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Background
Recently, I have been writing up some notes on derived functors and I came across the Eilenberg-Watts theorems [1], which essentially explain why it is hard to find derived functors besides Ext and Tor. Before asking my question, allow me to briefly state these theorems.
Let $R$ and $S$ be rings and $M$ and $S-R$ bimodule. A basic property of the functor $M\otimes_R-:R-\mathrm{Mod}\to S-\mathrm{Mod}$ is that it is an additive covariant right-exact functor. In fact, it also commutes with direct sums. Curiously these properties are enough to characterize it: any covariant additive $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ that is right-exact and commutes with direct sums is in fact naturally equivalent to some $M\otimes_R -$ for some $S-R$ bimodule $M$. This is the statement of the Eilenberg-Watts theorem for tensor functors.
For completeness, I should state the other version for Hom. If $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ is an additive left-exact contravariant functor which converts direct sums into direct products (i.e. $T(\oplus M_i) \cong \prod T(M_i)$) then there is an $R-S$ bimodule $M$ such that $T$ is naturally equivalent to $\mathrm{Hom}_{R}(-,M)$.
Finally, if $T:R-\mathrm{Mod}\to \mathbb{Z}-\mathrm{Mod}$ is left-exact covariant that commutes with inverse limits then there is a left $R$-module $M$ such that $T$ is naturally equivalent to $\mathrm{Hom}_{R}(M,-)$.
The Prompting for the Question
From the above, any functor satisfying the hypotheses of these Eilenberg-Watts type theorems are going to be naturally equivalent to either a tensor or a Hom, and thus its derived functors will just be Tor or Ext respectively. For instance, if $G$ is a group then $G-\mathrm{Mod}\to \mathbb{Z}-\mathrm{Mod}$ given by $A\mapsto A_G$, where $A_G$ is the quotient of $A$ by the submodule generated by $ga - a$ for all $g\in G$ and $a\in A$ is just the usual coinvariant functor, whose left derived functors are the homology groups $H^i(G,A)$. By Eilenberg-Watts, $-_G$ must be equivalent to some tensor functor, and in fact it is easy to prove that $A\mapsto A_G$ is naturally equivalent to $A\mapsto \mathbb{Z}\otimes_{\mathbb{Z}G}A$
Incidentally, the proof given by C.E. Watts is explicit enough so that the above natural equivalence is apparent.
The Question
Notice that in each of these the hypothesis of playing nice with limits is required. I am actually interested in functors which do
not play nicely with limits. For instance,
What are some examples of covariant right-exact functors $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ that do
notcommute with all direct sums? [Edit: $T$ also should not be left exact in this case.]
Such a $T$ of course cannot be a left-adjoint for otherwise it would commute with direct sums. Such a $T$ could be interesting because its left-derived functors may not be "like" the Tor functor. The question also goes for dropping the playing-nice-with-limit hypotheses in the other forms of the theorem. I tried a Google search but could not seem to find anything relevant.
Since I am asking for a list of examples, I have made this a community wiki. Thanks!
[1] Watts, "Intrinsic Characterizations of Some Additive Functors". Proceedings of the American Mathematical Society, Vol. 11, No. 1 (Feb., 1960), pp. 5-8
Addendum (edit)
Thanks everyone for their answers; I think I should have been more precise and asked a question more along the lines of:
What are some derived functors that are not Ext or Tor?
Which I believe some of the existing answers are. In essence I wanted examples that were neither tensors nor Homs in disguise...
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Before I’ll present the exposition of this theory, I’ll speak a little bit about the Mereological concept it is meant to catpure.
The idea is to work in Atomic General Extensional Mereology
“AGEM”, one can think of it easily as a theory about collections of atoms, where atoms are indivisible objects, i.e. objects that do not have proper parts. The relation is an atomic part of is defined as:
$ \sf Definition:$ $ x P^a y \iff atom(x) \land x P y$ .
where $ P$ stands for “is a part of”, and atom(x) is defined as:
$ atom(x) \iff \not \exists y (y P x \land y \neq x)$
This atomic part-hood relation can be regarded, conceptually speaking, as an instance of set membership relation.
Now the following theory is a try to define a
set theory by a strategy of mimicking properties of this atomic part-hood relation with the background theory being AGEM.
Notation: let $ \phi^{P^a}$ denote a formula that only use the binary relation $ P^a$ or otherwise the equality relation, as predicate symbols. The notation $ \phi^{\in|P^a}$ denotes the formula obtained by merely replacing each occurrence of the symbol $ “P^a”$ in $ \phi^{P^a}$ by the symbol $ “\in”$ .
Comprehension axiom schema: if $ \phi^{\in|P^a}(y)$ doesn’t have the symbol $ x$ occurring free, then all closures of:
$ $ \forall A [\exists x \forall y (y \ P^a \ x \leftrightarrow \phi^{P^a}(y)) \to \exists x \forall y (y \in x \leftrightarrow \phi^{\in|P^a}(y))]$ $
are axioms.
In order to complete this theory we add axioms of Extensionality, Empty set and Singletons:
Extensionality: $ \forall xy [\forall z (z \in x \leftrightarrow z \in y) \to x=y]$ .
Empty set: $ \exists x \forall y (y \not \in x)$
Singletons:: $ \forall A \exists x \forall y (y \in x \leftrightarrow y=A)$
This theory has a
universal set, also has a set of all sets that are in themselves, however it doesn’t have complements; axioms of Set union and Power are there. There are separation axioms for formulas $ \phi^{\in|P^a}$ where $ \phi^{P^a}$ holds of at least one object. Similarily replacement axioms are granted if $ \phi^{P^a}$ formula replace atoms with atoms and of course is non empty.
The trick is that all formulas of the form $ x \not \in x$ , $ \exists x_1,..,x_n: \neg (x_1 \in x_2 \land…\land x_n \in x_1)$ ; $ x \text { is well founded }$ , $ x \text{ is a von Neumann ordinal }$ , etc.. all of those won’t have their $ \phi^{P^a}$ corresponding formulas hold of mereological atoms and so cannot be used in comprehension because there do not exist an object that has no atomic parts, since we are already working in
AGEM.
Question: if one attempts to prove the consistency of this theory, which of the known alternative set theories have in some sense the nearest structure to this theory, other than positive set theory?
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But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty).
And Chrome has a Personal Blocklist extension which does what you want.
: )
Of course you already have a Google account but Chrome is cool : )
Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies?
do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created.
@QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value.
I have problem with showing that the limit of the following function$$\frac{\sqrt{\frac{3 \pi}{2n}} -\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$equal to $1$, with $n \to \infty$.
@QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0.
@KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc.
In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people
in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results
@QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O
@NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that.
@NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment.
@QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h).
@KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$ by picking some correct L (somehow)
Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
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Impedance matching is tricky, but the role of a quarter wave transmission line is to map from one impedance to another. The actual impedance of the line will not match either the input or the output impedance - this is entirely expected.
However at a given frequency, when a correctly designed quarter wave line is inserted with the correct impedance, the output impedance will appear to the input as perfectly matched. In your case, the transformer will make the \$20\Omega\$ impedance appear as if it is a \$100\Omega\$ impedance meaning no mismatch. Essentially it guides the waves from one characteristic impedance to another.
The easiest way to visualise this is on a Smith chart, plot the two points 0.4 (\$20\Omega\$) and 2 (\$100\Omega\$). Then draw a circle centred on the resistive/real axis (line down the middle) which intersects both points. You will find that this point is located at 0.894 (\$44.7\Omega\$) if your calculations are correct. This is shown below at \$500\mathrm{MHz}\$, but the frequency is only important when converting the electrical length to a physical length.
What a quarter wave transformer does is rotate a given point by \$180^\circ\$ around its characteristic impedance on the Smith chart (that's \$\lambda/4 = 90^\circ\$ forward plus \$90^\circ\$ reverse).
Exactly why it does this is complex. But the end result of a long derivation is that for a transmission line of impedance \$Z_0\$ connected to a load of impedance \$Z_L\$ and with a length \$l\$, then the impedance at the input is given by:
$$Z_{in}=Z_0\frac{Z_L+jZ_0\tan\left(\beta l\right)}{Z_0+jZ_L\tan\left(\beta l\right)}$$
That is an ugly equation, but it just so happens if the electrical length \$\beta l\$ is \$\lambda/4\$ (\$90^\circ\$), the \$\tan\$ part goes to infinity which allows the equation to be simplified to:
$$Z_{in}=Z_0\frac{Z_0}{Z_L}=\frac{(Z_0)^2}{Z_L}\rightarrow Z_0=\sqrt{\left(Z_{in}Z_L\right)}$$
Which is where your calculation comes from.
With the quarter wave transformer in place, the load appears as matched to the source. In other words, the transformer matches
both of its interfaces, not just the input end.
You can also see from this equation why the transformer only works for a single frequency - because it relies on the physical length being \$\lambda/4\$. You can actually (generally using advanced design tools) achieve an approximate match over a range of frequencies - basically a close enough but not exact match.
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Step 1: Note down the given values
Step 2: Set up the formula for arc length.
NOTE: The formula is arc length= 2 \pi (r)(\frac{\theta }{360}) ,
where {\displaystyle r} equals the radius of the circle and {\displaystyle \theta } equals the measurement of the arc’s central angle, in degrees.
or
Arc length = r * \theta
Step 3: Plug the length of the circle’s radius into the formula.
Step 4: Plug the value of the arc’s central angle into the formula.
Step 5: Simplify the equation to find the arc length
NOTE: Use multiplication and division to simplify the equation.
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Given my blog title, how have I gone this long without discussing triangle geometry? I will rectify this gross negligence in the next few weeks. Let’s begin with a seemingly impossible fact due to Frank Morley.
Start with a triangle
ABC, with any shape you wish. Now cut its angles into three equal parts, and extend these angle trisectors until they meet in pairs as illustrated below, forming triangle PQR. Morley’s amazing theorem says that this Morley triangle, PQR, will always be equilateral!
Why would this be true? Well, Morley’s theorem tells us that this diagram has three nice 60-degree angles in the middle, but we may suspect that, in fact,
all of the angles are nice! This key insight lets us piece together the following argument, where we build up the diagram backwards from its constituent pieces. Draw seven separate “puzzle piece” triangles with angles and side-lengths as shown below, where the original triangle ABC has angles \(3\alpha,3\beta,3\gamma\) respectively. (Be sure to check that puzzle pieces with these specifications actually exist! Hint: use the Law of Sines.)
Now, fit the pieces together: all matching edge-lengths are equal (by design), and the angles around vertex
P add up to \(60+(\gamma+60)+(\alpha+120)+(\beta+60)=360\) and similarly for Q and R, so the puzzle fits together into a triangle similar to our original triangle ABC. But now these pieces must make up the Morley configuration, and since we started with an equilateral in the middle, we’re done with the proof! [1]
But there’s more to the Morley story. Let’s push
A and C to make them swap positions, dragging the trisectors and Morley triangle along for the ride (a process known as extraversion). We end up using some external angle trisectors instead of only internal ones, but the Morley triangle remains equilateral throughout. This gives us a new equilateral Morley triangle for our original ABC!
In fact, each vertex of
ABC has six angle trisectors (three pairs of two), and if you continue applying extraversions you’ll soon uncover that our original triangle ABC has 18 equilateral Morley triangles arranged in a stunning configuration! (Click for larger image.)
Here’s a challenge: the diagram above has 27 equilateral triangles, but only 18 of them are Morley triangles (i.e., are made from a pair of trisectors from each vertex). Which ones are they?
Notes This is a slight modification of a proof by Conway and Doyle. [↩]
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Suppose we have the following experimental values for $\eta' \rightarrow \eta \pi \pi$ decay width:
$\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-} = 0.086 \pm 0.004$
$\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0} = 0.0430 \pm 0.0022$
We are investigating this decay in the isospin invariant limit and want to compare our results with experimental rates. So we should average over these 2 values to find the experimental decay width in this limit.It is written in one paper that we should average
$\Gamma_{\eta' \rightarrow \eta \pi^+ \pi^-} = 0.086 \pm 0.004$
2$\Gamma_{\eta' \rightarrow \eta \pi^0 \pi^0} = 0.0860 \pm 0.0044$
in a specific way (which is not my question).I don't know why we should This post imported from StackExchange Physics at 2014-05-04 11:26 (UCT), posted by SE-user soodeh
twice the second value and then average?
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I'm a high school student who never studied any relativity before, but I'm just wondering what was THE question that Einstein asked himself before going into this field. I knew he has done lots of work such as Brownian motion, photoelectric effect,etc. What was the question that baffled and therefore motivated him to work on relativity?
Let's talk about special relativity (1905) first, then general relativity (1915).
The motivation for special relativity is stated clearly in the first sentence of Einstein's paper "On the Electrodynamics of Moving Bodies":
It is known that Maxwell's electrodynamics -- as usually understood at the present time -- when applied to moving bodies, leads to asymmetries which do not appear to be inherent in the phenomena.
Let me unpack that. Maxwell's equations differ from the equations of Newtonian mechanics in one crucial aspect: Maxwell's equations
seem at first glance to single out a particular reference frame. The clearest example is the speed of electromagnetic waves: this is given by the formula $c=1/\sqrt{\epsilon_0\mu_0}$, where $\epsilon_0$ and $\mu_0$ are universal physical constants. The speed of the source has nothing to do with the speed of the wave, according to this formula. If you turn on a flashlight while standing still, the light beam that comes out should have the same speed as a light beam from a flashlight on a train whizzing along as fast as you please. (The speed of both beams being measured in the same frame.)
Before special relativity, physicists invoked an invisible medium, the aether, to explain this. Sound waves have a certain speed in air, independent of the source of the wave; likewise water waves on a pond. If EM waves are waves in the so-called luminiferous ("light-carrying") aether, then the formula $c=1/\sqrt{\epsilon_0\mu_0}$ makes sense. Here, $\epsilon_0$ and $\mu_0$ are constants describing aspects of the aether. And $c$ then describes the speed of light
in the rest frame of the aether.
On the other hand, Maxwell's equations also contain hints that there
is no special frame of reference. Right after that first sentence, Einstein gives an example. Move a coil of wire through the field of a magnet. A current will be induced. (This is one of Faraday's most famous discoveries: electromagnetic induction. Also discovered independently by Joseph Henry.) You can calculate the current using Maxwell's equations in two ways: either pick a frame where the wire is at rest, or one in which the magnet is at rest. You get the same current either way!
Another famous example: the Michelson-Morley experiment. I won't go into details, but the upshot is that Michelson and Morley failed to detect the speed at which the earth was, supposedly, traveling through the aether. Einstein alludes to this briefly in his 1905 paper:
Examples of this sort [the wire and magnet], together with the unsuccessful attempts to discover any motion of the earth relatively to the "light medium", suggest that the phenomena of electrodynamics as well as of mechanics [i.e., Newton's laws] possess no properties corresponding to the idea of absolute rest.
I should say that historians disagree on how critical this experiment was to Einstein's thought.
And now the punchline:
They suggest rather that ... the same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good. We will raise this conjecture (the purport of which will hereafter be called the "Principle of Relativity") to the status of a postulate, and also introduce another postulate, which is only apparently irreconcilable with the former, namely that light is always propagated in empty space with a definite velocity $c$ which is independent of the state of motion of the emitting body.
The key phrase here is "apparently irreconcilable". I hope you see the apparent contradiction right away. How can a light beam appear to travel at the same speed $c$ to
all observers, regardless of how they are moving themselves?
That was Einstein's motivation for special relativity (SR).
Now let's turn to general relativity (GR). This started off as an attempt to reconcile Newton's law of gravity with special relativity. Newton's law says that two point masses $m$ and $M$ attract each other with force $F=GmM/r^2$, where $r$ is the distance between them. SR doesn't like this for several reasons. Writing in 1920, Sir Arthur Stanley Eddington described some of the difficulties:
The most serious objection against the Newtonian law as an exact law was that it had become ambiguous. The law refers to the product of the masses of the two bodies; but the mass depends on the velocity -- a fact unknown in Newton's day. Are we to take the variable mass, or the mass reduced to rest? Perhaps a learned judge, interpreting Newton's statement like a last will and testament, could give a decision; but that is scarcely the way to settle an important point in scientific theory.
Further
distance, also referred to in the law, is something relative to an observer. Are we to take the observer travelling with the sun or with the other body concerned, or at rest in the aether or in some gravitational medium? [from Space, Time, and Gravitation, Eddington's pop-sci treatment]
(Eddington was one of first scientists to master GR, and played a key role in its early history, post-1915.)
But the following point probably loomed as even more problematic:
According to Newton's law, if you move one mass to a new position, this affects the gravitational force on the other mass instantaneously, according to Newton's formula. In principle, you could use this to send signals faster than light, indeed, instantaneously.
Not only Einstein, but several other physicists saw these problems, and set about trying to find the correct law of gravity for SR. In 1921, looking back, Einstein described his motivation this way:
When, in 1907, I was working on a comprehensive paper on the special theory of relativity for the
Jahrbuch der Radioaktivität und Electronik, I had also to attempt to modify the Newtonian theory of gravitation in such a way that its laws would fit in the [special relativity] theory. Attempts in this direction did show that this could be done, but did not satisfy me because they were based on physically unfounded hypotheses. [quoted in Pais, Subtle is the Lord, p.178]
Einstein then described how there occurred to him "the happiest thought of my life":
The gravitational field has only a relative existence in a way similar to the electric field generated by magnetoelectric induction.
Because for an observer falling freely from the roof of a house there exists-- at least in his immediate surroundings -- no gravitational field[his italics; op cit.]
This led to the famous Principle of Equivalence. Roughly speaking, a
free-falling frame of reference in a gravitational field, is equivalent to a non-accelerating frame of reference in a gravity-free field. Also, an accelerating frame of reference in a gravity-free field is equivalent to a non-accelerating frame of reference in a gravitational field (again, roughly speaking).
You can see why Einstein regarded this as an essential clue. To study what gravity should look like according to SR, we can study accelerating frames of reference without gravity. It turns out that there are useful ways to approach the latter question.
One of the early successes of the Principle of Equivalence was an explanation of why so-called gravitational mass is equal to inertial mass. In Newtonian mechanics, this equality explains why all things fall with the same acceleration (ignoring air resistance). The Principle of Equivalence takes a different tack on this: it replaces the two falling objects in a gravitational field, with freely floating objects viewed from an accelerating frame of reference. So they appear to accelerate at the same rate. You can then argue from that result back to the equality of gravitational and inertial mass. This discovery must have reassured Einstein that he was on the right track.
Einstein wrote his first paper on this new approach in 1907. He did not arrive at the equations of GR until 1915. The "happiest thought of his life" provided his initial motivation (plus the need to reconcile gravity with SR), but the full tale is much longer, with many twists and turns.
I suppose you are asking about Special relativity, which Einstein proposed in 1905. (Special relativity is about light and kinematics, while General relativity is about gravity).
There was an apparent contradiction concerning the speed of light. On the one hand there was Maxwell theory which predicted that the speed of light must be the same in all frames of reference. Maxwell's theory was well established and experimentally tested (by Hertz and others. Wireless communication was based on Maxwell theory). On the other hand, this contradicted classical mechanics, which was even better established and tested.
