnets. Well, no, they still seem to be intact, and a lot of modern science depends on them!

• That an oscilloscope trace travels faster than the speed of light on expensive high speed analog scopes (~1-2GHz.) This is not quite true: although the beam may sweep the surface of the CRT faster than c (due to the relatively small movement at the neck of the CRT), the trace cannot communicate information faster than c.

• More related to chemistry, but the fact that water can have "memory" and all that homeopathic nonsense that con arti^H^H^H^H^H^H^H^H homeopaths spurt out.

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It's important to make the distinction about the speed of electrons, but electrons traveling slowly is only half the story. Electrons have a slow drift velocity, but they have a high thermal velocity in most situations. –  Mark Eichenlaub Dec 3 '10 at 9:19
@Mark Eichenlaub: you can say that they move fast, but they don't travel fast. It's probably a language-related thing, but traveling implies overall displacement. –  Sklivvz Dec 3 '10 at 21:11
Wait, what's wrong with the colliding cars? Shouldn't it be the same in all frames of reference? It certainly is the same in the case of an elastic collision. –  Greg Graviton Dec 4 '10 at 17:49
Ah, because the energy is used to deform both cars. So, that would be like a car with 120 mph against another car instead of a wall. Would it? Kinetic energy is quadratic in velocity, but something weird happens when you switch the frame of reference. –  Greg Graviton Dec 5 '10 at 13:09
@Greg: here's an analysis of the crash that might interest you. Also see this question. –  David Z Dec 15 '10 at 0:06

If you somehow manage to BREAK a law of physics, the universe will vanish!

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Fortunately if you do it will be replaced by a backup copy –  Martin Beckett Mar 19 '11 at 5:59

"Summer is when the Earth is closest to the sun, and winter is when it's furthest away."

It's true that the Earth's orbit is slightly elliptical, but the effect of this, as far as seasons, is very small. For one thing, this wouldn't explain why the sun rises and sets at different times in different seasons, and if this were true, the whole planet would have summer at the same time.

The seasons are actually caused by the tilt of the Earth relative to its orbit around the Sun.

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Why do I have to learn this law when they change it every few years?

This has to do with a fact that some (actually, a lot of) people believe that the progress in physics is done in form of revolution and (in particular) that one day we might find laws that will contradict everything we knew up till then.

Well, if one looks closely on history of physics, it should become apparent that progress was always just evolutionary. Even when some idea needed a revolution in the way people think (as with SR and QM) it always turned out to be just a generalization of our previous ideas (so both SR and QM have nice classical limits which coincide with Newtonian mechanics).

Barring the useless philosophical views (like we might live in Matrix or we don't know whether the sun will rise tomorrow for sure) it's pretty certain that our universe is a comprehensible place and our theories are just better and better approximations to the reality. So it will always be useful to learn Newtonian mechanics, even one million years from now.

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You need something more to get something more

This is about emergent macroscopic properties of microscopical laws. Some people can't understand that statistics is powerful enough to make seemingly random heap of molecules suddenly show macroscopic properties like being solid or being magnetic and they think some hand of god is required to make this happen.

The best illustration of a contradiction to this principle are living organisms. They consist of nothing else than few physical laws all the way down and large number of molecules. All that was needed was statistics and natural selection.

I will agree though that it is quite amazing all of our nature can spontaneously emerge just from particles given enough space and time.

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@Ron: I am not saying it's obvious, only that it is possible (i.e. you don't need God to create life). And it is certainly the only scientific explanation we have, so what needs to be done is "only" better quantification of relevant processes. –  Marek Aug 16 '11 at 17:15

Classical mechanics is boring and mostly solved

...especially in case of fluid dynamics (-;

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I never met a person who would think this. If you don't know anything about mechanics then you obviously can't think it is boring. And if you do know it, just double pendulum or three-body problem should convince you that it's far from simple. I wouldn't call fluid dynamics classical mechanics though. While it is classical, it is certainly not mechanics but rather a field theory. And that is the main reason it is hard. –  Marek Dec 3 '10 at 12:46
I would say most physics undergrads think this is so. This is closely related to "anything that can be solved analytically, already has" –  Pete Dec 3 '10 at 16:01

The most common misconceptions are about gravity:

(1) Gravity turns off at space-shuttle orbit distance because the astronauts are weightless

Gravity is at about 80% strength compared to the surface of the Earth. The astronauts are weightless because the shuttle is in free-fall (orbit). If there was no gravity, the shuttle could not orbit.

(2) Gravity is generated by the spinning Earth. If the Earth stopped spinning, gravity would turn off.

Gravity is generated by virtue of the Earth's mass and the mass of the object; the two exert a mutual pull. There are some smaller effects associated with the spinning Earth (e.g. the Coriolis effect), but gravity would still work fine if the Earth stopped spinning.

There is another great misconception about the force of impact between a truck and a small car in a collision:

(3) The truck exerts a greater force of impact on the car, than the car on the truck

While the damage can be certainly unequal, the forces are.

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(3) is not really a misconception if force is understood in a colloquial way (normal people don't really use words like force and work in the physical sense). In particular, truck will have a lot more momentum so the situation is certainly asymmetric. But otherwise I like these. +1 –  Marek Dec 3 '10 at 16:25

the belief that cartoon-universe rules apply to falling objects:

as embodied by the statement that, "if you are trapped in a falling elevator, you can avoid destruction by jumping at the last moment, so that when the elevator hits, you are in the air, and only fall the last inch or two."

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It really bugs me. I've even heard from some quantum mechanics lecturers that they think quantum entanglement implies faster than light communication!

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Quantum theory is not non-local (and I am not really sure what you mean by that intergalactic invasions, @Pavel)! This could be put as another misconception stemming from EPR paradox. What EPR (or rather Bell's inequalities) say is that quantum theory is either non-local or incomplete (in a sense of absence of hidden parameters). We have good reasons to think it is local (e.g. QFT has to obey locality if it is to make any sense) so what these experiments actually concluded is that there are no hidden variables (i.e. QM is not just statistics). –  Marek Dec 5 '10 at 12:34
@Pavel: okay, done. It's interesting but it's still just the good old correlation of entangled spins, nothing more. Calling it non-local (or second-best, or whatever) is just confusing and precisely the reason why people think there is superluminal communication going on. –  Marek Dec 5 '10 at 17:55

I hear from time to time people highly educated and skilled in Physics (unlike those believing that heavy objects fall faster than light ones) making the following claim:

... quantum mechanics indicates that certain physical quantities can take only a countable set of discrete values. Consequently, many current approaches to foundational questions in physics and cosmology advocate novel discrete or 'digital' pictures of nature.

The discrete spectra of some quantum observables do not imply/suggest that nature, in particular spacetime, is fundamentally discrete. The spectrum of a continuous operator acting on Hilbert spaces [which is a topological (vector) space, hence is continuous], often has a discrete part. This has nothing to do with spacetime being discrete. If it will (eventually) turn out to be discrete, it will be for other reasons.

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If you are riding a bicycle and you turn the front wheel to the left then the bicycle will steer to the left.

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This is tongue-in-cheek!

That an emptying tank with a nozzle pointing downwards would actually experience a force!

:-)

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That if an object is moving, there must be a force propelling it in that direction. Students very commonly think that forces cause objects to have velocity, rather than the fact that forces cause objects to change velocity.

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Related to this, it really seems to confuse a lot of intro-level physics students when acceleration and velocity vectors are pointed in opposite directions. One of my standard quizzes when I taught freshman physics was to toss a ball straight up and down in the air, and then hand out a position vs. time plot and ask them to sketch the acceleration and velocity vectors at several key points. (neglecting air resistance). Always gets an interesting mix of answers. –  Tim Goodman Feb 21 '11 at 17:24

Here's a false belief. Everything your physics teacher says is true. Because physics teachers would never mislead you over how many states of matter there are or how lift is generated on a wing.

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That the PI Team is the best group to design and implement their own data systems, rather than bring in people experienced with IT security, data modeling for reuse by other groups, and other data informatics issues.

Related misconception: that their data is so different from other data that a data system must be designed from the ground up for each new experiment.

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text:
IN  WHICH Ross Rheingans-Yoo—a sometime economist, artist, trader, expat, poet, EA, and programmer—writes on things of interest.

# Reading Feed (last update: July 5)

A collection of things that I was glad I read. Views expressed by linked authors are chosen because I think they’re interesting, not because I think they’re correct, unless indicated otherwise.

### (4)

Blog: Tyler Cowen @ Bloomberg View | The NBA’s Reopening Is a Warning Sign for the U.S. Economy — "If so many NBA players are pondering non-participation, how keen do you think those workers — none of whom are millionaire professional athletes — are about returning to the office?"

# A Liberal Critique

content warning: discussion of discussion of rape; discussion of critique of an instance of opposition to racism

"Charity over absurdity", 5 of $\infty$.

[ | ]

### (1)

A friend pointed out on Facebook (I'm not going to hyperlink.) that they were seriously troubled that the first time I chose to write about race[1], it was against the idea that we should be doing more to fight racism:

The fact that the first thing that you have ever said about [race] is in defense of structural protections of racism -- even if I, personally, in this case... agree with you -- is highly suspicious, and it is important... to consider carefully how we choose the things about which to be passionate, [especially] publicly passionate, and how those choices influence those around us.

I'm actually pretty sympathetic to this view, though I was heartened by another friend's comment:

I'm not sure if I agree with you that people should make statements that have already been said in order to communicate we are as

# Why So Few Male Feminists?

content warning: unrepentant naïveté, use and interrogation of the word "feminism" by a cishet white male, statistical mention of rape, sexual/domestic violence, and abuse

content note: in parts, speaking only to people who have the privilege of choosing, intentionally and with lightness, how they engage with issues of social justice. (more in a previous post)

### (1)

Ozy Frantz[read this] is one of those bloggers who has significantly and dramatically changed the way I think -- and in this respect shares a reference class with Leah Libresco and Eliezer Yudkowsky. Though I've only been reading Ozy since they started Thing of Things last November, they're very quickly stepping into position as Possibly My Favorite Blogger Right Now. A perfect example of why is their recently re-run post, Thing of Things | Who Cares About Men's Rights?:

I do.

I care about every boy that was ever called a fag or a pussy or a sissy for being emotional, or sensitive, or unathletic, or just not manly enough. I care about

format: plain


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## Tuesday, April 18, 2017

### make-theme-image: a script to make yourself an idea of a icon theme

I created some time ago a utils repository on github to publish miscellaneous scripts, but it's only recently that I have started to really populate it. One of my recent work is make-theme-image script, that downloads an icon them package, grabs relevant, user-specifiable, icons, and arrange them in a neat montage. The images displayed are the results of running

~ make-theme-image gnome-icon-theme moblin-icon-theme

This is quite useful to get a good idea of the icons available in a package. You can select the icons you want to display using the -w option. The following command should provide you with a decent overview of the icon themes present in Debian:

apt search -- -icon-theme | grep / | cut -d/ -f1 | xargs make-theme-image

I hope you find it useful ! In any case, it's on github, so feel free to patch and share.

## Monday, April 3, 2017

### Variations around ls -lrt

I have been using almost compulsively ls -lrt for a long time now. As per the ls man page, this command lists the files of the current directory, with the latest files at the end, so that they are the ones that show up just above your next command-line. This is very convenient to work with, hmmm, not-so-well-organized directories, because it just shows what you're working on, and you can safely ignore the rest. Typical example is a big Downloads directory, for instance, but I use it everywhere. I quickly used alias lrt="ls -lrt" to make it easier, but... I though I might have as well a reliable way to directly use what I saw. So I came up with the following shell function (zsh, but probably works with most Bourne-like shells):
lrt() {
ls -lrt "$@" lrt="$(ls -rt "$@" | tail -n1)" }  This small function runs ls -lrt as usual, but also sets the $lrt shell variable to the latest file, so you can use it in your next commands ! Especially useful for complex file names. Demonstration:
22:05 vincent@ashitaka ~/Downloads lrt
[...]
-rw-r--r-- 1 vincent vincent   1490027 Apr  2 15:44 k.zip
-rw-r--r-- 1 vincent vincent    668566 Apr  3 22:05 1-s2.0-S0013468617305947-main.pdf
22:06 vincent@ashitaka ~/Downloads cp -v \$lrt ~/nice-paper.pdf
'1-s2.0-S0013468617305947-main.pdf' -> '/home/vincent/nice-paper.pdf'

This saves typing the name of the 1-s2.0-S0013468617305947-main.pdf: in this case, automatic completion doesn't help much, since many files in my Downloads directory start with the same orefix... I hope this helps !


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Browse Questions

# If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is

$(a)\;\large\frac{Q}{2}\qquad(b)\;-\large\frac{Q}{2}\qquad(c)\;\large\frac{Q}{4}\qquad(d)\;- \large\frac{Q}{4}$

Answer : (d) $\;-\large\frac{Q}{4}$
Explanation :
Let's see the force on one Q
$F= \large\frac{k Q^2}{4 a^2} + \large\frac{k\; q\; Q}{a^2}$
For the system to be in equilibrium F =0
$\large\frac{k Q^2}{4 a^2} + \large\frac{k\; q\; Q}{a^2}=0$
$q=-\large\frac{Q}{4}\;.$
edited Feb 13, 2014 by yamini.v


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# Overview

Chapter overview

This section describes imaging single-dish data.

The data should already be T$_{sys}$ and T$_{sky}$ calibrated (at least into antenna temperature units, $T_A^*$ [K]) , according to the process described in the single-dish calibration pages.

The CASA task used for imaging single-dish data is sdimaging.  This task will return a four-dimensional array with two position axes, one frequency or velocity axis, and one polarization axis. The output of sdimaging is a CASA image that can be explored, analyzed, and manipulated in CASA, or exported into a versatile FITS image format via exportfits.

The pages below first describe the general process of gridding single-dish data followed by the actual procedures invoked with CASA.


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# Evaluating the Integral $\int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-(x+y)}}{x+y} dxdy$

I need to find , $$\int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-(x+y)}}{x+y} dxdy$$

My attempt:

I tried doing integration wrt x first then substitute $t = x+y$, $$\int_{0}^{\infty} e^{-y} \left( \int_{0}^{\infty} \frac{e^{-x}}{x+y} dx\right) dy = \int_{0}^{\infty} e^{-y} \left( \int_{t=y}^{\infty} \frac{e^{-(t-y)}}{t} dt\right) = \int_{0}^{\infty} \int_{t=y}^{\infty} \frac{e^{-t}}{t} dt$$

I don't know how to proceed further. Any hint on how should i tackle this kind of problem would be appreciated ...

• Try $$\int_{0}^{\infty} \int_{y=0}^{y=t} \frac{e^{-t}}{t} dy\,dt$$ – Did Apr 5 '17 at 8:55
• Okay. Change of integration it is. Got it. Thank you @Did Answer is $1$. – Raindeer Apr 5 '17 at 8:58


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POWSUMS - Editorial

easy

PREREQUISITES

dp, matrix exponentiation, newton’s identities, elementary symmetric polynomials

PROBLEM

Let f(x) = {a_1}^x + {a_2}^x + \dots + {a_n}^x \pmod {10^9 + 7}, where a_1, a_2, \dots, a_n are integers. You don’t know a_1, a_2, \dots, a_n. Instead you have the values of f(1), f(2), \dots, f(n). You are to find f(x) for some given x.

EXPLANATION

Take small Examples

Let us take small examples to understand definition of f(x).

Let n = 1, f(x) = {a_1}^x. i.e. f(1) = a_1.
We are given value of f(1), so know value of a_1. So we know that function f(x). Now, we can evaluate f for any x.

Let n = 2. f(x) = {a_1}^x + {a_2}^x.

f(1) = a_1 + a_2
f(2) = {a_1}^2 + {a_2}^2

f(1) and f(2) are known values. We want to evaluate f(x) for any x.

\text{Write } {a_1}^2 + {a_2}^2 = {(a_1 + a_2)}^2 - 2 \cdot a_1 \cdot a_2
f(2) = f(1)^2 - 2 \cdot a_1 \cdot a_2
a_1 \cdot a_2 = \frac{f(2) - f(1)^2} {2}

We know value of a_1 \cdot a_2 and a_1 + a_2 = f(1). Using this information, we can find values of a_1, a_2.

Elementary symmetric polynomials

Elementary symmetric polynomials are one of the important constructs in mathematics, any symmetric polynomial can be represented as a polynomial in elementary symmetric polynomials. In other words, any symmetric polynomial can be represented by an expression involving only additions and multiplications of elementary symmetric polynomials. There is one elementary symmetric polynomial of degree d in n variables (d \leq n), which is formed by adding together products of d distinct variables. For brevity, we will refer elementary symmetric polynomial of degree d in n variables as ESP(n, d).

For example, let us consider only two variables i.e. n = 2, a_1 and a_2.

ESP(2, 0) = 1
ESP(2, 1) = a_1 + a_2
ESP(2, 2) = a_1 a_2

Now let us consider only two variables n = 3, a_1, a_2 and a_3.

ESP(3, 0) = 1
ESP(3, 1) = a_1 + a_2 + a_3
ESP(3, 2) = a_1 a_2 + a_2 a_3
ESP(3, 3) = a_1 a_2 a_3

A more complex example

ESP(4, 3) = a_1 \, a_2 \, a_3 + a_1 \, a_2 \, a_4 + a_1 \, a_3 \, a_4 + a_2 \, a_3 \, a_4

In other words, each term will be sum of product of all d sized subset of distinct variables. In general, there will be total \binom{n}{d} terms in it.

Any symmetric polynomial can be represented in terms of ESP’s. Here are some examples.

a_1 + a_2 = ESP(2, 1)
{a_1}^2 + {a_2}^2 = {(a_1 + a_2)}^2 - 2 \cdot a_1 \cdot a_2
{a_1}^2 + {a_2}^2 = {ESP(2, 1)}^2 - 2 \, ESP(2, 2)

Expressing EPSs in terms of power sums

Let us denote ESP(n, k) as e_k.