Some of the most striking consequences of Maxwell theory (obtained by Lorenz and Poincare) were that such things as length of a rod and time interval between two events depends on the frame of reference. This is why it is called "relativity".
Einstein's theory of special relativity was essentially a new kinematics which removed all these apparent contradictions. The famous formula $E=mc^2$ is a consequence of the special relativity.
Later general relativity was a different theory which was not motivated by any experiment. There, the main motivation was the strange fact (well known since Galileo) that the "gravitational mass" (or "gravitational charge", the thing which stands in the gravity law) is identical to the "inertial mass", the thing which stands in the newton second law. This strange coincidence is responsible for the well-known fact that all bodies fall with the same acceleration (in vacuum). General relativity was designed to explain this strange fact. As far as I know, nobody before Einstein even tried. The fact was considered as "evident". Unlike special relativity, the effects of general relativity are small, and difficult to measure. But there were two consequences of the new theory which could be measured: the gravitational lensing, and shift of mercury perihelion. Gravitational lensing was measured and found to conform to general relativity by Eddington in 1918, and this immediately made Einstein famous.
References. E. Whittaker, A history of the theories of aether and electricity. This is not for a high school student. When I was a high school student I read Martin Gardner, Relativity for the million, but there are many other good popular books.
I suggest to see these sources:
O.Darrigol, The Electrodynamic Origins of Relativity Theory (1996)
as well as :
Olivier Darrigol, The Genesis of the Theory of Relativity (2005).
From this one, page 2 :
A first indication of the primary context of the early theory of relativity is found in the very title of Einstein's founding paper:
On the electrodynamics of moving bodies. This title choice may seem bizarre to the modern reader, who defines relativity theory as a theory of space and time. In conformity with the latter view, the first section of Einstein's paper deals with a new kinematics meant to apply to any kind of physical phenomenon.
Much of the paper nonetheless deals with the application of this kinematics to the electrodynamics and optics of moving bodies. Clearly, Einstein wanted to solve difficulties he had encountered in this domain of physics.
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As I was studying about sufficiency I came across your question because I also wanted to understand the intuition about From what I've gathered this is what I come up with (let me know what you think, if I made any mistakes, etc).
Let $X_1,\ldots,X_n$ be a random sample from a Poisson distribution with mean $\theta>0$.
We know that $T({\bf{X}})=\sum_{i=1}^{n} X_i$ is a sufficient statistic for $\theta$, since the conditional distribution of $X_1,\ldots,X_n$ given $T({\bf{X}})$ is free of $\theta$, in other words, does not depend on $\theta$.
Now, statistician $A$
knows that $X_1,\ldots,X_n \overset{i.i.d}{\sim} Poisson(4)$ and creates $n=400$ random values from this distribution:
n<-400
theta<-4
set.seed(1234)
x<-rpois(n,theta)
y=sum(x)
freq.x<-table(x) # We will use this latter on
rel.freq.x<-freq.x/sum(freq.x)
For the values statistician $A$ has created, he takes the sum of it and asks statistician $B$ the following:
"I have these sample values $x_1,\ldots,x_n$ taken from a Poisson distribution. Knowing that $\sum_{i=1}^{n} x_i = y = 4068$, what can you tell me about this distribution?"
So, knowing only that $\sum_{i=1}^{n} x_i = y = 4068$ (and the fact that the sample arose from a Poisson distribution) is sufficient for statistician $B$ to say anything about $\theta$? Since we know that this is a sufficient statistic we know that the answer is "yes".
To gain some intution about the meaning of this, let's do the following (taken from Hogg & Mckean & Craig's "Introduction to Mathematical Statistics", 7th edition, exercise 7.1.9):
"$B$ decides to create some fake observations, which he calls $z_1,z_2,\ldots,z_n$ (as he knows they will probably not be equal the original $x$-values) as follows. He notes that the conditional probability of independent Poisson random variables $Z_1,Z_2\ldots,Z_n$ being equal to $z_1,z_2,\ldots,z_n$, given $\sum z_i = y$, is
$$\cfrac{\frac{\theta^{z_1}e^{-\theta}}{z_1!} \frac{\theta^{z_2}e^{-\theta}}{z_2!} \cdots \frac{\theta^{z_n}e^{-\theta}}{z_n!}}{\frac{n \theta^{y}e^{-n\theta}}{y!}}=\frac{y!}{z_1!z_2! \cdots z_n!} \left(\frac{1}{n}\right)^{z_1} \left(\frac{1}{n}\right)^{z_2} \cdots \left(\frac{1}{n}\right)^{z_n}$$
since $Y=\sum Z_i$ has a Poisson distribution with mean $n \theta$. The latter distribution is multinomial with $y$ independent trials, each terminating in one of $n$ mutually exclusive and exhaustive ways, each of which has the same probability $1/n$. Accordingly, $B$ runs such a multinomial experiment $y$ independent trials and obtains $z_1,\ldots,z_n$."
This is what the exercise states. So, let's do exactly that:
# Fake observations from multinomial experiment
prob<-rep(1/n,n)
set.seed(1234)
z<-as.numeric(t(rmultinom(y,n=c(1:n),prob)))
y.fake<-sum(z) # y and y.fake must be equal
freq.z<-table(z)
rel.freq.z<-freq.z/sum(freq.z)
And let's see what $Z$ looks like (I'm also plotting the real density of Poisson(4) for $k=0,1,\ldots,13$ - anything above 13 is pratically zero -, for comparison):
# Verifying distributions
k<-13
plot(x=c(0:k),y=dpois(c(0:k), lambda=theta, log = FALSE),t="o",ylab="Probability",xlab="k",
xlim=c(0,k),ylim=c(0,max(c(rel.freq.x,rel.freq.z))))
lines(rel.freq.z,t="o",col="green",pch=4)
legend(8,0.2, legend=c("Real Poisson","Random Z given y"),
col = c("black","green"),pch=c(1,4))
So, knowing nothing about $\theta$ and knowing only the sufficient statistic $Y=\sum X_i$ we were able to recriate a "distribution" that looks a lot like a Poisson(4) distribution (as $n$ increases, the two curves become more similar).
Now, comparing $X$ and $Z|y$:
plot(rel.freq.x,t="o",pch=16,col="red",ylab="Relative Frequency",xlab="k",
ylim=c(0,max(c(rel.freq.x,rel.freq.z))))
lines(rel.freq.z,t="o",col="green",pch=4)
legend(7,0.2, legend=c("Random X","Random Z given y"), col = c("red","green"),pch=c(16,4))
We see that they are pretty similar, as well (as expected)
So, "for the purpose of making a statistical decision, we can ignore the individual random variables $X_i$ and base the decision entirely on the $Y=X_1+X_2+\cdots+X_n$" (Ash, R. "Statistical Inference: A concise course", page 59).
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I'm trying to reproduce Figure 2 from this paper. In summary, I have the regression model
$$Y \sim N(\mu, \sigma^2) \\ \mu = \alpha + \beta_1X_1 + \beta_2X_2 $$
The prior distributions for the parameters are
$$ \alpha, \beta_1 , \beta_2 \sim N(0, 10)\\ \sigma \sim \vert N(0, 1)\vert $$
The goal is to compute posterior distributions for parameters based on the (synthetically generated) data using vanilla MCMC (in contrast to the NUTS sampler as done in the paper).
I'm defining 'state' $s$ as the tuple $s=(\alpha, \beta_1, \beta_2, \sigma)$. The log-likelihood of the data given the parameters is
$$ \log P(Y \vert s) =-\frac{M}{2}\log(2\pi\sigma^2) - \frac{1}{2\sigma^2}\sum_{i=1}^M\left(y_i -w^Tx_i\right)^2 $$
where $w = [\alpha, \beta_1, \beta_2]$ is the parameter vector, $x_i$ is the $i$th row of the (synthetic) data, with the bias term. I.e. $x_i = [1, x_{1i}, x_{2i}]$. In the code below, you'll see the first term computed as $-M\log\sigma$ since the constants will cancel out in the eventual subtraction in the acceptance calculation.
The log-prior is simply the sum of the logarithms of the PDFs of the parameters as described above (computed in the code below).
I'm using the random-walk MCMC where the proposed next state is chosen with the current state as the mean and a fixed variance vector as the scale hyper parameter.
$$ s^{t+1} \sim N(s^t, \nu) $$
with $\nu = [0.07, 0.25, 0.05, 0.02]$. With these manually tweaked, current values, I'm getting incorrect results. The results are problematic in two respects: (1) They converge to wrong values and (2) I'm generating correlated samples. I've experimented with the $\nu$ parameter for the whole weekend and have not had great success getting it to work.
As can be seen all of my trace plots show severe sample correlation problem. The posterior modes also converge to incorrect values (true values should be $[\alpha=1, \beta_1=1, \beta_2=2.5, \sigma=1]$.
I feel like I'm missing something obvious since this 4-dimensional problem should be solvable using the Metropolis algorithm. I'm including the full code below. Any help is appreciated.
import numpy as npfrom scipy.stats import norm, halfnormimport matplotlib.pyplot as plfrom tqdm import tqdmdef trace(): import ipdb; ipdb.set_trace()def loglik(state, X, Y): M = X.shape[0] coefs, sig = state[:-1], state[-1] err = Y - np.dot(X, coefs) term1 = -M * np.log(sig) term2 = -np.sum(err * err) / (2.0 * sig * sig) return term1 + term2 def proposalfunc(state): return np.random.normal(loc=state, scale=[0.07, 0.25, 0.05, 0.02]) def logprior(state, distributions): return sum(np.log(D.pdf(x)) for x, D in zip(state, distributions))def generate_data(size): alpha, beta1, beta2, sig = 1.0, 1.0, 2.5, 1.0 X1 = np.linspace(0, 1.0, size) X2 = np.linspace(0, 0.2, size) ONES = np.ones(X1.shape) noise = np.random.normal(loc=0, scale=sig, size=size) X = np.vstack((ONES, X1, X2)).T Y = np.dot(X, [alpha, beta1, beta2]) + noise return X, Ydef plot(chain, skip): param_names = ['alpha', 'beta1', 'beta2', 'sigma'] fig, axes = pl.subplots(4, 2, figsize=(10, 6)) histargs = dict(normed=True, histtype='step', bins=50, color='r', alpha=0.7) for i, pname in zip(range(chain.shape[1]), param_names): param = chain[skip:, i] axes[i, 0].hist(param, **histargs) axes[i, 1].plot(param, alpha=0.5) axes[i, 0].set_title(pname) fig.tight_layout() pl.show()def mcmc_regression(): np.random.seed(123) # Uncomment to reproduce the plot exactly X, Y = generate_data(100) alpha_dist = beta1_dist = beta2_dist = norm(0, 10) sigma_dist = halfnorm(0, 1) dists = (alpha_dist, beta1_dist, beta2_dist, sigma_dist) nburnin = 10000 niter = 5000 + nburnin ncomps = 4 chain = np.zeros(shape=(niter, ncomps)) chain[0, :] = np.random.normal(size=ncomps) for i in tqdm(range(niter - 1)): v = chain[i] log_posterior_old = loglik(v, X, Y) + logprior(v, dists) proposal = proposalfunc(v) log_posterior_new = loglik(proposal, X, Y) + logprior(proposal, dists) a = min(0.0, log_posterior_new - log_posterior_old) if np.random.random() < np.exp(a): chain[i + 1, :] = proposal else: chain[i + 1, :] = v plot(chain, nburnin)if __name__ == '__main__': mcmc_regression()
EDIT: I forgot to mention a few things in the OP.
I'm able to get a univariate Bayesian MCMC linear regression to work. But am a bit lost on this bivariate case.
The results are
quitesensitive to the randomizer seed (which further indicates that something is wrong at a very basic level).
I'm aware that HMC and NUTS can help mitigate the sample correlation problem, but I want to get a good understanding of the simple case first.
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This is an old revision of the document!
In this tutorial, we write an Alphabets program, starting from a mathematical equation for LU decomposition. Then we will generate code to execute the alphabets program, and test the generated code for correctness.
The equation for LU Decomposition, derived from first principles using simple algebra in Foundations (pg.3), is as follows: $$ U_{i,j}=\begin{cases} 1=i\le j & A_{i,j}\\ 1<i\le j & A_{i,j}-\sum_{k=1}^{i-1}L_{i,k}U_{k,j} \end{cases}\\ L_{i,j}=\begin{cases} 1 = i\le j & \frac{A_{i,j}}{U_{j,j}}\\ 1< i\le j & \frac{1}{U_{j,j}}(A_{i,j}-\sum_{k=1}^{j-1}L_{i,k}U_{k,j}) \end{cases} $$
[Temp note due to : in the last case of L, the condition is “1 < j ⇐ i”]
Let's start from an empty alphabets file, with LUD as the name of the system, and a positive integer N as its parameter. A system (Affine System) takes its name from system of affine recurrence equations, and represents a block of computation. An Alphabets program may contain multiple systems.
Caveat: Remember the phrase, “It's not a bug, it's a feature”? Well, in a tutorial, a feature is called a “learning opportunity.”
Parameters are runtime constants represented with some symbol in the code. In this example, parameter N will be used to define the size of the matrices, which is not known until runtime.
affine LUD {N|N>0} .
In most cases, a computation uses some inputs and produces outputs. Such variables must be declared with a name, a data type, and a shape/size. In Alphabets, the shape/size is represented with polyhedral domains.For this example, the
A matrix is given, and we are computing two triangular matrices
L and
U.
A is an NxN square matrix. The declaration for
A looks as follows:
float A {i,j|1<=(i,j)<=N}; //starting from 1 to be consistent with the equation in the notesSimilarly,
L is a lower triangular matrix of size N (with unit diagonals, implicit) and
U is an upper triangular matrix of size N. The declarations should look like the following:
// The convention is that i is the vertical axis going down, and j is the horizontal axis float L {i,j|1<i<=N && 1<=j<i}; // Note that the diagonal elements of L are not explicitly declared float U {i,j|1<=j<=N && 1<=i<=j};Now these variable declarations need to be placed at appropriate places to specify whether they are input/output/local.
input/
given is the keyword for input,
output/
returns is the keyword for output, and
local/
using is the keyword for local variables.
affine LUD {N|N>0} input float A {i,j|1<=(i,j)<=N}; output float L {i,j|1<=j<i<=N}; float U {i,j|1<=i<=j<=N}; .
Polyhedral domains are represented as { “index names” | “affine constraints using indices and parameters” }, where constraints can be intersected with
“&&”. Sometimes constraints can be expressed with short-hand notation like
“a<b<c” or
“(b,c)<0”.
Unions of such domains can be expressed as “{ a,b | constraints on a and b } || { c,d | constraints on c and d }”. One important point about Alphabets domains is that the names given to indices are only for textual representation. Internally, all analysis/transformation/code generation tools only care about which dimension the constraint applies to.For example, a domain { i,j | 0⇐i<j<N } is equivalent to { x,y | 0⇐x<y<N }, because
i and
x are both names given to the first dimension, and
j and
y are names given to the second dimension.
Now the only remaining step before a complete Alphabets program is writing the equations. After a little experience, the connection from mathematical equations (of a certain form) to Alphabets equations should become increasingly clear. There are two slightly different syntactic conventions for writing equations, one is called the “Show syntax” and the other is called “AShow syntax”. Show syntax is closer to the internal representation of Alphabets programs, and is more expressive when writing complex programs. AShow syntax uses “array notation” so that it is easier for people used to imperative programs.
We will first write the equation for
U in AShow syntax, and then move on to Show as we write the equation for
L.
In this equation,
U is on the left hand side, and the right hand side should define
U for each point in the declared domain of the
U variable.In AShow syntax, the names for indices used appear on the LHS of the equations.
For this example, the following LHS for
U gives
i, j as the names for the first and second dimensions to be used when writing the expressions in the RHS.These names do not have to match the names used in variable declaration. You could use
x,y instead of
i,j if desired.
U[i,j] = RHSexpr;
The first thing you notice in the definition of
U in the mathematical equation is the branch based on values of
i and
j. This branching is expressed with CaseExpression in Alphabets.A CaseExpression starts with the keyword “
case”, ends with keyword “
esac”, and has list of “
;”-delimited expressions, called “clauses” as its children.Often (but not always), each child of a case is a RestrictExpression (whose syntax is “domain : expr”), which restricts the domain to the specified domain.
Using the above expressions, the branching of the definition of
U is as follows :
U[i,j] = case {|1==i} : expr1; {|1<i} : expr2; esac;Note that because index names are already declared in the context (equation LHS), there is nothing to the left of the
| in the AShow syntax.
Moving on to the definitions in each case, the first case is . This is written as
A[i,j] in AShow syntax, similar to accessing an array. A variable without a square bracket, is treated either as a scalar variable (as in,
X[i,j] = 0) or as an access with the identity dependence function, (i.e.,
X[i] = A[i] would be the same as
X[i] = A).
The last piece missing before completing the definition of
U is the summation in the second branch.Mathematically, a reduction is an operation that applies an associative-comutative operator (in general, the operator may only be associative, but In Alphabets, we have only associative-comutative operators) to a set of values, such as summation (sum over a set of numbers).
Reductions are expressed with the following syntax :
reduce(operator, projection, expr);operator: operator to be applied (+, *, max, min, and, or)
In the mathematical equation, summation with one new index
k is used. For each value of
k, the expression
L[i,k]*U[k,j] is computed and added up to produce the result
U[i,j]. Thus, the projection function is
(i,j,k → i,j). (from the three dimensional space indexed by
i,j,k, all values computed at
[i,j,k] are used to compute
U[i,j] in the two dimensional space indexed by
i,j – i.e., the
k is 'projected out')
When the projection function is canonic (e.g.,
(i,j,k→i,j)), then the projection function can be replaced with a simpler syntax (AShow syntax for reductions) that specifies the names of new dimensions surrounded by square brackets.For example, the projection
(i,j,x,y→i,j) can be expressed as
[x,y].
Using the above, summation in the original equation can be written as the following Alphabets fragment.
reduce(+, [k], L[i,k]*U[k,j]);
Putting all this together, the final equation for
U is:
U[i,j] = case {|1==i} : A[i,j]; {|1<i} : A[i,j] - reduce(+, [k], L[i,k]*U[k,j]); esac;
This is exactly like the original equation Caveat
Now we will write the equation for
L, but this time in Show syntax. Unlike the AShow syntax, Show syntax does not rely on the context for naming of indices. Index names can be different in every (sub)expression if it makes sense to do so.Because of this, the LHS does not have square brackets, all we need is the variable name.
L = RHSexpr; //Show syntax
CaseExpression and RestrictExpression are same as AShow syntax. However, since index names are no longer deduced from the context where they occur, they must be explicitly named everywhere. While this may seem cumbersome, it allows expressions to have compositional semantics. In our example, the index names used in the domain of RestrictExpression have to be made explicit.The branch in the definition of
L becomes the following Alphabets :
L = case {i,j|1==j} : expr1; {i,j|1<i} : expr2; esac;
In the array notation in AShow syntax, a DependenceExpression was implicit: just add expressions within square brackets to access variables). In the Show syntax
DependenceExpression is used to explicitly specify which value of a variable is required for a computation. The syntax of
DependenceExpression is “(affine_function)@expr”, where
affine_function is of the form
(list_of_indices → list_of_affine_expressions). For example, the dependence
(i,j→i-1,i+j)@A means that at index point
(i,j) this computation evaluates to the value of
A at index point
(i-1,i+j).
The child of
DependenceExpression can be any Alphabets expression, possibly another DependenceExpression. For example,
(i,j→i,j,i+j,0)@(a,b,c,d→a,c-a)@A is a perfectly legal Alphabets expression.
Reductions in Show syntax are exactly like in the AShow syntax, except that the projection function is specified in the dependence syntax. This is all you need in order to write the rest in of the equation in Show syntax.
L = case {i,j|1==j} : (A / (i,j->j,j)@U); {i,j|1<i} : (A - reduce(+, (i,j,k->i,j), (i,j,k->i,k)@L*(i,j,k->k,j)@U))/(i,j->j,j)@U; esac;
Combine all of the above, and you will get the Alphabets program for LU decomposition. Don't forget the keyword
let/
through before equations the period at the end (since our example has no local variables). Notice how we can mix and match Show and AShow syntax within the program, but each equation must obviously, be consistent.
affine LUD {N|N>0} input float A {i,j|1<=(i,j)<=N}; output float L {i,j|1<i<=N && 1<=j<i}; float U {i,j|1<=j<=N && 1<=i<=j}; let U[i,j] = case {|1==i} : A[i,j]; {|1<i} : A[i,j] - reduce(+, [k], L[i,k]*U[k,j]); esac; L = case {i,j|1==j} : A / (i,j->j,j)@U; {i,j|1<j} : (A - reduce(+, (i,j,k->i,j), (i,j,k->i,k)@L*(i,j,k->k,j)@U))/(i,j->j,j)@U; esac; .
Analyses, transformations, and code generation of Alphabets programs are performed using the AlphaZ system. The normal interface for using AlphaZ is the scripting interface called compiler scripts. Given below is an example script for that does several things using the LUD program we wrote above.
# read program and store the internal representation in variable prog prog = ReadAlphabets("./LUD.ab"); # store string (corresponding to system name) to variable system system = "LUD"; # store output directory name to variable outDir outDir = "./test-out/"+system; # print out the program using Show syntax Show(prog); # print out the program using AShow syntax AShow(prog); # prints out the AST of the program (commented out) #PrintAST(prog); # generate codes (this is demand-driven, memoized code) generateWriteC(prog, system, outDir); generateWrapper(prog, system, outDir); generateMakefile(prog, system, outDir);Save this script with .cs extension, place the alphabets file in the same directory as the script, and then right click on the editor and select
“Run As → Compiler Script” to run the script.
If you get some error message, try looking at the first line of the error messages to find out what it is about. Common problems are:
FileNotFoundException in this case)
xxx does not exist)
In this tutorial, we use two basic code generators, without going into too much detail. The two types of codes generated are
WriteC and
Wrapper.
WriteC code may not be efficient, but it can be generated without any additional specification beyond the program.
Wrapper code is a wrapper around other generated codes that allocates/frees memory for input and output variables, and it also have different options for testing.
Note:Current implementation of the Wrapper prints out the bounding box of the domain of the output variable.
generateMakefile produces a Makefile that should compile the generated codes. You can make with different options.
Congratulations!! You are nearly at the end. Now, you will actually make and execute the code (in a separate terminal window).
compiles the code and produces an executable
xxx (where
xxx is the system name) that executes the program with default input that is 1 everywhere. Compiling with this option does not test very much, but it will test if it compiles and runs and produces no errors.
Compiles the code and produces an executable
xxx.check (
xxx is the system name) that prompts the user for all values of input variables.After executing, it prints out all values of the output variables.This option should be used for testing small to mid-sized input data.
Compiles the code with another code named
xxx_verify.c that defines a function
xxx_verify (
xxx is the system name).Users can provide different program as
xxx_verify to compare outputs.
Same as verify, except the inputs are generated randomly.
You will see that when you execute the code,
. You may be able to easily fix the error in your Alpha program and regenerate correctly executing C code, or you may want a bit of help. In either case, we would like to know. Please email Sanjay.Rajopadhye@colostate.edu with the error message that is produced. it will produce an error
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Let's put the succinct answer by @TheAlmightyBob into an abstract model:
We want to model the labor market.
Markets' structure assumptions: goods market and labor markets are perfectly competitive. All participants are "too small" economically, and they cannot affect equilibrium price through their quantities demanded/supplied - they are "price takers". Markets "clear" - i.e. prices adjust so that quantity actually supplied equals quantity actually bought.
Agents assumption: There are $n$ identical workers, and $m$ identical firms, that participate in the market. Both populations are fixed.
Other assumptions: a) deterministic environment, b) one perishable good produced, c) model in "real terms" (real wage etc, scaled by the price of the good produced).
The typical firm produces according to the technology$$Y_j = F_j(K_j,L_j;\mathbf q) \tag{1}$$
where $\mathbf q$ is a vector of parameters. Perfect competition in the goods market, and a perishable good imply that all output produced is sold.The goal of the firm is maximization of capital returns over the choice of labor.
$$\max_{L_j} \pi_j = F_j(K_j,L_j;\mathbf q) - wL_j$$
We are modelling the labor market, so we are interested in the first-order condition
$$\frac {\partial \pi_j}{\partial L_j} = 0 \tag{2}$$and the corresponding input demand schedule
$$L_j^* = L_j^*\left(K_j, \mathbf q, w\right) \tag{3}$$
Total Labor demand is $L_d = m\cdot L_j^*$. The labor market equilibrium assumption implies
$$ L_d = L_s \Rightarrow m\cdot L_j^*\left(K_j, \mathbf q, w\right) = L_s \tag{4}$$
which implicitly expresses the equilibrium wage as a function of technology constants, of per-firm capital, and of labor supplied. In order to fully characterize the labor market, we need to derive also the optimal labor supply.
Each identical worker derives utility from consumption and leisure, subject to a biological limit of available time, $T$, and the budget constraint that consumption equals wage income:
$$\max_{L_i} U(C_i, T-L_i;\mathbf \gamma),\;\; \text{s.t.} \;C_i= wL_i$$
where $\gamma$ is a vector of preference parameters, indicating the relative weight between utility from consumption, and from leisure.This will give us individual labor supply as
$$L_i^* = L_i^*(T,w, \mathbf \gamma) \tag{5}$$
and total labor supply is $L_s = n\cdot L_i^*$. Plugging this into $(4)$ we obtain
$$mL_j^*\left(K_j, \mathbf q, w\right) =n L_i^*(T,w, \mathbf \gamma) \tag{6}$$
If we stop here, we have a
model that examines the labor market. We have fully described the market, and the goals and the constraints of the participants in it (firms and workers), partial equilibrium related to the specific market. We can perform comparative statics in order to see how the various components of $(6)$ affect the equilibrium wage. Among them, there is the capital-per-firm term, whose effects on wage we can also consider based on $(6)$, by treating it as varying arbitrarily.
In order to turn this model into a
model: general equilibrium a) We need to specify things about capital: who owns it/controls it/makes decisions on it. What is the objective functions of these decision makers. This will lead us to an optimal $K_j^*$ as a function of the structure we will impose here. Then, comparative statics with respect to $K_j$ will turn into comparative statics with respect to the factors that affect the determination of $K_j^*$, which may very well prove to involve also $\mathbf q, w$ and even the other parameters in $(6)$, changing in this way the comparative statics results obtained in a partial equilibrium setting.
b) We also need to take into account any
macroeconomic identities that characterize this economy, something along the lines of $mY_j \equiv ...$ where the right hand side will be determined by the assumptions we make related to capital, but also, for example, by whether we will assume that the economy is closed or open, or partially open to the outside economic system.
So, apart from being more complicated as a model, it may also lead us to different conclusions than partial equilibrium analysis.
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[This is the 6th post in the current series about Wythoff’s game: see posts #1, #2, #3, #4, and #5.
Caveat lector: this post is a bit more difficult than usual. Let me know what you think in the comments!]
Our only remaining task from last week was to prove the mysterious Covering Theorem: we must show that there is exactly one dot in each row and column of the grid (we already covered the diagonal case). Since the rows and columns are symmetric, let’s focus on columns.
The columns really only care about the
x-coordinates of the points, so let’s draw just these x-coordinates on the number-line. We’ve drawn \(\phi,2\phi,3\phi,\ldots\) with small dots and \(\phi^2,2\phi^2,3\phi^2,\ldots\) with large dots. We need to show that there’s exactly one dot between 1 and 2, precisely one dot between 2 and 3, just one between 3 and 4, and so on down the line. For terminology’s sake, break the number line into length-1 intervals [1,2], [2,3], [3,4], etc., so we must show that each interval has one and only one dot:
Why is this true? One explanation hinges on a nice geometric observation: Take any small dot
s and large dot t on our number-line above, and cut segment st into two parts in the ratio \(1:\phi\) (with s on the shorter side). Then the point where we cut is always an integer! For example, the upper-left segment in the diagram below has endpoints at \(s=2\cdot\phi\) and \(t=1\cdot\phi^2\), and its cutting point is the integer 3:
In general, if
s is the jth small dot—i.e., \(s=j\cdot\phi\)—and \(t=k\cdot\phi^2\) is the kth large dot, then the cutting point between s and t is \(\frac{1}{\phi}\cdot s+\frac{1}{\phi^2}\cdot t = j+k\) (Why?! [1]). But more importantly, this observation shows that no interval has two or more dots: a small dot and a large dot can’t be in the same interval because they always have an integer between them! [2]
So all we have to do now is prove that no interval is
empty: for each integer n, some dot lies in the interval [ n, n+1]. We will prove this by contradiction. What happens if no dot hits this interval? Then the sequence \(\phi,2\phi,3\phi,\ldots\) jumps over the interval, i.e., for some j, the jth dot in the sequence is less than n but the ( j+1)st is greater than n+1. Likewise, the sequence \(\phi^2,2\phi^2,3\phi^2,\ldots\) jumps over the interval: its kth dot is less than n while its ( k+1)st dot is greater than n+1:
By our observation above on segment \(s=j\phi\) and \(t=k\phi^2\), we find that the integer
j+ k is less than n, so \(j+k\le n-1\). Similarly, \(j+k+2 > n+1\), so \(j+k+2 \ge n+2\). But together these inequalities say that \(n\le j+k\le n-1\), which is clearly absurd! This is the contradiction we were hoping for, so the interval [ n, n+1] is in fact not empty. This completes our proof of the Covering Theorem and the Wythoff formula!
It was a long journey, but we’ve finally seen exactly why the Wythoff losing positions are arranged as they are. Thank you for following me through this!
A Few Words on the Column Covering Theorem
Using the
floor function \(\lfloor x\rfloor\) that rounds x down to the nearest integer, we can restate the Column Covering Theorem in perhaps a more natural context. The sequence of integers $$\lfloor\phi\rfloor = 1, \lfloor 2\phi\rfloor = 3, \lfloor 3\phi\rfloor = 4, \lfloor 4\phi\rfloor = 6, \ldots$$ is called the Beatty sequence for the number \(\phi\), and similarly, $$\lfloor\phi^2\rfloor = 2, \lfloor 2\phi^2\rfloor = 5, \lfloor 3\phi^2\rfloor = 7, \lfloor 4\phi^2\rfloor = 8,\ldots$$ is the Beatty sequence for \(\phi^2\). Today we proved that these two sequence are complementary, i.e., together they contain each positive integer exactly once. We seemed to use very specific properties of the numbers \(\phi\) and \(\phi^2\), but in fact, a much more general theorem is true: Beatty’s Theorem: If \(\alpha\) and \(\beta\) are any positive irrational numbers with \(\frac{1}{\alpha}+\frac{1}{\beta}=1\), then their Beatty sequences \(\lfloor\alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor,\ldots\) and \(\lfloor\beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\ldots\) are complementary sequences.
Furthermore, our same argument—using \(\alpha\) and \(\beta\) instead of \(\phi\) and \(\phi^2\)—can be used to prove the more general Beatty’s Theorem!
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The expansion of alcohol in a thermometer is one of many commonly encountered examples of
thermal expansion, which is the change in size or volume of a given system as its temperature changes. The most visible example is the expansion of hot air. When air is heated, it expands and becomes less dense than the surrounding air, which then exerts an (upward) force on the hot air and makes steam and smoke rise, hot air balloons float, and so forth. The same behavior happens in all liquids and gases, driving natural heat transfer upward in homes, oceans, and weather systems, as we will discuss in an upcoming section. Solids also undergo thermal expansion. Railroad tracks and bridges, for example, have expansion joints to allow them to freely expand and contract with temperature changes, as shown in Figure \(\PageIndex{1}\).
Figure \(\PageIndex{1}\): (a) Thermal expansion joints like these in the (b) Auckland Harbour Bridge in New Zealand allow bridges to change length without buckling. (credit: “ŠJů”/Wikimedia Commons).
What is the underlying cause of thermal expansion? As previously mentioned, an increase in temperature means an increase in the kinetic energy of individual atoms. In a solid, unlike in a gas, the molecules are held in place by forces from neighboring molecules; as we saw in Oscillations, the forces can be modeled as in harmonic springs described by the Lennard-Jones potential. Energy in Simple Harmonic Motion shows that such potentials are asymmetrical in that the potential energy increases more steeply when the molecules get closer to each other than when they get farther away. Thus, at a given kinetic energy, the distance moved is greater when neighbors move away from each other than when they move toward each other. The result is that increased kinetic energy (increased temperature) increases the average distance between molecules—the substance expands.
For most substances under ordinary conditions, it is an excellent approximation that there is no preferred direction (that is, the solid is “isotropic”), and an increase in temperature increases the solid’s size by a certain fraction in each dimension. Therefore, if the solid is free to expand or contract, its proportions stay the same; only its overall size changes.
Definition: Thermal Expansion in One Dimension
According to experiments, the dependence of thermal expansion on temperature, substance, and original length is summarized in the equation
\[\dfrac{dL}{dT} = \alpha L\]
where \(L\) is the original length \(\frac{dL}{dT}\) is the change in length with respect to temperature, and \(\alpha\) is the
coefficient of linear expansion, a material property that varies slightly with temperature. As \(\alpha\) is nearly constant and also very small, for practical purposes, we use the linear approximation:
\[\Delta L \approx \alpha L \Delta T.\]
Table \(\PageIndex{1}\) lists representative values of the coefficient of linear expansion. As noted earlier, \(\Delta T\) is the same whether it is expressed in units of degrees Celsius or kelvins; thus, \(\alpha\) may have units of \(1/^oC\) or 1/K with the same value in either case. Approximating \(\alpha\) as a constant is quite accurate for small changes in temperature and sufficient for most practical purposes, even for large changes in temperature. We examine this approximation more closely in the next example.
Material Coefficient of Linear Expansion \(\alpha (1/^oC)\) Coefficient of Volume Expansion \(\beta(1/^oC)\) Solids Aluminum \(25 \times 10^{-6}\) \(75 \times 10^{-6}\) Brass \(19 \times 10^{-6}\) \(56 \times 10^{-6}\) Copper \(17 \times 10^{-6}\) \(51 \times 10^{-6}\) Gold \(14 \times 10^{-6}\) \(42 \times 10^{-6}\) Iron or steel \(12 \times 10^{-6}\) \(35 \times 10^{-6}\) Invar (nickel-iron alloy) \(0.9 \times 10^{-6}\) \(2.7 \times 10^{-6}\) Lead \(29 \times 10^{-6}\) \(87 \times 10^{-6}\) Silver \(18 \times 10^{-6}\) \(54 \times 10^{-6}\) Glass (ordinary) \(9 \times 10^{-6}\) \(27 \times 10^{-6}\) Glass (Pyrex®) \(3 \times 10^{-6}\) \(9 \times 10^{-6}\) Quartz \(0.4 \times 10^{-6}\) \(1 \times 10^{-6}\) Concrete, brick \(-12 \times 10^{-6}\) \(-36 \times 10^{-6}\) Marble (average) \(2.5 \times 10^{-6}\) \(7.5 \times 10^{-6}\) Liquids Ether \(1650 \times 10^{-6}\) Ethyl alcohol \(1100 \times 10^{-6}\) Gasoline \(950 \times 10^{-6}\) Glycerin \(500 \times 10^{-6}\) Mercury \(180 \times 10^{-6}\) Water \(210 \times 10^{-6}\) Gases Air and most other gases at atmospheric pressure \(3400 \times 10^{-6}\)
Bimetallic Strip as thermometers
Thermal expansion is exploited in the bimetallic strip (Figure \(\PageIndex{2}\)). This device can be used as a thermometer if the curving strip is attached to a pointer on a scale. It can also be used to automatically close or open a switch at a certain temperature, as in older or analog thermostats.
Figure \(\PageIndex{2}\): The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting temperature, where its two components have the same length. (b) At a higher temperature, this strip bends to the right, because the metal on the left has expanded more than the metal on the right. At a lower temperature, the strip would bend to the left.
Example \(\PageIndex{1}\): Calculating Linear Thermal Expansion
The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from \(-15^oC\) to \(40^oC\). What is its change in length between these temperatures? Assume that the bridge is made entirely of steel.
Strategy
Use the equation for linear thermal expansion \(\Delta L = \alpha L \Delta T\) to calculate the change in length, \(\Delta L\). Use the coefficient of linear expansion α for steel from Table \(\PageIndex{1}\), and note that the change in temperature \(\Delta T\) is \(55^oC\).
Solution
Substitute all of the known values into the equation to solve for \(\Delta L\):
\[ \begin{align*} \Delta L &= \alpha L \Delta T \\[5pt] &= \left(\dfrac{12 \times 10^{-6}}{^oC}\right) (1275 \, m)(55^oC) \\[5pt] &= 0.84 \, m. \end{align*}\]
Significance
Although not large compared with the length of the bridge, this change in length is observable. It is generally spread over many expansion joints so that the expansion at each joint is small.
Thermal Expansion in Two and Three Dimensions
Unconstrained objects expand in all dimensions, as illustrated in Figure \(\PageIndex{3}\). That is, their areas and volumes, as well as their lengths, increase with temperature. Because the proportions stay the same, holes and container volumes also get larger with temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the piece you removed were still in place. The piece would get bigger, so the hole must get bigger too.
Figure \(\PageIndex{3}\): In general, objects expand in all directions as temperature increases. In these drawings, the original boundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area increases because both length and width increase. The area of a circular plug also increases. (b) If the plug is removed, the hole it leaves becomes larger with increasing temperature, just as if the expanding plug were still in place. (c) Volume also increases, because all three dimensions increase.
Definition: Thermal Expansion in Two Dimensions
For small temperature changes, the change in area \(\Delta A\) is given by
\[\Delta A = 2 \alpha A \Delta T\]
where \(\Delta A\) is the range area \(A\), \(\Delta T\) is the change in temperature, and \(\alpha\) is the coefficient of linear expansion, which varies slightly with temperature.