We see

e_0 = f(0)
e_1 = f(1)
e_2 = \frac{e_1 \, f(1) - e_0 \, f(2)}{2}
e_3 = \frac{e_2 \, f(1) - e_1 \, f(2) + e_0 \, f(3)}{3}
e_4 = \frac{e_3 \, f(1) - e_2 \, f(2) + e_1 \, f(3) - e_0 \, f(4)}{4}

In general,

e_k = \frac{e_{k - 1} \, f(1) - e_{k - 2} \, f(2) + \dots \pm e_0 \, f(k)}{k}

Using these relations, we can obtain values of e_0, e_1, \dots e_n by the values of f(1), f(2), \dots, f(n) in total \mathcal{O}(n^2) time.

Expressing power sums in terms of ESPS

Rewriting the above equations used in previous section, we get

f(0) = e_0
f(1) = e_1
f(2) = e_1 \, f(1) - 2 \, e_2 \, f(0)
f(3) = e_1 \, f(2) - e_2 \, f(1) + 3 \, e_3 \, f(0)
f(4) = e_1 \, f(3) - e_2 \, f(2) + e_3 \, f(1) - 4 \, e_4 \, f(0)

In general,

f(n) = e_1 \, f(n - 1) - e_{2} \, f(n - 2) + \dots \pm n \, e_n \, f(0)

As we know that e_i for i > n will be zero.

So, we get

f(n + 1) = e_1 \, f(n) - e_{2} \, f(n - 1) + \dots \pm (n + 1) \, e_{n + 1} \, f(0)

As e_{n + 1} = 0.

f(n + 1) = e_1 \, f(n) - e_{2} \, f(n - 1) + \dots \pm (n) \, e_{n} \, f(1)

We see that it has only n terms.

In general f(X) for even X > n will contain n terms in its representation of ESPs.

Solving recurrence relations

Now, we see that f(X) for X > n can be represented as a recurrence relation of n terms. Also coefficients in this recurrence relation are constant or linearly related with i. So the recurrence relation is linear.

Direct matrix multiplication for each query
Using the direct matrix exponentation, we will get \mathcal{O}(n^3 \, log X) time complexity for a single query of f(X). There are Q queries, so, the total time will be \mathcal{O}(Q \, n^3 \, log X). This can solve up to three subtasks.

Slightly faster solution
We can precompute M, M^2, M^4, M^8, M^{16}, \dots, M^{2^p}, such that p is largest number satisfying 2^p \leq 10^{18}.

As mulitplying a matrix with a vector takes \mathcal{O}(n^2) time. The query f(X) can be answered in at most log(X) multiplications of the precomputed matrices and a constant vector, to answer each query in \mathcal{O}(n^2) log X time.

Overall time taken will be dominated by precomputing as number of queries are small. So, we get \mathcal{O}(n^3 \, log X) time.

Let us take a small example to describe this in detail. Let us take the simplest example of finding fibonacci number.

f(0) = 0
f(1) = 1
f(n) = f(n - 1) + f(n - 2)
\begin{pmatrix} f(1) \\ f(0) \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{matrix} f(n) \\ f(n - 1) \end{matrix} = \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} . \begin{matrix} f(n - 1) \\ f(n - 2) \end{matrix} \begin{matrix} f(n) \\ f(n - 1) \end{matrix} = { \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} }^{n - 1} . \begin{matrix} 1 \\ 1 \end{matrix}

Let

M = { \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} }

and the vector v be

v = { \begin{pmatrix} 1\\ 0 \end{pmatrix} }

Assume we computed, M, M^2, M^4, M^8 etc.

We want to compute f(8). As we know that f(8) = M^7 v. We can write M^7 as M^1 * M^2 * M^3.

f(8) = M^1 * M^2 * M^3 * v

Multiply the matrices in the order from right to left instead of left to right. So, each time we multiply the matrix with a vector. So, we do total 3 multiplications of matrix with vector when we go from right to left. If we go from left to right, we need to 2 multiplication of matrices and one multiplication with constant vector.


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# LOG#023. Math survey.

What is a triangle? It is a question of definition in Mathematics. Of course you could disagree, but it is true. Look the above three “triangles”. Euclidean geometry is based in the first one. The second “triangle” is commonly found in special relativity. Specially, hyperbolic functions. The third one is related to spherical/elliptical geometry.

Today’s summary: some basic concepts in arithmetics, complex numbers and functions. We are going to study and review the properties of some elementary and well known functions. We are doing this in order to prepare a better background for the upcoming posts, in which some special functions will appear. Maybe, this post can be useful for understanding some previous posts too.

First of all, let me remember you that elementary arithmetics is based on seven basic “operations”: addition, substraction, multiplication, division, powers, roots, exponentials and logarithms. You are familiar with the 4 first operations, likely you will also know about powers and roots, but exponentials and logarithms are the last kind of elementary operations taught in the school ( high school, in the case they are ever explained!).

Let me begin with addition/substraction of real numbers (it would be also valid for complex numbers $z=a+bi$ or even more general “numbers”, “algebras”, “rings” or “fields”, with suitable extensions).

$a+b=b+a$

$(a+b)+c=a+(b+c)$

$a+(-a)=0$

$a+0=a$

Multiplication is a harder operation. We have to be careful with the axioms since there are many places in physics where multiplication is generaliz loosing some of the following properties:

$kA=\underbrace{A+..+A}_\text{k-times}$

$AB=BA$

$(AB)C=A(BC)=ABC$

$1A=A1=A$

$(A+B)C=AC+BC$

$A(B+C)=AB+AC$

$A^{-1}A=AA^{-1}=1$

Indeed the last rule can be undestood as the “division” rule, provided $A\neq 0$ since in mathematics or physics there is no sense to “divide by zero”, as follows.

$A^{-1}=\dfrac{1}{A}$

Now, we are going to review powers and roots.

$x^a=\underbrace{x\cdots x}_\text{a-times}$

$(x^a)^b=x^{ab}$

$x^{-a}=\dfrac{1}{x^a}$

$x^ax^b=x^{a+b}$

$\sqrt[n]{x}=x^{1/n}$

Note that the identity $x^ay^a=(x+y)^a$ is not true in general. Moreover, if $x\neq 0$ then $x^0=1$ as well, as it can be easily deduced from the previous axioms. Now, the sixth operation is called exponentiation. It reads:

$\exp (a+b)=\exp (a)\exp (b)$

Sometimes you can read $e^{a+b}=e^{a}e^{b}$, where $\displaystyle{e=\lim_{x\to\infty}\left(a+\dfrac{1}{n}\right)^n}$ is the so-called “e” number. The definitions is even more general, since the previous property is the key feature for any exponential. I mean that,

$a^x=\underbrace{a\cdots a}_\text{x-times}$

$a^xb^x=(ab)^x$

$\left(\dfrac{a}{b}\right)^x=\dfrac{a^x}{b^x}$

We also get that for any $x\neq 0$, then $0^x=0$. Finally, the 7th operation. Likely, the most mysterious for the layman. However, it is very useful in many different places. Recall the definition of the logarithm in certain base “a”:

$\log_{(a)} x=y \leftrightarrow x=a^y$

Please, note that this definition has nothing to do with the “deformed” logarithm of my previous log-entry. Notations are subtle, but you must always be careful about what are you talking about!

Furthermore, there are more remarks:

1st. Sometimes you write $\log_e=\ln x$. Be careful, some books use other notations for the Napier’s logarithm/natural logarithm. Then, you can find out there $\log_e=L$ or even $\log_e=\log$.

2nd. Whenever you are using a calculator, you can generally find $\log_e=\ln$ and $\log_{(10)}=\log$. Please, note that in this case $\log$ is not the natural logarithm, it is the decimal logarithm.

Logarithms (caution: logarithms of real numbers, since the logarithms of  complex numbers are a bit more subtle) have some other cool properties:

$\log_{(a)}(xy)=\log_{(a)}x+\log_{(a)}y$

$\log_{(a)}\dfrac{x}{y}=\log_{(a)}x-\log_{(a)}y$

$\log_{(a)}x^y=y\log_{(a)}x$

$\log_{(a)}=\dfrac{\log_{(b)}}{\log_{(b)}a}$

Common values of the logarithm are:

$\ln 0^+=-\infty;\; \ln 1=0;\; \ln e=1;\; \ln e^x=e^{\ln x}=x$

Indeed, logarithms are also famous due to a remarkable formula by Dirac to express any number in terms of 2’s as follows:

$\displaystyle{N=-\log_2\log_2 \sqrt{\sqrt{\underbrace{\cdots}_\text{(N-1)-times}2}}=-\log_2\log_2\sqrt{\underbrace{\cdots}_\text{(N)-times}2}}$

However, it is quite a joke, since it is even easier to write $N=\log_a a^N$, or even $N=\log_{(1/a)}a^{-N}$

Are we finished? NO! There are more interesting functions to review. In particular, the trigonometric functions are the most important functions you can find in the practical applications.

EUCLIDEAN TRIGONOMETRY

Triangles are cool! Let me draw the basic triangle in euclidean trigonometry.

The trigonometric ratios/functions you can define from this figure are:

i)The function (sin), defined as the ratio of the side opposite the angle to the hypotenuse:
$\sin A=\dfrac{\textrm{opposite}}{\textrm{hypotenuse}}=\dfrac{a}{\,c\,}$
ii) The function (cos), defined as the ratio of the adjacent leg to the hypotenuse.
$\cos A=\dfrac{\textrm{adjacent}}{\textrm{hypotenuse}}=\dfrac{b}{\,c\,}$
iii) The function (tan), defined as the ratio of the opposite leg to the adjacent leg.
$\tan A=\dfrac{\textrm{opposite}}{\textrm{adjacent}}=\dfrac{a}{\,b\,}=\dfrac{\sin A}{\cos A}$

The hypotenuse is the side opposite to the 90 degree angle in a right triangle; it is the longest side of the triangle, and one of the two sides adjacent to angle ”A”. The ”’adjacent leg”’ is the other side that is adjacent to angle ”A”. The ”’opposite side”’ is the side that is opposite to angle ”A”. The terms ”’perpendicular”’ and ”’base”’ are sometimes used for the opposite and adjacent sides respectively. Many English speakers find it easy to remember what sides of the right triangle are equal to sine, cosine, or tangent, by memorizing the word SOH-CAH-TOA ( a mnemonics rule whose derivation and meaning is left to the reader).

The multiplicative inverse or reciprocals of these functions are named the cosecant (csc or cosec), secant(sec), and cotangent (cot), respectively:
$\csc A=\dfrac{1}{\sin A}=\dfrac{c}{a}$
$\sec A=\dfrac{1}{\cos A}=\dfrac{c}{b}$
$\cot A=\dfrac{1}{\tan A}=\dfrac{\cos A}{\sin A}=\dfrac{b}{a}$

The inverse trigonometric functions/inverse functions are called the arcsine, arccosine, and arctangent, respectively. These functions are what in common calculators are given by $\sin^{-1},\cos^{-1},\tan^{-1}$. Don’t confuse them with the multiplicative inverse trigonometric functions.

There are arithmetic relations between these functions, which are known as trigonometric identities.  The cosine, cotangent, and cosecant are so named because they are respectively the sine, tangent, and secant of the complementary angle abbreviated to “co-“. From the goniometric circle (a circle of radius equal to 1) you can read the Fundamental Theorem of (euclidean) Trigonometry:

$\cos^2\theta+\sin^2\theta=1$

Indeed, from the triangle above, you can find out that the pythagorean theorem implies

$a^2+b^2=c^2$

or equivalently

$\dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}=\dfrac{c^2}{c^2}$

So, the Fundamental Theorem of Trigonometry is just a dressed form of the pythagorean theorem!
The fundamental theorem of trigonometry can be rewritten too as follows:

$\tan^2\theta+1=\sec^2\theta$

$\cot^2+1=\csc^2\theta$

These equations can be easily derived geometrically from the goniometric circle:

The trigonometric ratios are also related geometrically to this circle, and it can seen from the next picture:

Other trigonometric identities are:

$\sin (x\pm y)=\sin x \cos y\pm \sin y\cos x$

$\cos (x\pm y)=\cos x \cos y -\sin x \sin y$

$\tan (x\pm y)=\dfrac{\tan x\pm \tan y}{1\mp \tan x \tan y}$

$\cot (x\pm y)=\dfrac{\cot x \cot y\mp 1}{\cot x\pm \cot y}$

$\sin 2x=2\sin x \cos x$

$\cos 2x=\cos^2 x-\sin^2 x$

$\tan 2x=\dfrac{2\tan x}{1-\tan^2 x}$

$\sin \dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}}$

$\cos \dfrac{x}{2}=\sqrt{\dfrac{1+\cos x}{2}}$

$\tan \dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{1+\cos x}}$

$\sin x\sin y=\dfrac{\cos(x-y)-\cos(x+y)}{2}$

$\sin x\cos y=\dfrac{\sin(x+y)+\sin(x-y)}{2}$

$\cos x\cos y=\dfrac{\cos (x+y)+\cos(x-y)}{2}$

The above trigonometric functions are also valid for complex numbers with care enough. Let us write a complex number as either a binomial expression $z=a+bi$ or like a trigonometric expression $z=re^{i\theta}$. The famous Euler identity:

$e^{i\theta}=\cos \theta + i\sin \theta$

allows us to relate both two expressions for a complex number since

$z=r(\cos \theta + i\sin \theta)$

implies that $a=r\cos\theta$ and $b=r\sin\theta$. The Euler formula is also useful to recover the identities for the sin and cos of a sum/difference, since $e^{iA}e^{iB}=e^{i(A+B)}$

The complex conjugate of a complex number is $\bar{z}=a-bi$, and the modulus is

$z\bar{z}=\vert z\vert^2=r^2$

with $\theta =\arctan \dfrac{b}{a}, \;\; \vert z \vert=\sqrt{a^2+b^2}$

Moreover, $\overline{\left(z_1\pm z_2\right)}=\bar{z_1}\pm\bar{z_2}$, $\vert \bar{z}\vert=\vert z\vert$, and if $z_2\neq 0$, then

$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar{z_1}}{\bar{z_2}}$

We also have the so-called Moivre’s formula

$z^n=r^n(\cos n\theta+i\sin n\theta)$

and for the complex roots of complex numbers with $w^n=z$ the identity:

$w=z^{1/n}=r^{1/n}\left(cos\left(\dfrac{\theta +2\pi k}{n}\right)+i\sin\left(\dfrac{\theta +2\pi k}{n}\right)\right)\forall k=0,1,\ldots,n-1$

The complex logarithm (or the complex power) is a multivalued functions (be aware!):

$\ln( re^{i\theta})=\ln r +i\theta +2\pi k,\forall k\in \mathbb{Z}$

The introduction of complex numbers and complex values of trigonometric functions are fun. You can check that

$\cos z=\dfrac{\exp (iz)+\exp (-iz)}{2}$ and $\sin z=\dfrac{\exp (iz)-\exp (-iz)}{2i}$ and $\tan z=\dfrac{\exp (iz)-\exp (-iz)}{i(\exp (iz)+\exp(-iz))}$

thanks to the Euler identity.

In special relativity, the geometry is “hyperbolic”, i.e., it is non-euclidean. Let me review the so-called hyperbolic trigonometry. More precisely, we are going to review the hyperbolic functions related to special relativity now.

HYPERBOLIC TRIGONOMETRY

We define the functions sinh, cosh and tanh ( sometimes written as sh, ch, th):

$\sinh x=\dfrac{\exp (x) -\exp (-x)}{2}$

$\cosh x=\dfrac{\exp (x) +\exp (-x)}{2}$

$\tanh x=\dfrac{\exp (x) -\exp (-x)}{\exp (x)+\exp (-x)}$

The fundamental theorem of hyperbolic trigonometry is

$\cosh^2 x-\sinh^2 x=1$

The hyperbolic triangles are objects like this:

The hyperbolic inverse functions are

$\sinh^{-1} x=\ln (x+\sqrt{x^2+1})$

$\cosh^{-1} x=\ln (x+\sqrt{x^2-1})$

$\tanh^{-1} x=\dfrac{1}{2}\ln\dfrac{1+x}{1-x}$

Two specially useful formulae in Special Relativity (related to the gamma factor, the velocity and a parameter called rapidity) are:

$\boxed{\sinh \tanh^{-1} x=\dfrac{x}{\sqrt{1-x^2}}}$

$\boxed{\cosh \tanh^{-1} x=\dfrac{1}{\sqrt{1-x^2}}}$

In fact, we also have:

$\exp(x)=\sinh x+\cosh x$

$\exp(-x)=-\sinh x+\cosh x$

$\sec\mbox{h}^2x+\tanh^2 x=1$

$\coth ^2 x-\csc\mbox{h}^2 x=1$

There are even more identities to be known. The most remarkable and important are likely to be:

$\sinh (x\pm y)=\sinh (x)\cosh (y)\pm\sinh (y)\cosh (x)$

$\cosh (x\pm y)=\cosh (x)\cosh (y)\pm\sinh (x)\sinh (y)$

$\tanh (x\pm y)=\dfrac{\tanh x\pm \tanh y}{1\pm \tanh x\tanh y}$

$\coth (x\pm y)=\dfrac{\coth x\coth y\pm 1}{\coth y\pm \coth x}$

You can also relate euclidean trigonometric functions with hyperbolic trigonometric functions with the aid of complex numbers. For instance, we get

$\sinh x=-i\sin ix$

$\cosh x=\cos ix$

$\tanh x= -i\tan ix$

and so on. The hyperbolic models of geometry/trigonometry are also very known in arts. Escher’s drawings are very beautiful and famous:

or the colorful variation of this theme

I love Escher’s drawings. And I also love Mathematics, Physics, Physmatics, and Science. Equations are cool. And hyperbolic functions, and other functions we have reviewed here today, will arise naturally in the next posts.

$\displaystyle{\pi/4 = 1 - 1/3 + 1/5 - 1/7 + 1/9 -\ldots=\sum_{n=0}^\infty \dfrac{(-1)^n}{2n+1}}$


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Given an n x n binary grid, in one step you can choose two adjacent rows of the grid and swap them.

A grid is said to be valid if all the cells above the main diagonal are zeros.

Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.

The main diagonal of a grid is the diagonal that starts at cell (1, 1) and ends at cell (n, n).

Example 1:

Input: grid = [[0,0,1],[1,1,0],[1,0,0]]
Output: 3

Example 2:

Input: grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]]
Output: -1
Explanation: All rows are similar, swaps have no effect on the grid.