Definition: Thermal Expansion in Three Dimensions
The relationship between volume and temperature \(\frac{dV}{dT}\) is given by \(\frac{dV}{dT} = \beta V \Delta T\), where \(\beta\) is the
coefficient of volume expansion. As you can show in Exercise, \(\beta = 3\alpha\). This equation is usually written as
\[\Delta V = \beta V \Delta T.\]
Note that the values of \(\beta\) in Table \(\PageIndex{1}\) are equal to \(3\alpha\) except for rounding.
Volume expansion is defined for liquids, but linear and area expansion are not, as a liquid’s changes in linear dimensions and area depend on the shape of its container. Thus, Table \(\PageIndex{1}\) shows liquids’ values of \(\beta\) but not \(\alpha\).
In general, objects expand with increasing temperature. Water is the most important exception to this rule. Water does expand with increasing temperature (its density
decreases) at temperatures greater than \(4^oC (40^oF)\). However, it is densest at \(+4^oC\) and expands with decreasing temperature between \(+4^oC\) and \(0^oC\) (\(40^oF \, to \, 32^oF\)), as shown in Figure \(\PageIndex{4}\). A striking effect of this phenomenon is the freezing of water in a pond. When water near the surface cools down to \(4^oC\), it is denser than the remaining water and thus sinks to the bottom. This “turnover” leaves a layer of warmer water near the surface, which is then cooled. However, if the temperature in the surface layer drops below \(4^oC\), that water is less dense than the water below, and thus stays near the top. As a result, the pond surface can freeze over. The layer of ice insulates the liquid water below it from low air temperatures. Fish and other aquatic life can survive in \(4^oC\) water beneath ice, due to this unusual characteristic of water.
Figure \(\PageIndex{4}\): This curve shows the density of water as a function of temperature. Note that the thermal expansion at low temperatures is very small. The maximum density at \(4^oC\) is only \(0.0075 \%\) greater than the density at \(2^oC\), and \(0.012 \%\) greater than that at \(0^oC\). The decrease of density below \(4^oC\) occurs because the liquid water approachs the solid crystal form of ice, which contains more empty space than the liquid.
Example \(\PageIndex{2}\): Calculating Thermal Expansion
Suppose your 60.0-L (\(15.9\)-gal) steel gasoline tank is full of gas that is cool because it has just been pumped from an underground reservoir. Now, both the tank and the gasoline have a temperature of \(15.0^oC\). How much gasoline has spilled by the time they warm to \(35.0^oC\)?
Strategy
The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank. (The gasoline tank can be treated as solid steel.)
Solution Use the equation for volume expansion to calculate the increase in volume of the steel tank: \[\Delta V_s = \beta_s V_s \Delta T. \nonumber\] The increase in volume of the gasoline is given by this equation: \[\Delta V_{gas} = \beta_{gas} V_{gas} \Delta T. \nonumber\] Find the difference in volume to determine the amount spilled as \[V_{spill} = \Delta V_{gas} - \Delta V_s. \nonumber\]
Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.)
\[ \begin{align*} V_{spill} &= (\beta_{gas} - \beta_s) V \Delta T \\[5pt] &= [(950 - 35) \times 10^{-6}/^oC](60.0 \, L)(20.0 \, ^oC) \\[5pt] &= 1.10 \, L. \end{align*} \]
Significance
This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed later in this chapter.
If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist compression with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them.
Exercises \(\PageIndex{1}\)
Does a given reading on a gasoline gauge indicate more gasoline in cold weather or in hot weather, or does the temperature not matter?
Answer
The actual amount (mass) of gasoline left in the tank when the gauge hits “empty” is less in the summer than in the winter. The gasoline has the same volume as it does in the winter when the “add fuel” light goes on, but because the gasoline has expanded, there is less mass.
Thermal Stress
If you change the temperature of an object while preventing it from expanding or contracting, the object is subjected to stress that is compressive if the object would expand in the absence of constraint and tensile if it would contract. This stress resulting from temperature changes is known as
thermal stress. It can be quite large and can cause damage.
To avoid this stress, engineers may design components so they can expand and contract freely. For instance, in highways, gaps are deliberately left between blocks to prevent thermal stress from developing. When no gaps can be left, engineers must consider thermal stress in their designs. Thus, the reinforcing rods in concrete are made of steel because steel’s coefficient of linear expansion is nearly equal to that of concrete.
To calculate the thermal stress in a rod whose ends are both fixed rigidly, we can think of the stress as developing in two steps. First, let the ends be free to expand (or contract) and find the expansion (or contraction). Second, find the stress necessary to compress (or extend) the rod to its original length by the methods you studied in Static Equilibrium and Elasticity on static equilibrium and elasticity. In other words, the \(\Delta L\) of the thermal expansion equals the \(\Delta L\) of the elastic distortion (except that the signs are opposite).
Example \(\PageIndex{3}\): Calculating Thermal Stress
Concrete blocks are laid out next to each other on a highway without any space between them, so they cannot expand. The construction crew did the work on a winter day when the temperature was \(5^oC\). Find the stress in the blocks on a hot summer day when the temperature is \(38^oC\). The compressive Young’s modulus of concrete is \(Y = 20 \times 10^9 \, N/m^2\).
Strategy
According to the chapter on static equilibrium and elasticity, the stress
F/ A is given by \[\dfrac{F}{A} = Y \dfrac{\Delta L}{L_0},\] where Y is the Young’s modulus of the material—concrete, in this case. In thermal expansion, \(\Delta L = \alpha L_0 \delta T\). We combine these two equations by noting that the two \(\Delta L\)'s are equal, as stated above. Because we are not given \(L_0\) or A, we can obtain a numerical answer only if they both cancel out. Solution
We substitute the thermal-expansion equation into the elasticity equation to get
\[ \begin{align*} \dfrac{F}{A} &= Y\dfrac{\alpha L_0 \Delta T}{L_0} \\[5pt] &= Y \alpha \Delta T, \end{align*} \]
and as we hoped, \(L_0\) has canceled and
A appears only in F/ A, the notation for the quantity we are calculating.
Now we need only insert the numbers:
\[ \begin{align*} \dfrac{F}{A} = (20 \times 10^6 \, N/m^2)(12 \times 10^{-6} /^oC)(38^oC - 5^oC) \\[5pt] &= 7.9 \times 10^6 \, N/m^2. \end{align*} \]
Significance The ultimate compressive strength of concrete is \(20 \times 10^6 \, N/m^2\), so the blocks are unlikely to break. However, the ultimate shear strength of concrete is only \(2 \times 10^6 \, N/m^2\), so some might chip off.
Exercise \(\PageIndex{2}\)
Two objects
A and B have the same dimensions and are constrained identically. A is made of a material with a higher thermal expansion coefficient than B. If the objects are heated identically, will A feel a greater stress than B? Answer:
Not necessarily, as the thermal stress is also proportional to Young’s modulus.
Contributors
Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).
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I am struggling with question about possible outcomes of momentum measurement and their probability. I know I can calculate it with momentum operator, but a wavefunction is of form
$$\psi (x)=3\cos\pi x+\cos3\pi x$$
and I am unsure how to deal with it, as the derivative consists of sines.
I know that $\cos{kx}=\frac{e^{ikx}+e^{−ikx}}{2}$ and $p_{x}=\hbar k$, but does it mean that the momentum is sum of eigenvalues of individual exponentials?
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$$ q \propto \Delta{T} $$
Heat is energy transferred due to a difference in temperature:
There are three modes of heat transfer:
In a general heating process, all modes of heat transfer may occur at the same time. A good example is the heating of a tin can of water by using a Bunsen burner:
The heat transferred by conduction is given by
Fourier's law. Considering only the simplest case of heat transfer by a single direction (i.e. along a homogeneous longitudinal plane), the Fourier's law of heat transfer can be written as:
where:
The negative sign is required to make the heat-flow positive when $T_1$ is higher than $T_2$.
The heat transfer by conduction can be easily examplified by looking at the cross section of a wall, where at the two sides there is a difference of temperature:
The
thermal resistance that create the gradient of temperature shown in the diagram above is given by:
The Fourier's law becomes:$$ \frac{dq}{dt} = - \frac{dT}{R} $$
Materials can be classified as insulators (high thermal resistance) or conductive (low thermal resistance). Examples of highly conductive materials are:
Material Thermal conductivity (W/ m $\cdot$ K) Application Copper 400 Distilleries Alluminium 200 Packaging Silver 430 Electronics Stainless steel 15 Heat exchangers
Instead, examples of insulating materials are:
Material Thermal resistance (W/K) Application Cork 0.04 Building Wood oak 0.20 Building Wool 0.04 Insulating walls Urethane foams 0.02 Insulating walls Air 0.03 Building
Convection is the transfer of energy by conduction and radiation inmoving, fluid media. The motion of the fluid is an essential part ofconvective heat transfer. In many cases, heat is transferred from one fluid, through a solid wall, to another fluid. Such transfer occurs typically in a heat exchanger. The amount of heat transferred by convection between a surface and a fluid is given by
Newton's law of cooling:
where:
The heat transfer coefficient depends on the movement of the fluid. Examples of $h$ values are shown below:
Movement Laminar heat transfer (W/ m$^2 \cdot$ K) Application Natural convection of air 5-10 Drier Forced convection of air 25-300 Convective oven Boiling water 1.500 - 57.000 Reactors Condensing steam, film 5.700 - 17.000 Evaporator Condensing steam, spray 30.000 - 100.000 Spray drier
This case is typical of the plate heat exchangers used for heating or cooling fluids. The problem is exemplified with the following diagram:
The case represented by the diagram above is a typical example of heat transfer during cooking. We can observe four temperatures (from $T_1$ to $T_4$):
Often, manufacturers of food machinery provides the
overall heat transfer coefficient (U), which is the value at the denominator of the previous equation:
With the knowledge of such parameter, it is possible to use again he Fourier's law of heat transfer:$$ \dfrac{dq}{dt} = A \cdot U \cdot \Delta{T} $$
where:
Heat exchangers are devices used to heat or cool fluids. It is possible to classify heat exchangers in two classes:
Indirect heat exchangers are the most common. The product is heated thanks to heat transfer from a hot fluid (generally hot water or steam), which is not in direct contact, but separated by a wall material.
Examples of indirect heat exchangers are:
Direct heat exchangers heat the product by direct mixing with the hot fluid (generally steam). This causes a flash heating of the product, but also a dilution with the condensed water.
Instead, the diagram below shows an example of tubular heat exchanger that is running with a co-current flow:
In general, the use of one or the other flow mode is determined by the type of application. Counter current heat exchangers offer the highest efficiency. With such configuration the heat exchange is always maximized. Moreover, the variation of temperature between the hot fluid and the product is minimized. This means that the heating is uniform along the process and as gentle as possible. Gentle heating is important to minimize the fouling mechanism.
However, co-curent heat exchangers are very important for applications where you need to quickly heat the product. With this mode of flow, as soon as the product enters into the heat exchanger, it is heated up very quickly thanks to the large difference of temperature between the hot fluid and the product. This condition is desired for evaporation process, where large amount of water must be removed from the product. The fact that along the heat exchanger the difference of temperature is reduced (i.e. lower heat transfer capacity) is also desired since, during evaporation, the product decresaes its heat capacity coefficient $C_p$. This means, in practice, that to rise its temperature along the evaporation process is required a progressively lower amount of energy. Conversely, if the same difference of temperature between the hot fluid and the product is maintained along the evaporation process, the drier product will easily burn at the contact with the metal plate, causing fouling.
For design purposes, the mode of flow affects the $\Delta{T}$ parameters.In particular for co-current heat exchangers, the changes of $\Delta{T}$ along the heating process can be relevant. Accordingly, to continue using the equation of heat transfer described before, it is often useful to express an average $\Delta{T}$. In case of counter-current heat exchangers, it is possible to estimate the $\Delta{T}$ with the arithmetic mean. Alternatively, it is very common to use a logarithmic mean temperature difference, often called as
log mean temperature difference (LMTD).This is expressed as follow:
where:
Very often, the amount of energy required for heating or cooling a product can be reused. For instance, the product that exits from the het exchanger has a high temperature that can be conveniently used to heat the product entering to the heat exchanger. A typical example of this situation is displayed in the following diagram, where three heat exchangers are used to heat and cool the product:
Design a sterilization process based on scheme shown above. The juice has a $C_p$ of 3.8 kJ/kg$\cdot$°C. Flow rate is 10 ton per hour. Initial temperature of the juice is 5°C. Sterilization temperature is 140°C. At the outlet of step (I), the temperature of the juice is cooled down to 45°C and, then, down to 5°C thanks to the use of ice-water. Ice-water enters into the heat exchanger at 0°C and exits at 20°C. The heating medium during the sterilization process is steam at 145°C. The overall heat transfer (U) is assumed as 3.000 W/m$^2 \cdot$°C
Determine:
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Update: I went over this answer and clarified some parts. Most importantly I expanded the Forces section to connect better with the question.
I like your reasoning and you actually come to the right conclusions, so congratulations on that! But understanding the relation between forces and particles isn't that simple and in my opinion the best one can do is provide you with the bottom-up description of how one arrives to the notion of force when one starts with particles. So here comes the firmware you wanted. I hope you won't find it too long-winded.
Particle physics
So let's start with particle physics. The building blocks are particles and interactions between them. That's all there is to it. Imagine you have a bunch of particles of various types (massive, massless, scalar, vector, charged, color-charged and so on) and at first you could suppose that all kinds of processes between this particles are allowed (e.g. three photons meeting at a point and creating a gluon and a quark; or sever electrons meeting at a point and creating four electrons a photon and three gravitons). Physics could indeed look like this and it would be an incomprehensible mess if it did.
Fortunately for us, there are few organizing principles that make the particle physics reasonably simple (but not
too simple, mind you!). These principles are known as conservation laws. After having done large number of experiments, we became convinced that electric charged is conserved (the number is the same before and after the experiment). We have also found that momentum is conserved. And lots of other things too. This means that processes such as the ones I mentioned before are already ruled out because they violate some if these laws. Only processes that can survive (very strict) conservation requirements are to be considered possible in a theory that could describe our world.
Another important principle is that we want our interactions simple. This one is not of experimental nature but it is appealing and in any case, it's easier to start with simpler interactions and only if that doesn't work trying to introduce more complex ones. Again fortunately for us, it turns out basic interactions are very simple. In a given interaction point there is always just a small number of particles. Namely:
two: particle changing direction three: particle absorbing another particle, e.g. $e^- + \gamma \to e^-$ or one particle decaying to two other particles $W^- \to e^- + \bar\nu_e$ four: these ones don't have as nice interpretation as the above ones; but to give an example anyone, one has e.g. two gluons going in and two gluons going out
So one example of such a simple process is electron absorbing a photon. This violates no conservation law and actually turns out to be the building block of a theory of electromagnetism. Also, the fact that there is a nice theory for this interaction is connected to the fact that the charge is conserved (and in general there is a relation between conservation of quantities and the way we build our theories) but this connection is better left for another question.
Back to the forces
So, you are asking yourself what was all that long and boring talk about, aren't you? The main point is: our world (as we currently understand it) is indeed described by all those different species of particles that are omnipresent everywhere and interact by the bizarre interactions allowed by the conservation laws.
So when one wants to understand electromagnetic force all the way down, there is no other way (actually, there is one and I will mention it in the end; but I didn't want to over-complicate the picture) than to imagine huge number of photons flying all around, being absorbed and emitted by charged particles all the time.
So let's illustrate this on your problem of Coulomb interaction between two electrons. The complete contribution to the force between the two electrons consists of all the possible combination of elementary processes. E.g. first electron emits photon, this then flies to the other electron and gets absorbed, or first electron emits photon, this changes to electron-positron pair which quickly recombine into another photon and this then flies to the second electron and gets absorbed. There is huge number of these processes to take into account but actually the simplest ones contribute the most.
But while we're at Coulomb force, I'd like to mention striking difference to the classical case. There the theory tells you that you have an EM field also when one electron is present. But in quantum theory this wouldn't make sense. The electron would need to emit photons (because this is what corresponds to the field) but they would have nowhere to fly to. Besides, electron would be losing energy and so wouldn't be stable. And there are various other reasons while this is not possible.
What I am getting at is that a single electron doesn't produce any EM field until it meets another charged particle! Actually, this should make sense if you think about it for a while. How do you detect there is an electron if nothing else at all is present? The simple answer is: you're out of luck, you won't detect it. You always need some test particles. So the classical picture of an electrostatic EM field of a point particle describes only what
would happen if another particle would be inserted in that field.
The above talk is part of the bigger bundle of issues with measurement (and indeed even of the very definition of) the mass, charge and other properties of system in quantum field theory. These issues are resolved by the means of renormalization but let's leave that for another day.
Quantum fields
Well, turns out all of the above talk about particles (although visually appealing and technically very useful) is just an approximation to the more precise picture of there existing just one quantum field for every particle type and the huge number of particles everywhere corresponding just to sharp local peaks of that field. These fields then interact by the means of quite complex interactions that reduce to the usual particle stuff when once look what those peaks are doing when they come close together.
This field view can be quite enlightening for certain topics and quite useless for others. One place where it is actually illuminating is when one is trying to understand to spontaneous appearance of so-called virtual particle-antiparticle pairs. It's not clear where do they appear from as particles. But from the point of view of the field, they are just local excitations. One should imagine quantum field as a sheet that is wiggling around all the time (by the means of inherent quantum wigglage) and from time to time wiggling hugely enough to create a peak that corresponds to the mentioned pair.This post imported from StackExchange Physics at 2014-04-01 16:26 (UCT), posted by SE-user Marek
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Here is a problem I don't even know how to approach: For arbitrary integers a,b,c which satisfy $(a,b)=(b,c)=1$ and $b>1$ there exist infinitely many integers n such that $ab^n+c$ is not prime. Any help with this problem will be appreciated.
closed as off-topic by Matthew Conroy, C. Falcon, Juniven, user91500, user223391 Apr 11 '17 at 21:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
" This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Matthew Conroy, C. Falcon, Juniven, user91500, Community
It's easy to show $\gcd(b, ab+c)=1$, from which using Euler's theorem $$b^{\varphi(ab+c)}\equiv 1 \pmod{ab+c} \Rightarrow b^{\varphi(ab+c)+1}\equiv b \pmod{ab+c} \tag{1}$$ or $$ab^{\varphi(ab+c)+1}+c\equiv ab+c \equiv 0 \pmod{ab+c}$$ which means $$ab+c \mid ab^{\varphi(ab+c)+1}+c$$ but, using $(1)$, this is also true for $$ab^{2\varphi(ab+c)+1}+c\equiv ab^{{\varphi(ab+c)+1}}+c \equiv ab+c \equiv 0 \pmod{ab+c}$$ and by induction $$ab^{k\cdot\varphi(ab+c)+1}+c \equiv 0 \pmod{ab+c}, \forall k \geq 1$$ i.e. there will be infinitely many $ab^n+c$ divisible by $ab+c$ and thus not primes.
Suppose $p$ is a prime factor of $ab+c$, then by Fermat's little Theorem, $p$ is a prime factor of $ab^{(p-1)N+1}+c$.
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A
closed-loop transfer function in control theory is a mathematical expression ( algorithm) describing the net result of the effects of a closed ( feedback) loop on the input signal to the circuits enclosed by the loop. Overview [ edit ]
The closed-loop
transfer function is measured at the output. The output signal waveform can be calculated from the closed-loop transfer function and the input signal waveform.