Example 3:

Input: grid = [[1,0,0],[1,1,0],[1,1,1]]
Output: 0

Constraints:

• n == grid.length
• n == grid[i].length
• 1 <= n <= 200
• grid[i][j] is 0 or 1

## Solution: Bubble Sort

Counting how many tailing zeros each row has.
Then input
[0, 0, 1]
[1, 1, 0]
[1, 0, 0]
becomes [0, 1, 2]
For i-th row, it needs n – i – 1 tailing zeros.
We need to find the first row that has at least n – i – 1 tailing zeros and bubbling it up to the i-th row. This process is very similar to bubble sort.
[0, 1, 2] -> [0, 2, 1] -> [2, 0, 1]
[2, 0, 1] -> [2, 1, 0]
Total 3 swaps.

Time complexity: O(n)
Space complexity: O(n)

## C++

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text:
# Greek In Latex

2010 Jun 30

## Summary

Method latin input greek input
babel \latintext Latin \greektext Ellhnik'a
babel+inputenc \latintext Latin \greektext Ελληνικά
auto-greek Latin Ελληνικά
greektex Latin Ελληνικά
xelatex Latin Ελληνικά
Method Preamble
babel \usepackage[english,greek]{babel}
babel+inputenc \usepackage[english,greek]{babel} \usepackage[iso-8859-7]{inputenc}
auto-greek Latin
greektex Latin
xelatex Latin

## Babel

The “standard” way of writing greek in LaTeX, is by using the babel package. You simply insert \usepackage[greek]{babel} in the preamble of your document and you are set. For example:

\documentclass{article}
\usepackage[greek]{babel}

\begin{document}
Ellhniko keimeno.
\end{document}

Using this method, you have to write latin characters to produce the greek text. So, writing

abgdezhjiklmnxoprs(c)tufqyw
you get:

and

'a a "i "i
gives you:

If you want to use more than one languages, you include them as options of the babel package (the last being the main language): \usepackage[english, greek]{babel} (main language is greek, so “Contents” will appear as “Περιεχόμενα” etc). To switch between greek and latin text, you have to use the commands \latintext and \greektext. (You may also use \selectlanguage{language} instead.) Putting all of the above together, the following code:

\documentclass{article}
\usepackage[english, greek]{babel}
\title{Titlos}
\author{Nikos Augoc}
\begin{document}
\maketitle
abgdezhjiklmnxoprs(c)tufqyw\\
'a a "i "i
\latintext Latin text \greektext Ellhniko keimeno
\end{document}

will give you:

## Babel + inputenc

If you do not want (as is usually the case) to write latin-text for greek, you’ll have to use the inputenc package. Insert the command \usepackage[iso-8859-7]{inputenc} (or utf-8) in the preamble of your document. So this:

\documentclass{article}
\usepackage[english, greek]{babel}
\usepackage[iso-8859-7]{inputenc}

\title{Titlos}
\author{Nikos Augoc}

\begin{document}
\maketitle abgdezhjiklmnxoprs(c)tufqyw\\
'a a "i "i

\latintext Latin text
\greektext Ελληνικό κείμενο

\end{document}

gives the same output as above.

## auto-greek

Assuming that you are writing a greek document (so you only need greek headings), you can avoid continuously typing \latintext, \greektext, or any other command of the kind, by using the package auto-greek. Putting \usepackage{auto-greek} in the preamble, you can write both greek and latin text directly. So,

\documentclass{article}
\usepackage{auto-greek}
\title{Τίτλος}
\author{Νίκος Αυγός}
\begin{document}
\maketitle

Ελληνικό και latin κείμενο.
\end{document}

gives you

## XeLaTeX

Xe(La)TeX is a modern (La)Tex engine which supports unicode by default. It avoids most of the pitfalls of all the other methods (on the “fonts” front) but is not as yet as advanced in microtypography. If font problems are your main issue with LaTeX then this should be your first choice.


text:
# Why is the helmholtz free energy minimized?

I don't quite understand the principle of minimum energy despite having read the derivation on Wikipedia.

I think I got lost when the free energy was defined as $A= \max_S{\left(U-TS\right)}$, because I don't know why is the max there.

• The wikipedia page has $\max_S$, not the maximum of the expression. – Quantumness Jun 13 '18 at 19:30
• @Quantumness I know, but I still don't even know what's that and why is that. – 545941st user Jun 13 '18 at 19:33
• It is referring to maximum entropy, the second law of thermodynamics. – Quantumness Jun 13 '18 at 19:43
• @Quantumness I still don't get the derivation :( – 545941st user Jun 13 '18 at 19:50
• Could you refer to the specific aspect where you are confused? I'm not sure I understand what you are looking for. – Quantumness Jun 13 '18 at 20:05

Here's an alternative derivation to show that Helmholtz energy will be minimized.

Consider the fact that by the second law of thermodynamics, the total entropy $S$ of the universe must increase. That is, the sum of all entropies has the relationship

$$dS_{\rm universe} = dS_{\rm sys} + dS_{\rm surr} \geq 0,$$

where 'sys' and 'surr' represent ours system and surrounds respectively. Hold this result for the moment. Recall the thermodynamic identity $dU = TdS-PdV+\mu dN$, and recognize that for constant volume and number of particles,

$$dS_{\rm surr} = \frac{dU_{\rm surr}}{T} = - \frac{dU_{\rm sys}}{T}.$$

Note that the last equality comes from the first law: energy is conserved, so $dU_{\rm surr} = -dU_{\rm sys}$. Then, plugging this into our earlier result and multiplying by $T$, we have

$$dS_{\rm universe} = dS_{\rm sys} + dS_{\rm surr} = dS_{\rm sys} - \frac{dU_{\rm sys}}{T}$$ $$\rightarrow TdS_{\rm universe} = TdS_{\rm sys} - dU_{\rm sys}.$$

Now the proper definition for Helmholtz free energy is $F = U - TS$, so for constant temperature, $dF = dU - TdS = -(TdS-dU)$. We can plug this into our last result as

$$T dS_{\rm universe} = - dF_{\rm sys}$$

and finally

$$dS_{\rm universe} = - \frac{dF_{\rm sys}}{T}.$$

Note that in order to maximize the energy of the universe, you have make the Helmholtz free energy as negative as possible (negative, but large magnitude).

• @545941st user does this answer your question? – Zack Hutchens Jun 15 '18 at 17:32


text:
# How to solve this inequation without plotting the graph?

Is there any way to solve the inequation: $${e^{2x - 1}} \ge 2x$$ without plotting the graph?

-

All you need to know is $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$

Equality holds when $x=0$

To prove $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$, consider $f(x) = e^x - 1 -x$

$f'(x) = e^x - 1 = 0 \Rightarrow$, $x=0$ is an extremum.

Further $f''(x) = e^x > 0$, $\forall x \in \mathbb{R}$

Hence, $x=0$ is a minimum

Hence, we get $f(x) \geq f(0)$, $\forall x \in \mathbb{R}$

Hence, $e^x - 1 - x \geq 0$, $\forall x \in \mathbb{R}$

Hence, $e^x \geq 1 + x$, $\forall x \in \mathbb{R}$

Replacing $x$ by $2x-1$, we get $e^{2x-1} \geq 1 + (2x-1) = 2x$, $\forall x \in \mathbb{R}$

Equality holds when $2x-1=0$ i.e. when $x = \frac{1}{2}$

-


text:
# If temperature is dependent on mass, how did the early universe have a temperature?

I am near the end of the book The First Three Minutes by Steven Weinberg. I am reading it in order to get a better picture of the early universe in the Big Bang model. But one thing I am having trouble with and wanted to get some clarity. Namely:

Question

What is meant by "temperature" if the very early universe in the Big Bang model was mostly radiation, or pre-inflation, something more basic than radiation?

Explanation

Below I will try to explain why I am getting confused. Please let me know if it needs clarification (but that's also part of the question, I haven't been able to clearly understand temperature in this context yet).

Since the temperature equation is about the movement of massive particles:

$$T = {m\overline{v^2}\over 3 k_B}$$

How do you think of temperature when there is no mass, or when there is mostly radiation in the very early universe? Here is where I get confused.. Radiation is moving at the speed of light $c$, and since the speed of light is constant, and since radiation doesn't have mass, it seems like the universe would be "maximally" hot (particles moving at the speed of light). But intuitively that's not true because the temperature formula requires mass. But then if the early universe was mostly radiation, how does it have a temperature? If mass slowly increased to be the majority ~400,000 years after the Big Bang, it seems the temperature would increase rather than decrease. But then taking into account the expansion, it makes sense that even if more matter was forming, the temperature would decrease. But how could it decrease if there was less matter before?

To summarize, in trying to pinpoint temperature, I seem to be stuck in cyclic reasoning and am looking for a better way of understanding it.

I am working toward learning the purely mathematical form of these theories, but have a ways to go. For now, looking for a description to broadly make sense of "temperature" during the evolution of the few moments after the Big Bang.

Examples of how "temperature" is used

In the book above, Steven Weinberg talks a lot about the temperature and radiation of the early universe, and how the early universe was dominated by photons/radiation, and ~400,000 years later became dominated by mass-particles when the temperature cooled.