An example of a closed-loop transfer function is shown below:
The summing node and the
G( s) and H( s) blocks can all be combined into one block, which would have the following transfer function: Y ( s ) X ( s ) = G ( s ) 1 + G ( s ) H ( s ) {\displaystyle {\dfrac {Y(s)}{X(s)}}={\dfrac {G(s)}{1+G(s)H(s)}}} Derivation [ edit ]
We define an intermediate signal Z shown as follows:
Using this figure we write:
Y ( s ) = G ( s ) Z ( s ) {\displaystyle Y(s)=G(s)Z(s)} Z ( s ) = X ( s ) − H ( s ) Y ( s ) {\displaystyle Z(s)=X(s)-H(s)Y(s)} Y ( s ) = G ( s ) ( X ( s ) − H ( s ) Y ( s ) ) = G ( s ) X ( s ) − G ( s ) H ( s ) Y ( s ) {\displaystyle Y(s)=G(s)(X(s)-H(s)Y(s))=G(s)X(s)-G(s)H(s)Y(s)} Y ( s ) + G ( s ) H ( s ) Y ( s ) = G ( s ) X ( s ) {\displaystyle Y(s)+G(s)H(s)Y(s)=G(s)X(s)} Y ( s ) ( 1 + G ( s ) H ( s ) ) = G ( s ) X ( s ) {\displaystyle Y(s)(1+G(s)H(s))=G(s)X(s)} ⇒ Y ( s ) X ( s ) = G ( s ) 1 + G ( s ) H ( s ) {\displaystyle \Rightarrow {\dfrac {Y(s)}{X(s)}}={\dfrac {G(s)}{1+G(s)H(s)}}} See also [ edit ] References [ edit ]
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So, this curiosity has arisen for a fun project I thought I'd tackle, where I'm attempting create a crude simulation of an internal combustion cylinder.
I wanted to explore the particle level interactions of a 4 step methane reaction, and see how they effect macroscopic properties like cylinder wall and force on the piston head. So the initial thought was simulate a set of particles modeled as hard spheres bouncing around, which I've found, are the general assumptions in kinetic theory. My background is with the macroscopic (engineering) and I'm trying to understand how these particle interactions of these relate to my existing macro knowledge.
I've read up on the Maxwell-Boltzmann distribution, which makes sense that for a collection of particles at a given temperature there is a probability of finding particles at a given speed in a small box.
$$ f(v) = \sqrt{\frac{m}{2\pi KT}^{3}} 4\pi v^2 e^{\frac{-mv^2}{2kT}} $$
But how is temperature defined for an individual particle, is there such a thing? I was also looking into other definitions of temperature which don't seem to apply to individual particles such as:
$$ 1/T = \frac{\delta S}{\delta U} $$
If anything this even more confusing because, again I'm not sure how these terms are defined for individual particles. So is temperature some type of scalar property in addition to velocity information? My understanding is particle velocity and temperature are independent properties.
Finally, I was hoping to take some properties of my particle and on collision have some model to determine if a reaction would or wouldd not occur. How is activation energy related to velocity and temperature on a particle level, or can individual reactions not be modeled from a Newtonian stand point?
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I have a question regarding capacitor values within a power distribution system. Unfortunately I believe that I have to give a bit of background before I can ask this question.
As stated in both the forum post and the app note the physical geometry of a capacitor dictates the self-inductance. In the case of decoupling the capacitor can be modeled as a small power supply with internal resistance, inductance and capacitance. In the frequency domain the view of the internal impedance of the capacitor is a "trough" where the beginning (zero) of the trough is dictated by the capacitance value and the end (pole) is from the parasitic-inductance. The lowest point of the trough is set by either the parasitic resistance or the lowest value of the resonance frequency of the LC combination of the capacitor / parasitic inductance value (whichever produces a higher impedance).
The following is an image illustrating the characteristics of a capacitor
here is the equation for the resonance frequency. $$ \frac{1}{2\pi \sqrt{L \times C}} $$ -Thanks for catching that Olin
By this reasoning one can choose the largest size capacitor in the given package size, for example 0402, and the properties of the pole will not change and only the zero will be moved to a lower frequency (in the image, the downward slope would be moved to the left for large capacitor values) allowing a wider bandwidths of frequency to be bypassed. The resonant pole that defines the upper portion of the capacitor should encompass any higher value capacitor of the same package size.
Later on in the app note there is a section called "Capacitor Placement" where, as described in Olin's response, the effectiveness of the capacitor doesn't just concern the inductance of the cap, but also has to do with the placement of the cap. In colloquial terms the problem is this: As an IC begins to draw more power the voltage begins to sag, the time it takes for that sag to be seen by the decoupling capacitor is determined by the propagation speed of the material that the signal (voltage drop) must travel, basically closer is better. An example is done within the app note which is as follows
0.001uF X7R ceramic chip capacitor, 0402 package Lis = 1.6 nH (theoretical inductance of both parasitic self-inductance, and board inductance)
The resonance frequency at which the capacitor has the lowest impedance is given as $$ Fris = \frac{1}{2\pi \sqrt{L \times C}} $$ $$ Fris = \frac{1}{2\pi \sqrt{1.6\times10^-9 \times 0.001\times10^-6}} = 125.8MHz $$
The period of this frequency is Tris
$$ Tris = \frac{1}{Fris} $$ $$ Tris = \frac{1}{125.8\times10^6} = 7.95ns$$
In order for a capacitor to be effective it needs to be able to respond faster than the voltage can sag on a pin. If the voltage sag were to happen faster than 7.95ns than there would be some time between the dip on the pin and the capacitors capability to respond to that dip manifesting in voltage spikes the can possible drop the voltage down to a point of brown out, or reset. In order for the capacitor to remain effective the voltage change must happen at a slower rate then some fraction of the resonant period (Tris). To quantize this statement an accepted effective response time of a capacitor is 1/40th of the resonance frequency, so the effective frequency of this capacitor is really
$$ Effective Fris = \frac{125.8\times10^6}{40} = 3.145MHz $$
or the capacitor will be able to cover a dip that occurs over a .318uS period.
$$ Effective Tris = \frac{1}{3.145\times10^6} = .318us $$
Unfortunately a capacitor cannot usually be placed on top of a pin so there is another delay contributed by the material the PCB is composed of. This delay can be modeled as a propagation speed of the material. In the app note the propagation speed of a standard FR4 dielectric is 166ps per inch.
Using the effective resonance period (Tris) from above and the propagation speed of the material we can find the distance at which the capacitor remains effective at the Effective Fris.
$$ Distance(x) = \frac{time(t)}{speed(\frac{t}{x})} $$ $$ Distance(x) = \frac{.318\times10^-6}{1.66\times10^-12} = 1.20in$$ or about 3.0cm
Finally I can ask my question!
Since the package size is the part of the cap that mitigates the pole or the upper bound of the impedance of the modeled power supply, then it shouldn't matter if I were to use a 0.001uF cap 0402 package, or a 0.47uF capacitor 0402 package. A better method to determine the Fris of the cap is to find the frequency at which either the internal resistance or the effective capacitance intersects with the pole (whichever point is higher). Is this correct? or is there some other factor that I have not taken into consideration?
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Let F be a field of characteristic $p > 0$. Let $g$ be a linear lie algebra, that is there exists a natural number $n$ such that $g \subset M_n(F)$. Does there exist a condition involving $n$ and $p$ such that $g$ is semisimple if and only if its killing form is non-degenerate?
The short answer is no. In prime characteristic, the Killing form sometimes behaves badly even for simple Lie algebras. If "semisimple" means that the solvable radical is zero, there is no way to obtain the classical equivalences with non-degeneracy of the Killing form and with the direct sum decomposition into simples. Moreover, the simple Lie algebras have only recently been classified when $p=5$ (Premet-Strade), while for $p=2,3$ little is known and for $p>5$ the classification takes an enormous amount of work (by Block-Wilson and others). Conditions on the dimension of a faithful representation or on the prime are not enough to sort out the concept of semisimplicity. Still, a lot is known. For example, Seligman and others explored in the 1960s the class of modular Lie algebras for which the Killing form is non-degenerate.
ADDED: Much more could be said along these lines, but for older results see the book
Modular Lie Algebras by G.B. Seligman (Springer, 1967). Modular Lie algebras have been of much less importance overall than linear algebraic groups, whose Lie algebras are "restricted" (have a nice $p$th power operation) but still don't reflect precisely the group structure or representation theory. Moreover, the representations of simple or more generally semisimple Lie algebras in prime characteristic are poorly understood even for those arising from Lie algebras of semisimple algebraic groups. So working with a fixed $p$ and fixed $n$ will usually not be illuminating.
If $g\subset M_n(F)$ and $n\le p-2$ then $g$ is semisimple if and only if the Killing form of g is non-degenerate. This statement is clear (and empty) if $p\in\{2,3\}$, while for $p>3$ one can use various results of the theory of modular Lie algebras. First one may assume that $g$ is semisimple as the converse statement is an easy exercise. One may also assume that $F$ is algebraically closed. Using Block's theorem one the structure of semisimple Lie algebras we then observe that $g$ is sandwiched between $\bigoplus_i S_i$ and $\bigoplus_i{\rm Der}(S_i)$ where the $S_i$'s are simple Lie algebras. As $n\le p-2$ the Classification Theorem implies that each $S_i$ is isomorphic to the Lie algebra of a simple algebraic $F$-group. Looking at the ranks of these groups more closely (and again using the inequality $n\le p-2$) we deduce that each $S_i$ has a non-degenerate Killing form and ${\rm Der}(S_i)=\mathrm{ad}\, S_i$ (this is discussed in Seligman's book quoted earlier). Hence the Killing form of $g\cong\bigoplus_i S_i$ is non-degenerate as well. The assumption on $p$ cannot be relaxed as for $p>3$ the Witt algebra $g={\rm Der}\,\mathcal{O}$ where $\mathcal{O}=F[X]/(X^p)$ is simple, has the zero Killing form, and acts irreducibly on the vector space $\mathcal{O}/F1$ of dimension $p-1$.
This started out as a comment on Jim's answer, but it got too long (EDIT : It looks like Jim edited his answer to say some of this while I wrote this up).
First, I recommend Seligman's book "Modular Lie Algebras" for a nice account of what happens to Lie algebras in positive char if you assume that the Killing form is nondegenerate.
It's maybe also worth giving an example to show what can happen (there might be easier ones, but this came up in a paper I wrote a while ago and gave me no end of headaches). Consider $\mathfrak{sp}_{2g}(\mathbb{Z}/p\mathbb{Z})$. It is not hard to show that this is simple for $p > 2$. For instance, an argument is contained in the paper
N. Jacobson, Classes of restricted Lie algebras of characteristic p. I, Amer. J. Math. 63 (1941), 481–515.
By the way, this is false for $p=2$. Anyway, one can calculate the Killing form, and it turns out to be degenerate exactly when $p$ divides $g+1$. One interesting observation, however, is that we are really looking at the wrong bilinear form. Elements of $\mathfrak{sp}_{2g}(\mathbb{Z}/p\mathbb{Z})$ are matrices, so we can define a blinear form
$$(A,B) = \text{Tr}(AB).$$
This is NOT the Killing form, as we are not looking at the adjoint representation. One can then show that this bilinear form is nondegenerate for all $p>2$.
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Learning Objectives
By the end of this section, you will be able to:
Calculate the Reynolds number for an object moving through a fluid. Explain whether the Reynolds number indicates laminar or turbulent flow. Describe the conditions under which an object has a terminal speed.
A moving object in a viscous fluid is equivalent to a stationary object in a flowing fluid stream. (For example, when you ride a bicycle at 10 m/s in still air, you feel the air in your face exactly as if you were stationary in a 10-m/s wind.) Flow of the stationary fluid around a moving object may be laminar, turbulent, or a combination of the two. Just as with flow in tubes, it is possible to predict when a moving object creates turbulence. We use another form of the Reynolds number
N′ R, defined for an object moving in a fluid to be
[latex]{N′}_{\text{R}}=\frac{\rho vL}{\eta}\text{(object in fluid)}\\[/latex],
where
L is a characteristic length of the object (a sphere’s diameter, for example), ρ the fluid density, η its viscosity, and v the object’s speed in the fluid. If N′ R is less than about 1, flow around the object can be laminar, particularly if the object has a smooth shape. The transition to turbulent flow occurs for N′ R between 1 and about 10, depending on surface roughness and so on. Depending on the surface, there can be a turbulent wake behind the object with some laminar flow over its surface. For an N′ R between 10 and 10 6, the flow may be either laminar or turbulent and may oscillate between the two. For N′ R greater than about 10 6, the flow is entirely turbulent, even at the surface of the object. (See Figure 1.) Laminar flow occurs mostly when the objects in the fluid are small, such as raindrops, pollen, and blood cells in plasma. Example 1. Does a Ball Have a Turbulent Wake?
Calculate the Reynolds number
N′ R for a ball with a 7.40-cm diameter thrown at 40.0 m/s. Strategy
We can use [latex]{N′}_{\text{R}}^{}=\frac{\rho \text{vL}}{\eta}\\[/latex] to calculate
N′ R, since all values in it are either given or can be found in tables of density and viscosity. Solution
Substituting values into the equation for
N′ R yields
[latex]\begin{array}{lll}{N′}_{R}^{}& =& \frac{\rho \text{vL}}{\eta }=\frac{\left(1\text{.}\text{29}{\text{ kg/m}}^{3}\right)\left(\text{40.0 m/s}\right)\left(\text{0.0740 m}\right)}{1.81\times {\text{10}}^{-5}1.00 \text{ Pa}\cdot \text{s}}\\ & =& 2.11\times 10^{5}\end{array}\\[/latex]
Discussion
This value is sufficiently high to imply a turbulent wake. Most large objects, such as airplanes and sailboats, create significant turbulence as they move. As noted before, the Bernoulli principle gives only qualitatively-correct results in such situations.
One of the consequences of viscosity is a resistance force called
viscous drag F V that is exerted on a moving object. This force typically depends on the object’s speed (in contrast with simple friction). Experiments have shown that for laminar flow ( N′ R less than about one) viscous drag is proportional to speed, whereas for N′ R between about 10 and 10 6, viscous drag is proportional to speed squared. (This relationship is a strong dependence and is pertinent to bicycle racing, where even a small headwind causes significantly increased drag on the racer. Cyclists take turns being the leader in the pack for this reason.) For N′ R greater than 10 6, drag increases dramatically and behaves with greater complexity. For laminar flow around a sphere, F V is proportional to fluid viscosity η, the object’s characteristic size L, and its speed v. All of which makes sense—the more viscous the fluid and the larger the object, the more drag we expect. Recall Stoke’s law F S = 6 πrηv. For the special case of a small sphere of radius R moving slowly in a fluid of viscosity η, the drag force F S is given by F S = 6 πRηv.
An interesting consequence of the increase in
F V with speed is that an object falling through a fluid will not continue to accelerate indefinitely (as it would if we neglect air resistance, for example). Instead, viscous drag increases, slowing acceleration, until a critical speed, called the terminal speed, is reached and the acceleration of the object becomes zero. Once this happens, the object continues to fall at constant speed (the terminal speed). This is the case for particles of sand falling in the ocean, cells falling in a centrifuge, and sky divers falling through the air. Figure 2 shows some of the factors that affect terminal speed. There is a viscous drag on the object that depends on the viscosity of the fluid and the size of the object. But there is also a buoyant force that depends on the density of the object relative to the fluid. Terminal speed will be greatest for low-viscosity fluids and objects with high densities and small sizes. Thus a skydiver falls more slowly with outspread limbs than when they are in a pike position—head first with hands at their side and legs together. Take-Home Experiment: Don’t Lose Your Marbles
Knowledge of terminal speed is useful for estimating sedimentation rates of small particles. We know from watching mud settle out of dirty water that sedimentation is usually a slow process. Centrifuges are used to speed sedimentation by creating accelerated frames in which gravitational acceleration is replaced by centripetal acceleration, which can be much greater, increasing the terminal speed.
Section Summary When an object moves in a fluid, there is a different form of the Reynolds number [latex]{N′}_{\text{R}}^{}=\frac{\rho \text{vL}}{\eta }\text{(object in fluid)}\\[/latex], which indicates whether flow is laminar or turbulent. For N′ Rless than about one, flow is laminar. For N′ Rgreater than 10 6, flow is entirely turbulent. Conceptual Questions
1. What direction will a helium balloon move inside a car that is slowing down—toward the front or back? Explain your answer.
2. Will identical raindrops fall more rapidly in 5º C air or 25º C air, neglecting any differences in air density? Explain your answer.
3. If you took two marbles of different sizes, what would you expect to observe about the relative magnitudes of their terminal velocities?
Glossary viscous drag: a resistance force exerted on a moving object, with a nontrivial dependence on velocity terminal speed: the speed at which the viscous drag of an object falling in a viscous fluid is equal to the other forces acting on the object (such as gravity), so that the acceleration of the object is zero
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I was studying Pascal's Law which states(According to Wikipedia):-
Pascal's law is a principle in fluid mechanics that states that a pressure change occurring anywhere in a confined
"incompressible" fluid is transmitted throughout the fluid such that the same change occurs everywhere.
This principle is stated mathematically as:
${\displaystyle \Delta P=\rho g(\Delta h)\,}$
This made me think what would the pressure be if the fluid was compressible. Now as I did study(but do not have a good command on) about elasticity of materials in the chapter before fluids, so I thought Bulk modulus must come into play here, and knowing the following about the Bulk modulus I set a task for myself to calculate the pressure at a depth in a compressible liquid.
According to Wikipedia(again):-
The bulk modulus ( ${\displaystyle K}$ or ${\displaystyle B}$) of a substance is a measure of how incompressible/resistant to compressibility that substance is. It is defined as the ratio of the
"infinitesimal pressure increase" to the resulting relative decrease of the volume.
The infinitesimal pressure increase(or presure differential) part is posing a problem for me as you will see ahead in the post.
Consider a cuboid(made of water) of differential height $dh$ and cross sectional area $A(=lb)$ at a depth of $h$ below the free surface of water as shown in the diagram below:-
And as stated in Wikipedia:-
Following the basic premises of continuum mechanics, stress is a macroscopic concept. Namely, the particles considered in its definition and analysis should be just small enough to be treated as homogeneous in composition and state, but still large enough to ignore quantum effects and the detailed motions of molecules.
So, consider the height $dh$ of the cuboid so small such that we can consider the pressure to be constant for that height.
If $\rho(h)$ denotes the density of the liquid at a height $h$ then the force exerted on all the faces of the cuboid is given by
$$\int_{F_0}^{F}{dF}=\int_{0}^{y}{\rho(y)Ag\cdot{dy}}$$
As, cross-sectional area for all such cuboids will be same so we can say that the pressure exerted on the faces of all such cuboids will be
$$\int _{P_0}^{P}{dP}=\int_{0}^{y}{\rho(y)g\cdot{dy}}\tag{1}$$ (where $P_0$ is the atmospheric pressure at the surface of the fluid)
Now, as we know that the bulk modulus(isothermal) is given as:-
$$B=-V\dfrac{dP}{dV}$$
Now, going according to the Wikipedia statement, what I make out of the statement "infinitesimal change in the pressure that results in the infinitesimal decrease in volume" is that that the pressure that acts to compress the element is $\displaystyle\int_{P_0}^{P}{dP}$ (summing all the pressure differences) which produces a net compression in the volume of element, given by $\Delta V$. So we have the relation
$$\int_{P_0}^{P}{dP}=-B\dfrac{\Delta V}{V}\\ \implies \int_{0}^{y}{\rho(y)g\cdot{dy}}=-B\dfrac{\Delta V }{V}$$
Now, after arriving at this erroneous equality for which I see no way to proceed further, I thought of searching for the same(hoping that it doesn't deal with something which leaves me empty handed) and figuring out where I erred. But, all I could find was the expression for barometric equation which I think, is based on statistical mechanics while what I am dealing with is continuum mechanincs.
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The most elegant way I've seen to describe this is described in the paper The river model of black holes. If we write the Schwarzschild metric in Gullstrand-Painlevé coordinates we get (in units where $c = G = 1$):
$$ ds^2 = -dt_{ff}^2 + \left(dr + \beta dt_{ff} \right)^2 + r^2 d\Omega^2 $$
where:
$$ \beta = \sqrt{\frac{2M}{r}} $$
This looks like the Minkowski metric except that the radial coordinate is replaced by $dr + \beta dt_{ff}$. The parameter $\beta$ has the dimensions of velocity, and in fact it's equal to the Newtonian escape velocity. $\beta < 1$ outside the event horizon, $\beta = 1$ at the event horizon and $\beta > 1$ inside the event horizon.
The physical interpretation of this is that the radial coordinate is flowing inwards at a velocity of $\beta$. In effect space is flowing inwards into the black hole and carries observers along with it in the same way a river carries along observers floating on it - hence the name
river model. A light beam moves with velocity $1$ with respect to the spacetime around it, so relative to the observer at infinity the net velocity of a radial outbound light beam is:
$$ v_{eff} = 1 - \beta $$
So outside the event horizon $v_{eff} > 0$ and the light beam moves outwards. At the event horizon $v_{eff} = 0$ and the light beam is frozen at the horizon unable to move outwards. Inside the event horizon $v_{eff} < 0$ i.e. even if you shine the light directly outwards it still moves inwards towards the singularity.
As it happens I have addressed this issue before, in the question Why is a black hole black?. However that was a more algebraic approach and I think the river model approach is far more intuitive (insofar as anything in GR can be intuitive!).
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It seems me that there is a "difference" (at least apparently) in how the Belinfante-Rosenfeld tensor is thought of in section 7.4 of Volume 1 of Weinberg's QFT book and in section 2.5.1 of the conformal field theory book by di Francesco et. al.
I would be glad if someone can help reconcile the issue.
Schematically the main issue as I see is this -
If the action and the lagrangian density is writable as $ I = \int d^4x L$ and $\omega_{\mu \nu}$ be the small parameter of Lorentz transformation then Weinberg is thinking of $\omega_{\mu \nu}$ to be space-time independent and he is varying the action to write it in the form, $\delta I = \int d^4x (\delta L = (A^{\mu \nu})\omega_{\mu \nu}) $ Then some symmetrized form of whatever this $A^{\mu \nu}$ comes out to be is what is being called the Belinfante tensor. ( Its conservation needs the fields to satisfy the equations of motion)
But following Francesco et.al's set-up I am inclined to think of $\omega_{\mu \nu}$ as being space-time dependent and then the variation of the action will also pick up terms from the Jacobian and the calculation roughly goes as saying, $\delta I = \int (\delta(d^4x)) L + \int d^4x (\delta L)$. Since under rigid Lorentz transformations the volume element is invariant the coefficient of $\omega_{\mu \nu}$ in the first variation will vanish but the second variation will produce a coefficient say $B^{\mu \nu}$. But both the variations will produce a coefficient for the derivative of $\omega_{\mu \nu}$ and let them be $C^{\mu \nu \lambda}$ and $D^{\mu \nu \lambda}$ respectively.
Now the argument will be that since the original action was to start off invariant under Lorentz transformations, when evaluated about them, the $B^{\mu \nu}$ should be $0$ and on shifting partial derivatives the sumof $C^{\mu \nu \lambda}$ and $D^{\mu \nu \lambda}$ is the conserved current (..and its not clear whether their conservation needs the field to satisfy the equation of motion..)
So by the first way $A^{\mu \nu}$ will be the conserved current and in the second the conserved current will be, $C^{\mu \nu \lambda} + D^{\mu \nu \lambda}$ (the C tensor will basically look like $-x^\nu \eta^{\lambda \mu}L$)
Is the above argument correct?
If yes then are the two arguments equivalent?
How or is Weinberg's argument taking into account the contribution to the conserved current from the variation of the Jacobian of the transformation?
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2-manifolds and inverse limits of set-valued functions on intervals
1.
University of Auckland, Private Bag 92019, Auckland, New Zealand
2.
University of Oxford, Andrew Wiles Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford, OX2 6GG, United Kingdom
Suppose for each $n\in\mathbb{N}$, $f_n \colon [0,1] \to 2^{[0,1]}$ is a function whose graph $\Gamma(f_n) = \left\lbrace (x,y) \in [0,1]^2 \colon y \in f_n(x)\right\rbrace$ is closed in $[0,1]^2$ (here $2^{[0,1]}$ is the space of non-empty closed subsets of $[0,1]$). We show that the generalized inverse limit $\varprojlim (f_n) = \left\lbrace (x_n) \in [0,1]^\mathbb{N} \colon \forall n \in \mathbb{N},\ x_n \in f_n(x_{n+1})\right\rbrace$ of such a sequence of functions cannot be an arbitrary continuum, answering a long-standing open problem in the study of generalized inverse limits. In particular we show that if such an inverse limit is a 2-manifold then it is a torus and hence it is impossible to obtain a sphere.
Keywords:Generalized inverse limit, upper semicontinuous function, continuum, 2-manifold, closed relation. Mathematics Subject Classification:Primary: 54C08, 54E45; Secondary: 54F15, 54F65. Citation:Sina Greenwood, Rolf Suabedissen. 2-manifolds and inverse limits of set-valued functions on intervals. Discrete & Continuous Dynamical Systems - A, 2017, 37 (11) : 5693-5706. doi: 10.3934/dcds.2017246
References:
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E. Akin,
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R. Engelking,
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K. P. Hart, J. Nagata and J. E. Vaughan (eds.),
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R. Engelking,
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K. P. Hart, J. Nagata and J. E. Vaughan (eds.),
[10]
W. Hurewicz and H. Wallman,
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A. R. Pears,
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Tamara Fastovska.
Upper semicontinuous attractor for 2D Mindlin-Timoshenko thermoelastic model with memory.
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Dingheng Pi.
Limit cycles for regularized piecewise smooth systems with a switching manifold of codimension two.
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Agust Sverrir Egilsson.
On embedding the $1:1:2$ resonance space in a Poisson manifold.
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Armengol Gasull, Hector Giacomini.
Upper bounds for the number of limit cycles of some
planar polynomial differential systems.
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Seung Jun Chang, Jae Gil Choi.
Generalized transforms and generalized convolution products associated with Gaussian paths on function space.
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Zhuchun Li, Yi Liu, Xiaoping Xue.
Convergence and stability of generalized gradient systems by Łojasiewicz inequality with application in continuum Kuramoto model.
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Naeem M. H. Alkoumi, Pedro J. Torres.
Estimates on the number of limit cycles
of a generalized Abel equation.
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A. Pankov.
Gap solitons in periodic discrete nonlinear Schrödinger equations II: A generalized Nehari manifold approach.
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Duanzhi Zhang.
$P$-cyclic symmetric closed characteristics on compact convex $P$-cyclic symmetric hypersurface in
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Luca Dieci, Cinzia Elia.
Piecewise smooth systems near a co-dimension 2 discontinuity manifold: Can one say what should happen?.
2018 Impact Factor: 1.143
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Meshing Considerations for Linear Static Problems
In this blog entry, we introduce meshing considerations for linear static finite element problems. This is the first in a series of postings on meshing techniques that is meant to provide guidance on how to approach the meshing of your finite element model with confidence.
About Finite Element Meshing
The finite element mesh serves two purposes. It first subdivides the CAD geometry being modeled into smaller pieces, or
elements, over which it is possible to write a set of equations describing the solution to the governing equation. The mesh is also used to represent the solution field to the physics being solved. There is error associated with both the discretization of the geometry as well as discretization of the solution, so let’s examine these separately. Geometric Discretization
Consider two very simple geometries, a block and a cylindrical shell:
There are four different types of elements that can be used to mesh these geometries — tetrahedra (tets), hexahedra (bricks), triangular prismatics (prisms), and pyramid elements:
The grey circles represent the corners, or
nodes, of the elements. Any combination of the above four elements can be used. (For 2D modeling, triangular and quadrilateral elements are available.) You can see by examination that both of these geometries could be meshed with as few as one brick element, two prisms, three pyramids, or five tets. As we learned in the previous blog post about solving linear static finite element problems, you will always arrive at a solution in one Newton-Raphson iteration. This is true for linear finite element problems regardless of the mesh. So let’s take a look at the simplest mesh we could put on these structures. Here’s a plot of a single brick element discretizing these geometries:
The mesh of the block is obviously a perfect representation of the true geometry, while the mesh of the cylindrical shell appears quite poor. In fact, it only appears that way when plotted. Elements are always plotted on the screen as having straight edges (this is done for graphics performance purposes) but COMSOL usually uses a second-order Lagrangian element to discretize the geometry (and the solution). So although the element edges always appear straight, they are internally represented as:
The white circles represent the midpoint nodes of these second-order element edges. That is, the lines defining the edges of the elements are represented by three points, and the edges approximated via a polynomial fit. There are also additional nodes at the center of each of these quadrilateral faces and in the center of the volume for these second-order Lagrangian hexahedral elements (omitted for clarity). Clearly, these elements do a better job of representing the curved boundaries of the elements. By default, COMSOL uses second-order elements for most physics, the two exceptions are problems involving chemical species transport and when solving for a fluid flow field. (Since those types of problems are convection dominated, the governing equations are better solved with first-order elements.) Higher order elements are also available, but the default second-order elements usually represent a good compromise between accuracy and computational requirements.
The figure below shows the geometric discretization error when meshing a 90° arc in terms of the number of first- and second-order elements:
The conclusion that can be made from this is that at least two second-order elements, or at least eight first-order elements, are needed to reduce the geometric discretization error below 1%. In fact, two second-order elements introduce a geometric discretization error of less that 0.1%. Finer meshes will more accurately represent the geometry, but will take more computational resources. This gives us a couple of good practical guidelines:
When using first-order elements, adjust the mesh such that there are at least eight elements per 90° arc When using second-order elements, use two elements per 90° arc
With these rules of thumb, we can now estimate the error we’ve introduced by meshing the geometry, and we can do so with some confidence before even having to solve the model. Now let’s turn our attention to how the mesh discretizes the solution.
Solution Discretization
The finite element mesh is also used to represent the solution field. The solution is computed at the node points, and a polynomial basis is used to interpolate this solution throughout the element to recover the total solution field. When solving linear finite elements problems, we are always able to compute a solution, no matter how coarse the mesh, but it may not be very accurate. To understand how mesh density affects solution accuracy, let’s look at a simple heat transfer problem on our previous geometries:
A temperature difference is applied to opposing faces of the block and the cylindrical shell. The thermal conductivity is constant, and all other surfaces are thermally insulated.
The solution for the case of the square block is that the temperature field varies linearly throughout the block. So for this model, a single, first-order, hexahedral element would actually be sufficient to compute the true solution. Of course, you will rarely be that lucky!
Therefore, let’s look at the slightly more challenging case. We’ve already seen that the cylindrical shell model will have geometric discretization error due to the curved edges, so we would start this model with at least two second-order (or eight first-order) elements along the curved edges. If you look closely at the above plot, you can see that the element edges on the boundaries are curved, while the interior elements have straight edges.
Along the axis of the cylinder, we can use a single element, since the temperature field will not vary in this direction. However, in the radial direction, from the inside to outside surface, we also need to have enough elements to discretize the solution. The analytic solution for this case goes as \ln(r) and can be compared against our finite element solution. Since the polynomial basis functions cannot perfectly describe the function, let’s plot the error in the finite element solution for both the linear and quadratic elements:
What you can see from this plot is that, as you increase the number of elements in the model, the error goes down. This is a fundamental property of the finite element method: the more elements, the more accurate your solution. Of course, there is also a cost associated with this. More computational resources, both time and hardware, are required to solve larger models. Now, you’ll notice that there are no units to the
x-axis of this graph, and that is on purpose. The rate at which error decreases with respect to mesh refinement will be different for every model, and depends on many factors. The only important point is that it will always go down, monotonically, for well-posed problems.
You’ll also notice that, after a point, the error starts to go back up. This will happen once the individual mesh elements start to get very small, and we run into the limits of numerical precision. That is, the numbers in our model are smaller than can be accurately represented on a computer. This is an inherent problem with all computational methods, not just the finite element method; computers cannot represent all real numbers accurately. The point at which the error starts to go back up will be around \sqrt{2^{-52}} \approx 1.5 \times 10^{-8} and to be on the safe and practical side, we often say that the minimal achievable error is 10
-6. Thus, if we integrate the scaled difference between the true and computed solution over the entire model:
We say that the error, \epsilon, can typically be made as small as 10
-6 in the limits of mesh refinement. In practice, the inputs to our models will anyways usually have much greater uncertainty than this. Also keep in mind that in general we don’t know the true solution, we will instead have to compare the computed solutions between different sized meshes and observe what values the solution converges toward. Adaptive Mesh Refinement
I would like to close this blog post by introducing a better way to refine the mesh. The plots above show that error decreases as all of the elements in the model are made smaller. However, ideally you would only make the elements smaller in regions where the error is high. COMSOL addresses this via
Adaptive Mesh Refinement, which first solves on an initial mesh and iteratively inserts elements into regions where the error is estimated to be high, and then re-solves the model. This can be continued for as many iterations as desired. This functionality works with triangular elements in 2D and tetrahedrals in 3D. Let’s examine this problem in the context of a simple structural mechanics problem — a plate under uniaxial tension with a hole, as shown in the figure below. Using symmetry, only one quarter of the model needs to be solved.
The computed displacement fields, and the resultant stresses, are quite uniform some distance away from the hole, but vary strongly nearby. The figure below shows an initial mesh, as well as the results of several adaptive mesh refinement iterations, along with the computed stress field.
Note how COMSOL preferentially inserts smaller elements around the hole. This should not be a surprise, since we already know there will be higher stresses around the hole. In practice, it is recommended to use a combination of adaptive mesh refinement, engineering judgment, and experience to find an acceptable mesh.
Summary of Main Points You will always want to perform a mesh refinement study and compare results on different sized meshes Use your knowledge of geometric discretization error to choose as coarse a starting mesh as possible, and refine from there You can use adaptive mesh refinement, or your own engineering judgment, to refine the mesh Comments (5) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science TAGS CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
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Quadratic irrationality A root of a quadratic trinomial with rational coefficients which is irreducible over the field of rational numbers. A quadratic irrationality is representable in the form $a+b\sqrt{d}$, where $a$ and $b$ are rational numbers, $b\ne 0$, and $d$ is an integer which is not a perfect square. A real number $\alpha$ is a quadratic irrationality if and only if it has an infinite periodic continued fraction expansion. References
[a1] A.Ya. Khinchin, "Continued fractions" , Phoenix Sci. Press (1964) pp. Chapt. II, §10 (Translated from Russian) How to Cite This Entry:
Quadratic irrationality.
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Quadratic_irrationality&oldid=29336
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Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a...
@Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well
@Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$.
However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1.
Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$
Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ?
Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son...
I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying.
UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton.
hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24*x_2 + 5cos(x_1)*x_2)/abs(x_2). The fixed point is x_1=0, x_2=0
Consider the following integral: $\int 1/4*(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2*\arctan{(u/2)}$. Or am I overseeing something?
*it should be du instead of dx in the integral
**and the solution is missing a constant C of course
Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$?
My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical.
My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction.
Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on.
"... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.)
Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
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I have a counting experiment:
Let's say I have N identical bees. I take one of them, expose it to $\gamma$-radiation and look if it has died or survives. If it survives, I count it as 1 count. If it dies, - 0 counts. I do that N times, until I run out of bees. I calculate the probability of death as:
$p=\frac{\sum_{i=1}^{N}n_i}{N}$
where $n_i$ can be either 1 or 0. Now one says, that the standard error on $p$ is
$\Delta p = \sqrt{\frac{p(1-p)}{N}}$
Why? Could somebody point me to some reading materials?
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In statistic, the Coefficient of variation formula or known as CV, also known as relative standard deviation (RSD) is a standardized measure of dispersion of a probability distribution or frequency distribution. When the value of coefficient of variation is lower, it means the data has less variability and high stability.
The formula for coefficient of variation is given below:
\[\LARGE Coefficient\;of\;Variation\;Formula = \frac{Standard\;Deviation}{Mean}\]
As per sample and population data type, the formula for standard deviation may vary –
\[\large Sample\;Standard\;Deviation=\frac{\sqrt{\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}}}{n-1}\]
\[\large Population\;Standard\;Deviation=\frac{\sqrt{\sum_{i=1}^{n}(X_{i}-\overline{X})^{2}}}{n}\]
x
i= Terms given in the data
$\overline{x}$ = Mean
n = Total number of terms.
Example: A researcher is comparing two multiple-choice tests with different conditions. In the first test, a typical multiple-choice test is administered. In the second test, alternative choices (i.e. incorrect answers) are randomly assigned to test takers. The results from the two tests are:
Regular Test Randomized Answer Mean 59.9 44.8 SD 10.2 12.7
Trying to compare the two test results is challenging. Comparing standard deviations doesn’t really work, because the means are also different. Calculation using the formula CV=(SD/Mean)*100 helps to make sense of the data:
Regular Test Randomized Answer Mean 59.9 44.8 SD 10.2 12.7 CV 17.03 28.35
Looking at the standard deviations of 10.2 and 12.7, you might think that the tests have similar results. However, when you adjust for the difference in the means, the results have more significance:
Regular test: CV = 17.03
Randomized answers: CV = 28.35
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[This is the 6th post in the current series about Wythoff’s game: see posts #1, #2, #3, #4, and #5.
Caveat lector: this post is a bit more difficult than usual. Let me know what you think in the comments!]
Our only remaining task from last week was to prove the mysterious Covering Theorem: we must show that there is exactly one dot in each row and column of the grid (we already covered the diagonal case). Since the rows and columns are symmetric, let’s focus on columns.