Some other quotes from the web include:

In the first moments after the big bang, the universe was extremely hot and dense. As the universe cooled, conditions became just right to give rise to the building blocks of matter – the quarks and electrons of which we are all made. (CERN)

...the temperature was so high that the four fundamental forces ... were one fundamental force. (Planck Epoch, Wikipedia)

...the temperature of the universe was still too high to allow quarks to bind together to form hadrons. (Quark Epoch, Wikipedia)

• – HDE 226868 Jul 12 '15 at 20:18
• Maybe need to ask a different question Lance. Because a) it's tricky to respond to a question about a pop-science book, b) "all ideas concerning the very early universe are speculative" and c) the speed of light isn't constant, see Einstein talking about it here. At the black hole event horizon we say the coordinate speed of light is zero, and we liken the early universe to a black hole. So you could argue that in the early universe the speed of light was zero. How does that have a temperature? – John Duffield Jul 12 '15 at 20:40
• What's the temperature of the surface of the sun? We can't stick a thermometer in the sun, but we can measure the temperature by the radiation the sun puts off. Likewise there were no thermometers in the early universe, but there was energy, and we can model the energy density and thus a temperature. – Asher Jul 12 '15 at 21:51

Really, your confusion is rooted in the fact that the equation you give relating temperature and particle mass assumes that the thing's whose temperature you're predicting is a gas of particular matter that obeys Newtonian mechanics.

Light, and really anything in the early universe, decidedly does not obey Newtonian mechanics. There are several ways one can arrive at a more fundamental definition of temperature, but at root, it amounts to "two systems are at the same temperature if they don't exchange energy when you put them into contact with each other. If they do exchange energy, the energy flow is from the hotter system to the colder system."

Mathematically, this can be made more precise if you introduce a definition of entropy and energy, but for the purposes of this question, it's good enough to say that you can define the temperature of radiation using this second definition combined with some knowledge of how radiation behaves (in particular, the knowledge that it's pressure is equal to its density).

The way to understand this is as follows.

Assume that the early-universe is radiation-dominated, and additionally assume that the early universe is of FLRW type with a single fluid that obeys a barotropic equation of state

$p = w \mu$ where:

• $w = 1/3$ for radiation
• $w=0$ for dust
• $w=1$ for a stiff fluid, etc...

Now, the Einstein field equations imply the Friedmann equation:

$\frac{\dot{a}^2}{a^2} = \frac{\mu}{3} - \frac{k}{a^2}$.

Note that $k=1,0,-1$ depending on whether the type of FLRW model you are considering is spatially closed, flat, or open.

Now, one integrates the Friedmann equation above and uses the Stefann-Boltzmann law

$\mu \approx T^4$ (This is incidentally how you get around the temperature-mass dependency from standard old thermodynamics)

to obtain the temperature of the universe.

Just for your information, if you take the specific example of a radiation-dominated early universe, going through the above calculations, we get something like:

$T \approx t^{-1/2}$.

That is, as the universe ages, the temperature falls, which is consistent with observations.

UPDATED: An example of how to understand this is as follows:

I will consider for simplicity a spatially flat spatially homogeneous and isotropic universe, i.e., a $k=0$ FLRW model. Now, for such universes, Einstein's field equations give: The Raychaudhuri equation:

$\dot{\theta} + \frac{1}{3}\theta^2 + \frac{1}{2} \left(\mu + 3p\right) = 0$,

The local energy conservation equation:

$\dot{\mu} + \theta \left(\mu + p\right) = 0$,

and

The Friedmann equation: $\frac{1}{3} \theta^2 = \mu$.

Note that in these equations, $\theta$ is the expansion scalar and is typically denoted as $\frac{\dot{a}}{a}$, where $a(t)$ is the universe scale factor, while $\mu$ and $p$ are the matter energy density and pressure respectively. As mentioned above, we will assume that the matter in the universe obeys a barotropic equation of state, such that $p = w \mu$, where $-1 \leq w \leq 1$.

Now, substituting $p = w\mu$ into the above equations, we see that we have the Raychaudhuri equation:

$\dot{\theta} + \frac{1}{3}\theta^2 + \frac{1}{2} \left(\mu + 3w \mu\right) = 0$,

The energy conservation equation: $\dot{\mu} + \theta \left(\mu + w \mu\right) = 0$.

We can use the Friedmann equation to solve for $\theta$, and substitute this into the conservation equation above to obtain:

$\boxed{\dot{\mu} = -\sqrt{3} \left(1+w\right) \mu^{3/2}}$.

Solving this ODE with initial condition $\mu(0) = c$, we obtain the solution: $\boxed{\mu(t) = \frac{4 c}{3 c t^2 (w+1)^2-4 \sqrt{3} \sqrt{c} t (w+1)+4}}$.

Now, from the Stefann-Boltzmann law, we have the relationship that: $\mu \approx T^4 \Rightarrow T \approx \mu^{1/4}$, so that we may write

$\boxed{T \approx \sqrt{2} \left[{\frac{c}{3 c t^2 (w+1)^2-4 \sqrt{3} \sqrt{c} t (w+1)+4}}\right]^{1/4}}$.

I have plotted this function for some values of $w$ and over a range of initial conditions. In particular, for a radiation-dominated universe $w=1/3$:

For $w = 0$, which is pressureless dust:

Either way, you see that, as time goes on and as the universe expands, the temperature falls, but this is the mathematics in detail for one such example.

Note that observations actually show that we live in a universe that is close to $k=0$ FLRW, so this example is actually quite relevant!!

I hope this helps.

• This looks interesting but I am not super familiar with the math yet. Mind elaborating on some of the parts? $μ$ in $p=wμ$, and showing how to integrate the Friedmann equation (or linking to example). – Lance Pollard Jul 12 '15 at 20:48
• Hi. Yes, I have attached an example calculation that goes into some details above. – Dr. Ikjyot Singh Kohli Jul 12 '15 at 21:03


text:
# Conditional probability question - given pdf

I am given that X is a continuous variable with the pdf, $f(x)= 2 e^{-2x}$ if $x \geq 0$ and asked to find the $P(X \geq 3 | X \geq 2)$.

I am having trouble with the intuition behind the answer, my understanding is that the conditional probability could be written as:

$P(X \geq 3 \cap X \geq 2)/P(X \geq 2)$

Wouldn't the intersection simply be the smaller set, namely $P(X \geq 2)$? But this doesn't make sense because then all you would have is:

$P(X\geq 2)/P(X \geq 2)$

Answers I am seeing online give the probability by:

$P(X \geq 3)/P(X \geq 2)$

But I don't understand how you arrive there. I feel like the answer is probably obvious and I'm just missing it.

Thank you.

• The event $X \gt 2$ is strictly larger than the event $X \gt 3$, not smaller.
– whuber
Sep 13, 2016 at 18:16
• oh duh - thank you so much, this was really bugging me!
– Edna
Sep 13, 2016 at 18:17
• I think your question needs a self-study tag as this is clearly an exercise you need to solve by yourself. Sep 13, 2016 at 19:15
• It is worth reading our wiki to see how we handle self-study questions - this is just the type of question that the self-study tag is designed for. Sep 13, 2016 at 19:17

The distribution is an exponential distribution:

$f(x;\lambda) = \lambda e^{-\lambda x } x \ge 0,$

with parameter $\lambda=2$.

An exponentially distributed random variable $x$ is a memory-less probability distributions that obeys the relation

$P( x > s + t | x > s) = P(x> t), \qquad \forall s, t \ge 0$.

Intuitively this means that if you have waited $s$ seconds for a event the probability that that event will occur $t$ seconds later is the same as if you spent no time waiting i.e $s=0$.

This is simple to demonstrate:

$P( x > s + t | x > s)=\frac{P( x > s + t ,x>s)}{P( x > s)},$

because $x > s + t>s$ then $P( x > s + t ,x>s)=P( x > s + t)$ we simplifies he expression to:

$\frac{P( x > s + t ,x>s)}{P( x > s)}=\frac{P( x > s + t)}{P( x > s)}$

$=\frac{1-P( x < s + t)}{1-P( x < s)} (1)$

where the term $P( x < u)=1-e^{-\lambda u } (2)$ and $P( x >u)=e^{-\lambda u } (3)$

substituting (2) into (1) you get :

$\frac{1-P( x < s + t)}{1-P( x < s)}=\frac{1-(1-e^{-\lambda (s + t)}) }{1-(1-e^{-\lambda s )})}=\frac{e^{-\lambda (s + t)} }{e^{-\lambda s }} =e^{-\lambda (t)}$

This expression is the same as (3) i.e $P(x>t)$.


text:
# A calculus of the absurd

### Chapter 6 Proof

We must not believe those, who today, with philosophical bearing and deliberative tone, prophesy the fall of culture and accept the ignorabimus [that we cannot know whether something is true or false]. For us there is no ignorabimus, and in my opinion none whatever in natural science. In opposition to the foolish ignorabimus our slogan shall be “Wir müssen wissen - wir werden wissen” [we must know — we will know]

— David Hilbert, RADIO ADDRESS IN 1930

The human mind is incapable of formulating (or mechanizing) all its mathematical intuitions, i.e., if it has succeeded in formulating some of them, this very fact yields new intuitive knowledge, e.g., the consistency of this formalism. This fact may be called the “incompletability” of mathematics. On the other hand, on the basis of what has been proved so far, it remains possible that there may exist (and even be empirically discoverable) a theorem proving machine which in fact is equivalent to mathematical intuition, but cannot be proved to be so, nor even be proved to yield only correct theorems of finitary number theory

Kurt Gödel, REMARKS ON HIS INCOMPLETENESS THEOREMS

#### 6.1 Direct proof

A direct proof is where we show that a statement in the form “if $$A$$, then $$B$$” is true by assuming that $$A$$ is true, and showing that $$B$$ must therefore also be true.

• Example 6.1.1 Show that if $$k$$ is an odd positive integer, then $$k^2$$ is also odd.

We start by assuming that $$k$$ is an odd positive integer. To do this, we want to find an algebraic way to represent $$k$$. As every odd number can be written in the form4343 This can be proved by induction (see Section 6.3) $$2p + 1$$ for suitable $$p$$ (e.g. $$3 = 2\cdot 1 + 1$$, $$5 = 2\cdot 2 + 1$$, etc.) we can write $$k$$ as $$2p + 1$$. From there we apply our assumption to show that the result is true

\begin{align} (2p + 1)^2 &= 4p^2 + 4p + 1 \\ &= 2(2p^2 + 2p) + 1 \end{align}

As $$2(2p^2 + 2p) + 1$$ is in the form4444 It’s not the specific expression or variable names which we chose that are important here - it’s the overarching structure of the expression - i.e. that it’s in the form $$2 \cdot \text {something} + 1$$ which is important. $$2x + 1$$, we have shown that this is true.

$$\Box$$


text:
## Final notes for ECE1254, Modelling of Multiphysics Systems

I’ve now finished my first grad course, Modelling of Multiphysics Systems, taught by Prof Piero Triverio.

I’ve posted notes for lectures and other material as I was taking the course, but now have an aggregated set of notes for the whole course posted.
This is now updated with all my notes from the lectures, solved problems, additional notes on auxillary topics I wanted to explore (like SVD), plus the notes from the Harmonic Balance report that Mike and I will be presenting in January.

This version of my notes also includes all the matlab figures regenerating using http://www.mathworks.com/matlabcentral/fileexchange/23629-export-fig, which allows a save-as pdf, which rescales much better than Matlab saveas() png’s when embedded in latex.  I’m not sure if that’s the best way to include Matlab figures in latex, but they are at least not fuzzy looking now.

All in all, I’m pretty pleased with my notes for this course.  They are a lot more readable than any of the ones I’ve done for the physics undergrad courses I was taking (http://peeterjoot.com/writing/).  While there was quite a lot covered in this course, the material really only requires an introductory circuits course and some basic math (linear algebra and intro calculus), so is pretty accessible.

This was a fun course.  I recall, back in ancient times when I was a first year student, being unsatisfied with all the ad-hoc strategies we used to solve circuits problems.  This finally answers the questions of how to tackle things more systematically.

Here’s the contents outline for these notes:

Preface
Lecture notes
1 nodal analysis
1.1 In slides
1.2 Mechanical structures example
1.3 Assembling system equations automatically. Node/branch method
1.4 Nodal Analysis
1.5 Modified nodal analysis (MNA)
2 solving large systems
2.1 Gaussian elimination
2.2 LU decomposition
2.3 Problems
3 numerical errors and conditioning
3.1 Strict diagonal dominance
3.2 Exploring uniqueness and existence
3.3 Perturbation and norms
3.4 Matrix norm
4 singular value decomposition, and conditioning number
4.1 Singular value decomposition
4.2 Conditioning number
5 sparse factorization
5.1 Fill ins
5.2 Markowitz product
5.3 Markowitz reordering
5.4 Graph representation
6.1 Summary of factorization costs
6.2 Iterative methods
6.4 Recap: Summary of Gradient method
6.6 Full Algorithm
6.7 Order analysis
6.9 Gershgorin circle theorem
6.10 Preconditioning
6.11 Symmetric preconditioning
6.13 Problems
7 solution of nonlinear systems
7.1 Nonlinear systems
7.2 Richardson and Linear Convergence
7.3 Newton’s method
7.4 Solution of N nonlinear equations in N unknowns
7.5 Multivariable Newton’s iteration
7.6 Automatic assembly of equations for nonlinear system
7.7 Damped Newton’s method
7.8 Continuation parameters
7.9 Singular Jacobians
7.10 Struts and Joints, Node branch formulation
7.11 Problems
8 time dependent systems
8.1 Assembling equations automatically for dynamical systems
8.2 Numerical solution of differential equations
8.3 Forward Euler method
8.4 Backward Euler method
8.5 Trapezoidal rule (TR)
8.6 Nonlinear differential equations
8.7 Analysis, accuracy and stability (Dt ! 0)
8.8 Residual for LMS methods
8.9 Global error estimate
8.10 Stability
8.11 Stability (continued)
8.12 Problems
9 model order reduction
9.1 Model order reduction
9.2 Moment matching
9.3 Model order reduction (cont).
9.4 Moment matching
9.5 Truncated Balanced Realization (1000 ft overview)
9.6 Problems
Final report
10 harmonic balance
10.1 Abstract
10.2 Introduction
10.2.1 Modifications to the netlist syntax
10.3 Background
10.3.1 Discrete Fourier Transform
10.3.2 Harmonic Balance equations
10.3.3 Frequency domain representation of MNA equations
10.3.4 Example. RC circuit with a diode.
10.3.5 Jacobian
10.3.6 Newton’s method solution
10.3.7 Alternative handling of the non-linear currents and Jacobians
10.4 Results
10.4.1 Low pass filter
10.4.2 Half wave rectifier
10.4.3 AC to DC conversion
10.4.4 Bridge rectifier
10.4.5 Cpu time and error vs N
10.4.6 Taylor series non-linearities
10.4.7 Stiff systems
10.5 Conclusion
10.6 Appendices
10.6.1 Discrete Fourier Transform inversion
Appendices
a singular value decomposition
b basic theorems and definitions
c norton equivalents
d stability of discretized linear differential equations
e laplace transform refresher
f discrete fourier transform
g harmonic balance, rough notes
g.1 Block matrix form, with physical parameter ordering
g.2 Block matrix form, with frequency ordering
g.3 Representing the linear sources
g.4 Representing non-linear sources
g.5 Newton’s method
g.6 A matrix formulation of Harmonic Balance non-linear currents
h matlab notebooks
i mathematica notebooks
Index
Bibliography

## Stability of discretized linear differential equations

In class today was a highlight of stability methods for linear multistep methods. To motivate the methods used, it is helpful to take a step back and review stability concepts for LDE systems.

By way of example, consider a second order LDE homogeneous system defined by

\label{eqn:stabilityLDEandDiscreteTime:20}
\frac{d^2 x}{dt^2} + 3 \frac{dx}{dt} + 2 = 0.

Such a system can be solved by assuming an exponential solution, say

\label{eqn:stabilityLDEandDiscreteTime:40}
x(t) = e^{s t}.

Substitution gives

\label{eqn:stabilityLDEandDiscreteTime:60}
e^{st} \lr{ s^2 + 3 s + 2 } = 0,

The polynomial part of this equation, the characteristic equation has roots $$s = -2, -1$$.

The general solution of \ref{eqn:stabilityLDEandDiscreteTime:20} is formed by a superposition of solutions for each value of $$s$$

\label{eqn:stabilityLDEandDiscreteTime:80}
x(t) = a e^{-2 t} + b e^{-t}.

Independent of any selection of the superposition constants $$a, b$$, this function will not blow up as $$t \rightarrow \infty$$.

This “stability” is due to the fact that both of the characteristic equation roots lie in the left hand Argand plane.

Now consider a discretized form of this LDE. This will have the form

\label{eqn:stabilityLDEandDiscreteTime:100}
\begin{aligned}
0 &=
\inv{\lr{\Delta t}^2}
\lr{ x_{n+2} – 2 x_{n-1} + x_n } + \frac{3}{\Delta t} \lr{ x_{n+1} – x_n } + 2
x_n \\
&=
x_{n+2} \lr{
\inv{\lr{\Delta t}^2}
}
+
x_{n+1} \lr{
\frac{3}{\Delta t}
-\frac{2}{\lr{\Delta t}^2}
}
+
x_{n} \lr{
\frac{1}{\lr{\Delta t}^2}
-\frac{3}{\Delta t}
+ 2
},
\end{aligned}

or

\label{eqn:stabilityLDEandDiscreteTime:220}
0
=
x_{n+2}
+
x_{n+1} \lr{
3 \Delta t – 2
}
+
x_{n} \lr{
1 – 3 \Delta t + 2 \lr{ \Delta t}^2
}.

Note that after discretization, each subsequent index corresponds to a time shift. Also observe that the coefficients of this discretized equation are dependent on the discretization interval size $$\Delta t$$. If the specifics of those coefficients are ignored, a general form with the following structure can be observed

\label{eqn:stabilityLDEandDiscreteTime:120}
0 =
x_{n+2} \gamma_0
+
x_{n+1} \gamma_1
+
x_{n} \gamma_2.

It turns out that, much like the LDE solution by characteristic polynomial, it is possible to attack this problem by assuming a solution of the form

\label{eqn:stabilityLDEandDiscreteTime:140}
x_n = C z^n.