The columns really only care about the
x-coordinates of the points, so let’s draw just these x-coordinates on the number-line. We’ve drawn \(\phi,2\phi,3\phi,\ldots\) with small dots and \(\phi^2,2\phi^2,3\phi^2,\ldots\) with large dots. We need to show that there’s exactly one dot between 1 and 2, precisely one dot between 2 and 3, just one between 3 and 4, and so on down the line. For terminology’s sake, break the number line into length-1 intervals [1,2], [2,3], [3,4], etc., so we must show that each interval has one and only one dot:
Why is this true? One explanation hinges on a nice geometric observation: Take any small dot
s and large dot t on our number-line above, and cut segment st into two parts in the ratio \(1:\phi\) (with s on the shorter side). Then the point where we cut is always an integer! For example, the upper-left segment in the diagram below has endpoints at \(s=2\cdot\phi\) and \(t=1\cdot\phi^2\), and its cutting point is the integer 3:
In general, if
s is the jth small dot—i.e., \(s=j\cdot\phi\)—and \(t=k\cdot\phi^2\) is the kth large dot, then the cutting point between s and t is \(\frac{1}{\phi}\cdot s+\frac{1}{\phi^2}\cdot t = j+k\) (Why?! [1]). But more importantly, this observation shows that no interval has two or more dots: a small dot and a large dot can’t be in the same interval because they always have an integer between them! [2]
So all we have to do now is prove that no interval is
empty: for each integer n, some dot lies in the interval [ n, n+1]. We will prove this by contradiction. What happens if no dot hits this interval? Then the sequence \(\phi,2\phi,3\phi,\ldots\) jumps over the interval, i.e., for some j, the jth dot in the sequence is less than n but the ( j+1)st is greater than n+1. Likewise, the sequence \(\phi^2,2\phi^2,3\phi^2,\ldots\) jumps over the interval: its kth dot is less than n while its ( k+1)st dot is greater than n+1:
By our observation above on segment \(s=j\phi\) and \(t=k\phi^2\), we find that the integer
j+ k is less than n, so \(j+k\le n-1\). Similarly, \(j+k+2 > n+1\), so \(j+k+2 \ge n+2\). But together these inequalities say that \(n\le j+k\le n-1\), which is clearly absurd! This is the contradiction we were hoping for, so the interval [ n, n+1] is in fact not empty. This completes our proof of the Covering Theorem and the Wythoff formula!
It was a long journey, but we’ve finally seen exactly why the Wythoff losing positions are arranged as they are. Thank you for following me through this!
A Few Words on the Column Covering Theorem
Using the
floor function \(\lfloor x\rfloor\) that rounds x down to the nearest integer, we can restate the Column Covering Theorem in perhaps a more natural context. The sequence of integers $$\lfloor\phi\rfloor = 1, \lfloor 2\phi\rfloor = 3, \lfloor 3\phi\rfloor = 4, \lfloor 4\phi\rfloor = 6, \ldots$$ is called the Beatty sequence for the number \(\phi\), and similarly, $$\lfloor\phi^2\rfloor = 2, \lfloor 2\phi^2\rfloor = 5, \lfloor 3\phi^2\rfloor = 7, \lfloor 4\phi^2\rfloor = 8,\ldots$$ is the Beatty sequence for \(\phi^2\). Today we proved that these two sequence are complementary, i.e., together they contain each positive integer exactly once. We seemed to use very specific properties of the numbers \(\phi\) and \(\phi^2\), but in fact, a much more general theorem is true: Beatty’s Theorem: If \(\alpha\) and \(\beta\) are any positive irrational numbers with \(\frac{1}{\alpha}+\frac{1}{\beta}=1\), then their Beatty sequences \(\lfloor\alpha\rfloor, \lfloor 2\alpha\rfloor, \lfloor 3\alpha\rfloor,\ldots\) and \(\lfloor\beta\rfloor, \lfloor 2\beta\rfloor, \lfloor 3\beta\rfloor,\ldots\) are complementary sequences.
Furthermore, our same argument—using \(\alpha\) and \(\beta\) instead of \(\phi\) and \(\phi^2\)—can be used to prove the more general Beatty’s Theorem!
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Solving for Acceleration In Terms of Displacement
\[a= \frac{d^2 x}{dt^2}\]so it might appear that we have to integrate twice and require two condition or cannot solve.{/jatex}. How can we find an expression for the velocity?
If the acceleration is given in terms of the displacement, there is a way forward. We can use the relationship
\[a=\frac{dv}{dt}= \frac{dx}{dt} \frac{dv}{dx}\]and integrate, requiring only one condition - for simultaneous values of
\[v\]and
\[x\].
Suppose a condition is
\[v=1, \: x=2\]in appropriate units.
\[a=2x^2\]and
\[v=1\]when
\[tx=2\](in the appropriate units), then we can write
\[v\frac{dv}{dx}=2x^2\]
Now separate variables.
\[v dv=2x^2dx\]
Now integrate in the usual way.
\[\int^v_1 vdv = \int^x_2 2x^2dx\]
\[[ \frac{v^2}{2}]^v_1 = [\frac{x^3}{3}]^x_2\]
\[\frac{v^2}{2}- \frac{1}{2}= \frac{x^3}{3}- \frac{8}{3}\]
Making
\[v\]the subject gives
\[v= \sqrt{\frac{1}{6}(2x^3-13}\].
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Today was another day of parallel talks. There was again six streams of talks where again the topics in each stream varied throughout the day. I would have loved to spend the day in the Higgs stream, listening to the new ATLAS, CMS, CDF and D0 results, however, I spent the day in the heavy flavour physics stream. This was because my talk was scheduled in this stream.
First off, I need to give you a bit of background regarding the \(B_s^0\) meson. As the \(B_s^0\) meson is neutral, it can transform, via the box Feynman diagrams on the left to its antimatter partner, the \(\overline{B}_s^0\) meson, and back again. If we look at and manipulate the equations governing the mixing and decay of the \(B_s^0-\overline{B}_s^0\) meson system we find that there are two \(B_s^0\) mass eigenstates, \(B_{s,H}^0\) and \(B_{s,L}^0\), with two different lifetimes, \(\tau_H = 1 / \Gamma_H\) and \(\tau_L = \Gamma_l\) with \(\Delta\Gamma_s = \Gamma_L – \Gamma_H\) and \(\Gamma_s = (\Gamma_L+\Gamma_H)/2\).
We can measure \(\Delta\Gamma_s\) and \(\Gamma_s\) through the analysis of the decay \(B_s^0 \to J/\psi \phi\) and access \(\tau_H\) and \(\tau_L\) from the measurement of the \(B_s^0\) lifetime in \(B_s^0 \to J/\psi f_0(980)\) and \(B_s^0 \to K^+K^-\) decays:
\(\Gamma_s = 0.6580 \pm 0.0054 \pm 0.0066\, {\rm ps}^{−1}\)
\(\Delta\Gamma_s = 0.116 \pm 0.018 \pm 0.006 \,{\rm ps}^{−1}\) \(\tau_H \simeq \tau_{J/\psi f_0} = 1.700 \pm 0.040 \pm 0.026 \,{\rm ps}\) \(\tau_L \simeq \tau_{KK} = 1.468 \pm 0.046 \pm 0.006 \,{\rm ps}\)
These results can all be shown as a function of \(\Delta\Gamma_s\) and \(\Gamma_s\) like below. You can see that all the results are fairly consistent, and the experimental combination overlaps all three individual experimental results. It is also consistent with the theoretical prediction of \(\Delta\Gamma_s\).
The measurement of \(B_s^0\) lifetimes and the information they provide regarding \(\Delta\Gamma_s\) is interesting as the value of \(\Delta\Gamma_s\) can be affected by physics beyond the Standard Model…
And that’s it for Day Three of ICHEP 2012 for me. Until Monday everybody!
Tags: ICHEP2012
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Last week we saw how to compute the area of a circle from first principles. What about spheres?
To compute the volume of a sphere, let’s show that a hemisphere (with radius \(r\)) has the same volume as the
vase shown in the figure below, formed by carving a cone from the circular cylinder with radius and height \(r\). Why this shape? Here’s why: if we cut these two solids at any height \(h\) (between 0 and \(r\)), the areas of the two slices match. Indeed, the slice—usually called cross section—of the sphere is a circle of radius \(\sqrt{r^2-h^2}\), which has area \(\pi(r^2-h^2)\). Similarly, the vase’s cross section is a radius \(r\) circle with a radius \(h\) circle cut out, so its area is \(\pi r^2-\pi h^2\), as claimed.
If we imagine the hemisphere and vase as being made from lots of tiny grains of sand, then we just showed, intuitively, that the two solids have the same number of grains of sand in every layer. So there should be the same number of grains in total, i.e., the volumes should match. This intuition is exactly right:
Cavalieri’s Principle: any two shapes that have matching horizontal cross sectional areas also have the same volume.
So the volumes are indeed equal, and all that’s left is to compute the volume of the vase. But we can do this! Recall that the cone has volume \(\frac{1}{3} (\text{area of base}) (\text{height}) = \frac{1}{3}\pi r^3\) (better yet, prove this too! Hint: use Cavalieri’s Principle again to compare to a triangular pyramid). Likewise, the cylinder has volume \((\text{area of base}) (\text{height}) = \pi r^3\), so the vase (and hemisphere) have volume \(\pi r^3 – \frac{1}{3} \pi r^3 = \frac{2}{3}\pi r^3\). The volume of the whole sphere is thus \(\frac{4}{3}\pi r^3\). Success!
The following visualization illustrates what we have shown, namely $$\text{hemisphere} + \text{cone} = \text{cylinder}.$$ The “grains of sand” in the hemisphere are being displaced horizontally by the stabbing cone, and at the end we have exactly filled the cylinder.
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I'm making a 12V power distribution box, and it's my first time I work with high currents. This is for a boat, and the idea is to run low-power up to the switch panel and having the high-power cables as short as possible.
Simplified schematic:
Input on X1-1, F1 is a 25A fast acting fuse, K1 is a relay switch and finally output X2-1. Ideally the X1-1 should be replaced with a copper bar soldered to the PCB to distribute power to the 10 similar on/off circuits I'm having. 250A is too much for any PCB I suppose... :)
The trace with calculator says I need a 25.5mm trace width. But the trace from the X1-1 to F1 is less than 10mm, and it will be a ugly blob of copper 10mm long and 25.5mm wide.
Hope someone can help me do the math here.
Ok, my math from here assumes a 2oz (\$70\mu m\$) board, single-sided and using through-hole components and still standing air as ambient:
$$R=\frac{\rho \times l}{w \times h} = \frac{1.7 \times 10^{-8}\times 10mm}{12.7mm \times 70\mu m}=191.23\mu\Omega$$ $$U_{trace}=RI=191.23\mu\Omega\times 25A=4.78mV$$ $$P_{trace}=IU=25A\times4.78mV=119.5mW$$ $$\Delta T=\frac{P}{Sa\times h}=\frac{119.5mW}{(1cm\times1.27cm)\times 0.001W/cm^2/^\circ C}=94.1^\circ C$$
It seems my \$10\times 12.7mm\$ blob is too small to dissipate the power adequately. If I use a \$35\mu m\$ copper board and double the area I get the \$\Delta T\$ down to about \$47^\circ C\$.
Are my calculations correct? If yes, will it also perform equally after I build it?
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Informally, a signature scheme with message recovery is one where some or all of the message is embedded in the signature, allowing to conserve bandwidth when transmitting a signed message, compared to a signature scheme with appendix.
Total message recovery
A
[some sources make signature scheme with total message recovery total implicit, e.g. the HAC section 11.2.3] can be formally defined as three functions [easily computable, with a compact and public definition]: a key generation function with inputs some parameters (in particular for key size) a random bitstring, with output a public/private key pair $(K_\text{pub},K_\text{priv})$, a signature function $\mathcal S$ with inputs same parameters $K_\text{priv}$ obtained from the output of the key generation function any message $M$ in some large subset $\mathbb M$ of the set of bitstrings $\{0,1\}^*$ an (optional) random bitstring [that is used in ISO/IEC 9796-2 scheme 2] with output a bitstring designated as the signed message $S=\mathcal S_{K_\text{priv}}(M)$, a verification function $\mathcal V$ with inputs same parameters $K_\text{pub}$ obtained from the output of the key generation function any bitstring $S\in\{0,1\}^*$ with output a pair $(b,M)=\mathcal V_{K_\text{pub}}(S)$ with $b$ a boolean, either $\text{false}$ [bad signature] or $\text{true}$ [good signature] a bitstring $M$
such that for all parameters and random seed(s), $\forall M\in\mathcal M, V_{K_\text{pub}}(S_{K_\text{priv}}(M))=(\text{true},M)$.
A signature scheme with total message recovery is defined to be existentially secure [or equivalently: secure in a chosen-messages setup] if a computationally bounded adversary, given $K_\text{pub}$ and access to an oracle producing $S_{K_\text{priv}}(M)$ when given any $M\in\mathbb M$, can not with sizable odds exhibit $S\in\{0,1\}^*$ with $V_{K_\text{pub}}(S)=(\text{true},M)$, and such that the oracle was not queried with $M$.
Notice that in a signature scheme with total message recovery, $M$ is an
output of the message verification function; when it is an input in a signature scheme with appendix. Recovering $M$ from the signed message $S$ requires evaluating $\mathcal V_{K_\text{pub}}(S)$, and in particular knowledge of $K_\text{pub}$.
It is trivial to transform a secure signature scheme with appendix [such as any of the signature schemes in PKCS#1] into a secure signature scheme with message recovery: construct the signed message of the new scheme by concatenating the message to the signature of the original scheme; and extract $M$ and that original signature from the signature of the new scheme, before applying the original verification procedure and giving its result as the boolean output of the new scheme. Practical signature schemes with message recovery are not constructed in this way, for they have an additional goal: make the signed message as short as possible. This is especially useful with RSA: signature schemes with message recovery can reduce the size of the signed message [by a proportion sometime approaching 2, e.g. for public-key certificates], while keeping fast signature verification.
For $k$-bit security, the size of signature and message must obey $|M|\le|S|-k$ (for odds of success of a trivial adversary not querying the oracle and trying $S$ at random are at least $2^{|M|-|S|}$, assuming $|S|$ is fixed for a given $|M|$ and all messages of size $|M|$ can be signed).
Partial message recovery
Some schemes also offer
: signature generation breaks the message $M$ into $M_1$ and $M_2$ in a public, scheme-defined manner, with $|M|=|M_1|+|M_2|$; it is assumed that $M_2$ [the non-recoverable part] is made an additional input of the signature verification function (as in signature with appendix); while $M_1$ [the recoverable part] replaces $M$ in the output of the verification function, which becomes $(b,M_1)$. partial message recovery
The correctness requirement becomes that for all parameters and random seed(s), $\forall M\in\mathcal M, V_{K_\text{pub}}(S_{K_\text{priv}}(M),M_2)=(\text{true},M_1)$, where $M_1$ and $M_2$ are extracted from $M$ in a prescribed manner, and can be recombined into $M$.
The security requirement becomes that a computationally bounded adversary, given $K_\text{pub}$ and access to an oracle producing $S_{K_\text{priv}}(M)$ when given any $M\in\mathbb M$, can not with sizable odds exhibit $M_2$ and $S\in\{0,1\}^*$ with $V_{K_\text{pub}}(S,M_2)=(\text{true},M_1)$, and such that the oracle was not queried with $M$ obtained by recombining $M_1$ and $M_2$.
Typically, $|S|$ is fixed for a given public key, but independent of $|M|$; and $|M_2|=0$ for $|M|$ below some threshold. For $k$-bit security, $|M_1|\le|S|-k$ must hold, or equivalently $|M_2|\ge|M|-|S|+k$
It is easy to turn any secure signature scheme with partial message recovery into a secure signature scheme with total message recovery (by appending $M_2$ to the signature of the original scheme). The reverse is also possible if the $M_2$ fraction of $M$ appears verbatim in $S$ at some unambiguous position, given the parameters (and possibly the public-key and the rest of the signature).
Practice
The first standard message scheme with message recovery was ISO/IEC 9796:1991 [later renamed ISO/IEC 9796-1, but never approved under this name]. It did not use a hash, because standard hashes where then a novelty. It featured only total message recovery, and had the constraint $|M|\le|S|/2$ (within a few bits). It was broken in a chosen-messages setup [first within 1 bit; then fully (password obtained by subscribing to a mailing list) but with many messages; then with another attack that obtains the signature of $k$ extra messages from that of $k+2$ chosen other messages for most practical arguments]; and was withdrawn in 2000.
The second standard message scheme was ISO/IEC 9796-2:1997, now known as scheme 1 of ISO/IEC 9796-2. It improves on ISO/IEC 9796:1991 by using a hash, which allows a more compact signed message, and partial message recovery. Scheme 1 of ISO/IEC 9796-2 was broken in a chosen-messages setup [the best known attack is that in the paper (alternate link) noted in the question], but remains secure in practice when the adversary is restricted to obtaining signature of messages not specifically constructed to enable an attack. ISO/IEC 9796-2:2002 enforced a minimum hash width of $h=160$ as a mitigating measure, added schemes 2 and 3 intended to fix security for good, defined the so-called alternative signature production and opening functions that practice demanded, and included as normative the previously informative non-alternative signature production and opening functions. Amendment 1:2008 to ISO/IEC 9796-2:2002 added an ASN.1 module. ISO/IEC 9796-2:2010 makes only purely editorial changes AFAIK, beyond consolidating that amendment.
In ISO/IEC 9796-2, the message $M$ to be signed is subdivided as $M=M_1\|M_2$. The recoverable part $M_1$ is at the beginning of the message, of some maximum size determined by the parameters; it is transmitted embedded into the signature $S$, which has size fixed by public key parameters. The non-recoverable part $M_2$ is empty for short $M$ [and always a byte-string even when $M$ is not, thus $|M_1|\equiv|M|\pmod 8$]; $M_2$ is often transmitted unmodified along the signed message, but could be implicit in whole or in part (e.g. a version number). When $M_2$ is empty [resp. non-empty], the scheme operates in total [resp. partial] recovery mode.
Scheme 1 of ISO/IEC 9796-2 is deterministic. For signature, a message representative is formed by concatenating prescribed padding, $M_1$, the hash of the whole message $M$, and other prescribed padding; the underlying [RSA-like] private-key function is then applied to obtain the signature, which is about the size of the public modulus. For verification of the signature and message recovery, after preliminary checks, the underlying public-key function is applied; the padding in that result is checked; the recovered $M_1$ is extracted from that result; and the hash extracted from that result is compared to the hash of the message reconstructed from the recovered $M_1$ and, for partial message recovery, the non-recovered $M_2$ obtained by other means.
I refer to EMV 4.3, Book 2, Annex A2.1, for a detailed exposition of ISO/IEC 9796-2 scheme 1, restricted to $n\equiv m\equiv 0\pmod 8$, SHA-1 (thus $h=160$), implicit hash identifier, $m\gt n+176$ [that is partial message recovery], and the so-called alternative signature production and opening functions RSASP1 and RSAVP1 in PKCS#1. This parameterization is common in Smart Cards, and directly supported [including with total message recovery] by Java Card 2.2.x/Classic.
Scheme 2 of ISO/IEC 9796-2 is randomized [and correspondingly has the signed message larger than scheme 1 in many use cases], with a security argument resembling that of RSASSA-PSS of PKCS#1, applicable including in a chosen-messages setup. Scheme 3 is a special case of scheme 2 with the random bitstring reduced to empty. The split of message into recoverable and non-recoverable portions is such that scheme 3 can be used as functional equivalent to scheme 1 with restored security in a chosen-messages setup [although I know no formal security argument].