A time shift index change $$x_n \rightarrow x_{n+1}$$ results in a power adjustment in this assumed solution. This substitution applied to \ref{eqn:stabilityLDEandDiscreteTime:120} yields

\label{eqn:stabilityLDEandDiscreteTime:160}
0 =
C z^n
\lr{
z^{2} \gamma_0
+
z \gamma_1
+
1 \gamma_2
},

Suppose that this polynomial has roots $$z \in \{z_1, z_2\}$$. A superposition, such as

\label{eqn:stabilityLDEandDiscreteTime:180}
x_n = a z_1^n + b z_2^n,

will also be a solution since insertion of this into the RHS of \ref{eqn:stabilityLDEandDiscreteTime:120} yields

\label{eqn:stabilityLDEandDiscreteTime:200}
a z_1^n
\lr{
z_1^{2} \gamma_0
+
z_1 \gamma_1
+
\gamma_2
}
+
b
z_2^n
\lr{
z_2^{2} \gamma_0
+
z_2 \gamma_1
+
\gamma_2
}
=
a z_1^n \times 0
+b z_2^n \times 0.

The zero equality follows since $$z_1, z_2$$ are both roots of the characteristic equation for this discretized LDE.
In the discrete $$z$$ domain stability requires that the roots satisfy the bound $$\Abs{z} < 1$$, a different stability criteria than in the continuous domain. In fact, there is no a-priori guarantee that stability in the continuous domain will imply stability in the discretized domain. Let's plot those z-domain roots for this example LDE, using $$\Delta t \in \{ 1/2, 1, 2 \}$$. The respective characteristic polynomials are $$\label{eqn:stabilityLDEandDiscreteTime:260} 0 = z^2 - \inv{2} z = z \lr{ z - \inv{2} }$$ $$\label{eqn:stabilityLDEandDiscreteTime:240} 0 = z^2 + z = z\lr{ z + 1 }$$ $$\label{eqn:stabilityLDEandDiscreteTime:280} 0 = z^2 + 4 z + 3 = (z + 3)(z + 1).$$ These have respective roots $$\label{eqn:stabilityLDEandDiscreteTime:300} z = 0, \inv{2}$$ $$\label{eqn:stabilityLDEandDiscreteTime:320} z = 0, -1$$ $$\label{eqn:stabilityLDEandDiscreteTime:340} z = -1, -3$$ Only the first discretization of these three yields stable solutions in the z domain, although it appears that $$\Delta t = 1$$ is right on the boundary.

## ECE1254H Modeling of Multiphysics Systems. Lecture 16: LMS systems and stability. Taught by Prof. Piero Triverio

November 17, 2014 ece1254 No comments , , ,

## Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

## Residual for LMS methods

### Mostly on slides:

12_ODS.pdf

Residual is illustrated in fig. 1, assuming that the iterative method was accurate until $$t_{n}$$

fig. 1. Residual illustrated

### Summary

• [FE]: $$R_{n+1} \sim \lr{ \Delta t}^2$$. This is of order $$p = 1$$.
• [BE]: $$R_{n+1} \sim \lr{ \Delta t}^2$$. This is of order $$p = 1$$.
• [TR]: $$R_{n+1} \sim \lr{ \Delta t}^3$$. This is of order $$p = 2$$.
• [BESTE]: $$R_{n+1} \sim \lr{ \Delta t}^4$$. This is of order $$p = 3$$.

## Global error estimate

Suppose $$t \in [0, 1] s$$, with $$N = 1/{\Delta t}$$ intervals. For a method with local error of order $$R_{n+1} \sim \lr{ \Delta t}^2$$ the global error is approximately $$N R_{n+1} \sim \Delta t$$.

## Stability

Recall that a linear multistep method (LMS) was a system of the form

\label{eqn:multiphysicsL16:20}
\sum_{j=-1}^{k-1} \alpha_j x_{n-j} = \Delta t \sum_{j=-1}^{k-1} \beta_j f( x_{n-j}, t_{n-j} )

Consider a one dimensional test problem

\label{eqn:multiphysicsL16:40}
\dot{x}(t) = \lambda x(t)

where as in fig. 2, $$\Re(\lambda) < 0$$ is assumed to ensure stability.

fig. 2. Stable system

Linear stability theory can be thought of as asking the question: “Is the solution of \ref{eqn:multiphysicsL16:40} computed by my LMS method also stable?”

Application of \ref{eqn:multiphysicsL16:20} to \ref{eqn:multiphysicsL16:40} gives

\label{eqn:multiphysicsL16:60}
\sum_{j=-1}^{k-1} \alpha_j x_{n-j} = \Delta t \sum_{j=-1}^{k-1} \beta_j \lambda x_{n-j},

or
\label{eqn:multiphysicsL16:80}
\sum_{j=-1}^{k-1} \lr{ \alpha_j – \Delta \beta_j \lambda }
x_{n-j} = 0.

With

\label{eqn:multiphysicsL16:100}
\gamma_j = \alpha_j – \Delta \beta_j \lambda,

this expands to
\label{eqn:multiphysicsL16:120}
\gamma_{-1} x_{n+1}
+
\gamma_{0} x_{n}
+
\gamma_{1} x_{n-1}
+
\cdots
+
\gamma_{k-1} x_{n-k} .

This can be seen as a

• discrete time system
• FIR filter

The numerical solution $$x_n$$ will be stable if \ref{eqn:multiphysicsL16:120} is stable.

A characteristic equation associated with \ref{eqn:multiphysicsL16:120} can be defined as

\label{eqn:multiphysicsL16:140}
\gamma_{-1} z^k
+
\gamma_{0} z^{k-1}
+
\gamma_{1} z^{k-2}
+
\cdots
+
\gamma_{k-1} = 0.

This is a polynomial with roots $$z_n$$ (poles). This is stable if the poles satisfy $$\Abs{z_n} < 1$$, as illustrated in fig. 3

Stability

Observe that the $$\gamma’s$$ are dependent on $$\Delta t$$.

FIXME: There’s a lot of handwaving here that could use more strict justification. Check if the text covers this in more detail.

### Example: Forward Euler stability

For $$k = 1$$ step.

\label{eqn:multiphysicsL16:180}
x_{n+1} – x_n = \Delta t f( x_n, t_n ),

the coefficients are $$\alpha_{-1} = 1, \alpha_0 = -1, \beta_{-1} = 0, \beta_0 =1$$. For the simple function above

\label{eqn:multiphysicsL16:200}
\gamma_{-1} = \alpha_{-1} – \Delta t \lambda \beta_{-1} = 1

\label{eqn:multiphysicsL16:220}
\gamma_{0} = \alpha_{0} – \Delta t \lambda \beta_{0} = -1 – \Delta t \lambda.

The stability polynomial is

\label{eqn:multiphysicsL16:240}
1 z + \lr{ -1 – \Delta t \lambda} = 0,

or

\label{eqn:multiphysicsL16:260}
\boxed{
z = 1 + \delta t \lambda.
}

This is the root, or pole.

For stability we must have

\label{eqn:multiphysicsL16:280}
\Abs{ 1 + \Delta t \lambda } < 1,

or
\label{eqn:multiphysicsL16:300}
\Abs{ \lambda – \lr{ -\inv{\Delta t} } } < \inv{\Delta t},

This inequality is illustrated roughly in fig. 4.

fig. 4. Stability region of FE

All poles of my system must be inside the stability region in order to get stable $$\gamma$$.


text:
American Institute of Mathematical Sciences

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October  2017, 10(5): 1165-1174. doi: 10.3934/dcdss.2017063

Stability and bifurcation analysis in a chemotaxis bistable growth system

 Y. Y. Tseng Functional Analysis Research Center and School of Mathematical Sciences, Harbin Normal University, Harbin, Heilongjiang 150025, China

* Corresponding author: Ping Liu

Received  September 2016 Revised  January 2017 Published  June 2017

Fund Project: Partially supported by NSFC grant 11571086,11471091 and Science Research Funds for Overseas Returned Chinese Scholars of Heilongjiang Province LC2013C01.

The stability analysis of a chemotaxis model with a bistable growth term in both unbounded and bounded domains is studied analytically. By the global bifurcation theorem, we identify the full parameter regimes in which the steady state bifurcation occurs.

Citation: Shubo Zhao, Ping Liu, Mingchao Jiang. Stability and bifurcation analysis in a chemotaxis bistable growth system. Discrete & Continuous Dynamical Systems - S, 2017, 10 (5) : 1165-1174. doi: 10.3934/dcdss.2017063
References:

show all references

References:
Basic phase portrait of (2)
Parameter space for Turing instability. The parameter values are $d_1=d_2=f=g=1, \nu=\frac{1}{4},$$M=\frac{(\sqrt{d_1g}+\sqrt{(1-\nu)d_2})^2}{f}$
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text:
# Meta-Reinforcement Learning¶

Warning

Meta-RL results are particularly finicky to compare. Different papers use different environment implementations, which in turn produce different convergence and rewards. The plots below only serve to indicate what kind of performance you can expect with learn2learn.

## MAML¶

The above results are obtained by running maml_trpo.py on HalfCheetahForwardBackwardEnv and AntForwardBackwardEnv for 300 updates. The figures show the expected sum of rewards over all tasks. The line and shadow are the mean and standard deviation computed over 3 random seeds.

Info

Those results were obtained in August 2019, and might be outdated.