Assuming an $n$-bit public modulus, when sending an $m$-bit message and its signature, the signed message of ISO/IEC 9796-2 schemes 1 or 3 has about $\min(n,m+h+16)$ bits where $h$ is the width of a hash [SHA-1 is popular, giving $h=160$]. This allows a significant [at least when considering proportion] size saving for messages with $m$ commensurate to $n$, such as public-key certificates [compare to $m+n$ bits using the schemes in PKCS#1]. For example: with $n=1984$, $h=160$, $m=2000$ [250 bytes], the signed message may use $2176$ bits [272 bytes, rather than 498 bytes with PKCS#1]: $1984$ bits [248 bytes] for the RSA cryptogram, conveying $1808$ useful bits [226 bytes] forming the recoverable part of the message; and $192$ bits [24 bytes] forming the non-recoverable part of the message.
ISO/IEC 9796-2 has a maze of features and parameters: three schemes; many hashes; optional hash identifier [intended to allow interoperability in contexts where the signer decides which hash is used, without requiring guesswork on the verifier side]; modulus of size not multiple of 8; messages of size not multiple of 8 when the hash allows; total or partial message recovery; either RSA or Rabin signature scheme [an analog with even public exponent]; and two slightly different options of RSA signature production and opening functions: those not designated as
alternative sign as $x\mapsto \min(x^d\bmod N, N-(x^d\bmod N))$ which saves one bit over the alternative: PKCS#1's RSASP1. EMV originally referred to ISO/IEC 9796-2:1997, but only uses partial message recovery, the alternative functions not alluded to in this standard, and now refers to ISO/IEC 9796-2:2010 without stating the scheme, functions, or other options. The European Digital Tachograph [annex 1B, appendix 11] does just as EMV when it refers to ISO/IEC 9796-2 in a signature context for certificates, but also refers to ISO/IEC 9796-2 in its authentication and key agreement scheme where it uses scheme 1 with total message recovery [thus different padding than EMV] and the non-alternative signature production and opening functions of that standard [because that enables re-encryption of the signature using a different private key with the same modulus size as the signature key, regardless of which modulus is smaller].
ISO/IEC 9796:1991, ISO/IEC 9796-2, and EMSR3 of P1363a (similar to ISO/IEC 9796-2) use RSA or Rabin as the underlying public-key cryposystem. There's also ISO/IEC 9796-3, the similar EMSR1, and EMSR2, that use the Discrete Logarithm (possibly over an Elliptic Curve group).
Regarding the note in ISO/IEC 9796-2:2010 quoted in that comment to the question:
For practical purposes, an application may wish to structure the message $M$ to ensure that data it wants to be explicitly stored or transmitted (e.g., address information) is allocated to the non-recoverable message part $M_2$
That's intended to draw attention to the fact that only the non-recoverable part of the signed message [if any] is intelligible before verification of the signature. As a consequence, information that needs to be readily available before or absent signature verification is better placed in that non-recoverable part of the message, rather than in the recoverable part.
As a real-life illustration: contrary to the recommendation in this note, in the aforementioned European Digital Tachograph specification, the message in a certificate [defined by Annex 1B, appendix 11, 3.3.1 CSM_017] has the identification [designated CAR] of the public key of the authority that produced that certificate placed near the beginning, thus in the recoverable part of the message, thus unintelligible before signature verification, which can be a problem to decide which public key is appropriate to verify that certificate, absent context. To avoid this, a copy of CAR, designated X.CAR, was appended at the end of the signed certificate [defined in 3.3.2 CSM_018], allowing not reading the certification authority's certificate if it is already known and its public key cached, and giving a chance to know what allegedly generated a certificate even if it fails to verify. That complication increases the certificate size by 8 bytes, and creates confusion about if an hypothetical certificate with its X.CAR not matching CAR is deemed valid, and what/who is deemed to have generated it; that could have been avoided if CAR had been at the end of the certificate data, thus in the non-recoverable portion.
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FLoC 2018 Selected Talks Dependency Quantified Boolean Formulas
QBF PSPACE-complete DQBF Dependencies explicitly specified NEXPTIME-complete
Every \(\exists\) depends on all \(\forall\) before it.
Dependency Quantified Boolean Formulas
\(\forall x_1...\forall x_n \exists y_1(D_{y1}).. y_n(D_{yn}) : \phi \)
with \(D_{yi} \subseteq \{x_i, ..., x_n\}\)
General Form:
Dependency Quantified Boolean Formulas
Example:
\(\forall x_1, x_2, x_3 \exists y_1 (x_1,x_2), y_2(x_1,x_3):\phi\)
\(\forall x_1\)
\(\forall x_2\)
\(\forall x_3\)
\(\exists y_1\)
\(\exists y_2\)
\(S_{y_1}(t,t)\)
\(S_{y_1}(t,t)\)
\(S_{y_1}(t,f)\)
\(S_{y_1}(t,f)\)
\(S_{y_2}(t,t)\)
\(S_{y_2}(t,f)\)
\(S_{y_2}(t,t)\)
\(S_{y_2}(t,f)\)
Dependency Quantified Boolean Formulas
Example:
\(\forall x_1, x_2, x_3 \exists y_1 (x_1,x_2), y_2(x_1,x_3):\phi\)
\(\forall x_1\)
\(\forall x_2\)
\(\forall x_3\)
\(\exists y_1\)
\(\exists y_2\)
Dependency Quantified Boolean Formulas
Example application: Partial Equivalence Checking
\(\forall x_1 x_2 \exists y_1 y_2\) \(\forall x_1 \exists y_1 \forall x_2 \exists y_2\) \(\forall x_1 \exists y_2 \forall x_2 \exists y_1\)
Linearizations cannot express the problem
Rabe: "DQBF can encode existential quantification over functions"
Dependency Quantified Boolean Formulas
Solved using:
Incomplete Approximations: implicit depencies \(\supseteq\) explicit Search + Skolem clauses Universal Expansion until QBF: Eliminate \(x_i\) from incomparable dependency sets (MAXSAT) until SAT: (potentially) exponential increase of variables (information fork resolution) Dependency Quantified Boolean Formulas
More expressive/harder problems
Can model \[\forall x : (\forall y : \phi(x,y)) \land (\forall z : \psi(x,z))\]
without loss of information/smaller search space
Approximation Fixpoint Theory and the Well-Founded Semantics of Higher-Order Logic Programs
See original slides
PySAT: A Python Toolkit forPrototyping with SAT Oracles Python toolkit for working with multiple SAT solvers Its MAXSAT implementation won the MAXSAT competition FLoC 2018
By krr
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But if you don't want to have a Google account: Chrome is really good. Much faster than FF (I can't run FF on either of the laptops here) and more reliable (it restores your previous session if it crashes with 100% certainty).
And Chrome has a Personal Blocklist extension which does what you want.
: )
Of course you already have a Google account but Chrome is cool : )
Guys, I feel a little defeated in trying to understand infinitesimals. I'm sure you all think this is hilarious. But if I can't understand this, then I'm yet again stalled. How did you guys come to terms with them, later in your studies?
do you know the history? Calculus was invented based on the notion of infinitesimals. There were serious logical difficulties found in it, and a new theory developed based on limits. In modern times using some quite deep ideas from logic a new rigorous theory of infinitesimals was created.
@QED No. This is my question as best as I can put it: I understand that lim_{x->a} f(x) = f(a), but then to say that the gradient of the tangent curve is some value, is like saying that when x=a, then f(x) = f(a). The whole point of the limit, I thought, was to say, instead, that we don't know what f(a) is, but we can say that it approaches some value.
I have problem with showing that the limit of the following function$$\frac{\sqrt{\frac{3 \pi}{2n}} -\int_0^{\sqrt 6}(1-\frac{x^2}{6}+\frac{x^4}{120})^ndx}{\frac{3}{20}\frac 1n \sqrt{\frac{3 \pi}{2n}}}$$equal to $1$, with $n \to \infty$.
@QED When I said, "So if I'm working with function f, and f is continuous, my derivative dy/dx is by definition not continuous, since it is undefined at dx=0." I guess what I'm saying is that (f(x+h)-f(x))/h is not continuous since it's not defined at h=0.
@KorganRivera There are lots of things wrong with that: dx=0 is wrong. dy/dx - what/s y? "dy/dx is by definition not continuous" it's not a function how can you ask whether or not it's continous, ... etc.
In general this stuff with 'dy/dx' is supposed to help as some kind of memory aid, but since there's no rigorous mathematics behind it - all it's going to do is confuse people
in fact there was a big controversy about it since using it in obvious ways suggested by the notation leads to wrong results
@QED I'll work on trying to understand that the gradient of the tangent is the limit, rather than the gradient of the tangent approaches the limit. I'll read your proof. Thanks for your help. I think I just need some sleep. O_O
@NikhilBellarykar Either way, don't highlight everyone and ask them to check out some link. If you have a specific user which you think can say something in particular feel free to highlight them; you may also address "to all", but don't highlight several people like that.
@NikhilBellarykar No. I know what the link is. I have no idea why I am looking at it, what should I do about it, and frankly I have enough as it is. I use this chat to vent, not to exercise my better judgment.
@QED So now it makes sense to me that the derivative is the limit. What I think I was doing in my head was saying to myself that g(x) isn't continuous at x=h so how can I evaluate g(h)? But that's not what's happening. The derivative is the limit, not g(h).
@KorganRivera, in that case you'll need to be proving $\forall \varepsilon > 0,\,\,\,\, \exists \delta,\,\,\,\, \forall x,\,\,\,\, 0 < |x - a| < \delta \implies |f(x) - L| < \varepsilon.$ by picking some correct L (somehow)
Hey guys, I have a short question a friend of mine asked me which I cannot answer because I have not learnt about measure theory (or whatever is needed to answer the question) yet. He asks what is wrong with \int_0^{2 \pi} \frac{d}{dn} e^{inx} dx when he applies Lesbegue's dominated convergence theorem, because apparently, if he first integrates and then derives, the result is 0 but if he first derives and then integrates it's not 0. Does anyone know?
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Correcting an Important Goertzel Filter Misconception
Recently I was on the Signal Processing Stack Exchange web site (a question and answer site for DSP people) and I read a posted question regarding Goertzel filters [1]. One of the subscribers posted a reply to the question by pointing interested readers to a Wikipedia web page discussing Goertzel filters [2]. I noticed the Wiki web site stated that a Goertzel filter:
"...is marginally stable and vulnerable to numerical error accumulation when computed using low-precision arithmetic and long input sequences."
That statement can often be misinterpreted. Because of that possible stability misconception, and the fact the literature of the Goertzel algorithm doesn't discuss Goertzel filter stability, I decided to write this blog.
This article is also available in PDF format
Goertzel Filter Descriptions Typical DSP textbook descriptions of Goertzel filters begin by discussing the network shown in Figure 1(a) that can be used to compute N-point discrete Fourier transform (DFT) X( k) spectral samples [3]–[5].
Figure 1 - Recursive network used to compute a singleDFT spectral sample: (a) implementation;(b) z-plane pole.
The Figure 1(a) complex
resonator's
-domain transfer
function is
z
$$ H_R(z) = {1\over {1 - e^{j2 \pi k/N} z^{-1}}}\tag{1} $$
which, with
infinite-precision arithmetic, has a single pole located on the
-plane's unit circle as shown in Figure
1(b).
z
While the Figure
1(a) resonator can be used to compute individual
N-point DFT spectral samples it has a major flaw. When finite-precision
arithmetic (as in our digital implementations) is used the network's
z-plane pole does not lie exactly on the
unit circle. This means the resonator can be unstable and a computed DFT sample
value will be incorrect. In the literature of DSP the resonator is referred to
as "marginally stable."
We show this in
Figure 2 for various
z-plane poles, in
the frequency range of $ 0 \le 2\pi k/N \le \pi /2 $ radians (zero to one quarter of the input sample rate in Hz), when
5-bit binary arithmetic is used.
Figure 2 - First quadrant z-plane pole locations of a complex resonator using 5-bit arithmetic.
DSP textbook
descriptions of Goertzel filters don't discuss the complex resonator's unstable
operation. That's because they move on to improve the computational efficiency
of the complex resonator by multiplying Eq. (1)'s numerator and denominator by $ (1 - e^{-j2 \pi k/N}) $
as:
$$ H_G(z) = {1\over {1 - e^{j2 \pi k/N} z^{-1}}} \cdot {{1 - e^{-j2 \pi k/N} z^{-1}} \over {1 - e^{-j2 \pi k/N} z^{-1}} } \tag{2}$$
$$ \qquad \qquad = {{1 - e^{-j2 \pi k/N} z^{-1}} \over {(1 - e^{j2 \pi k/N} z^{-1})(1 - e^{-j2 \pi k/N} z^{-1})}} \tag{3}$$
$$ \qquad \qquad = {{1 - e^{-j2 \pi k/N} z^{-1}} \over {1 - 2cos(2 \pi k/N)z^{-1} + z^{-2}}} \tag{4}$$
which is the
z-domain transfer function of the
standard Goertzel filter. The Eq. (4) Goertzel filter's block diagram, shown in
Figure in Figure 3(a), has two conjugate poles and a single pole-canceling zero
on the
z-plane as shown in Figure
3(b).
Figure 3 - Standard Goertzel filter: (a) filterimplementation; (b) z-plane poles and zero.
The Goertzel Filter Misconception The potential misconception, regarding finite-precision arithmetic, is the following: Misconception: Because the first ratio in Eq. (2), which is only marginally stable, times a unity-valued ratio equals Eq. (4), then the implementation of Eq. (4) must also be only marginally stable.
I'm familiar with this misunderstanding because I was painfully guilty of it myself years ago -- and this is the misconception stated on the Goertzel Wikipedia web site [2].
The Stability of a Goertzel Filter
As it turns out, a Goertzel filter, described by Eq. (4), is guaranteed stable! The filter's poles always lie on the z-plane's unit circle. (Proof of that statement is given in Appendix A.)
So, we can say:
The poles of the Figure 3(a) Goertzel filter always lie exactly on the z-plane's unit circle making the filter guaranteed stable regardless of the numerical precision of its coefficients.
We show this Goertzel filter stability behavior in Figure 4 for various first quadrant z-plane poles for 5-bit binary filter coefficients.
Figure 4 First quadrant z-plane pole locations of a
Goertzel filter using 5-bit arithmetic.
The Limitations of the Goertzel Filter In finishing this blog I'll mention two important characteristics of a Goertzel filter.
[1] While the Goertzel filter is guaranteed stable, quantized coefficients (a finite number of binary bits) limit the locations where the filter's z-plane poles can be placed on the unit circle. A small number of binary bits used to represent the filter's coefficients limits the precision with which we can specify the filter's resonant frequency.
[2] Quantized coefficients make it difficult to build Goertzel filters that have low resonant frequencies. The lower the desired filter resonent frequency the larger must be the number of coefficient binary bits used.
We illustrate both of those characteristics in Figure 5 and Figure 6 for 5- and 8-bit coefficients.
Figure 5 First quadrant z-plane pole locations of a
Goertzel filter using 5-bit arithmetic.
Figure 6 First quadrant z-plane pole locations of a
Goertzel filter using 8-bit arithmetic.
Conclusion Here we showed that Goertzel filters will never be unstable due to z-domain pole placement on the z-plane's unit circle. However as pointed out by Ollie Niemitalo, another form of instability is possible with Goertzel filters. A Goertzel filter can become unstable, due to finite precision arithmetic being used for the first feedback accumulator (adder) in Figure 3 to produce the v(n) sequence, for very large values of N. So the user must take care to ensure that, for large N (in the tens of thousands for 32-bit floating point numbers), all binary register word widths can accommodate all necessary arithmetic operations. APPENDIX A – Proof of Goertzel Filter Guaranteed Stability Goertzel filters, described by Eq. (4), are guaranteed stable. Proof of that statement was shown to me by the skilled DSP engineer Prof. Pavel Rajmic (Brno University of Technology, Czech Republic). Rajmic's straightforward proof of Goertzel filter stability goes like this:
For mathematical convenience, multiply the Goertzel filter's Eq. (4) numerator and denominator by $z^2$ as:
$$ {{1 - e^{-j2 \pi k/N} z^{-1}} \over {1 - 2cos(2 \pi k/N)z^{-1} + z^{-2}}} \cdot {{z^2} \over {z^2}} $$
$$ \qquad = {{z^2 - e^{-j2 \pi k/N} z^1} \over {z^2 - 2cos(2 \pi k / N)z + 1}}\tag{A-1} $$
If we substitute $ \alpha = cos(2 \pi k/N) $ into Eq. (A-1) then Eq. (A-1)'s denominator becomes:$$ z^2 = 2 \alpha z + 1\tag{A-2} $$
The roots of this polynomial will be the z-plane locations of the trtaditional Goertzel filter's poles.
The roots of the polynomial are:$$ z = {{2 \alpha \pm \sqrt{4 \alpha ^2 - 4}} \over 2} = {{2 \alpha \pm 2\sqrt{\alpha ^2 - 1}} \over 2} = {\alpha \pm \sqrt{\alpha^2 - 1}} \tag{A-3} $$
Because $ \alpha $ is in the range $ -1 \le \alpha \le 1 $, then $ \alpha^2 \le 1 $ and we can rewrite expression (A-3) as:
$$ z = \alpha \pm \sqrt{-(1 - \alpha^2)} = \alpha \pm j \sqrt{1 - \alpha^2}\tag{A-4} $$
The magnitudes of both of the two complex conjugate poles are:$$ \lvert z \rvert = \sqrt{\alpha^2 + (1 - \alpha^2)} = \sqrt{1} = 1 \tag{A-5}$$
which is what I set out to show. So, a Goertzel filter's poles always lie exactly on the z-plane's unit circle.
References
[1] Signal Processing Stack Exchange, http://dsp.stackexchange.com/
[2] Goertzel algorithm, https://en.wikipedia.org/wiki/Goertzel_algorithm
[3] S. Mitra, "Digital Signal Processing", Fourth Edition, McGraw Hill Publishing, 2011, pp. 618-620.
[4] J. Proakis and D. Manolakis, "Digital Signal Processing-Principles, Algorithms, and Applications", Third Edition, Prentice Hall, Upper Saddle River, New Jersey, 1996, pp. 480-481.
[5] A. Oppenheim, R. Schafer, and J. Buck, "Discrete-Time Signal Processing", Second Edition, Prentice Hall, Upper Saddle River, New Jersey, 1996, pp. 633-634.
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Four Ways to Compute an Inverse FFT Using the Forward FFT Algorithm
I just now saw your Comment. I don't understand your question of "what can be said of long sequences?" If you can rephrase your question I'll do my best to answer it. I took a look at the thread you referenced in your URL address. There's no way I have the time to plow through the details of that long thread.
Regards,
[-Rick-]
I suspect that most people wouldn't be terribly concerned from a practical standpoint given modest precision and input lengths. But in my case I was dealing with very long input sequences (up to 1e9 samples!) For a while I was looking at pre-decimating with a CIC+FIR as a mitigation. But after further consideration, I realized that the Reincsh modification would always work in my application. The reason is that for very large input sequences, my frequency is well away from pi/2 (where Goertzel performs numerically better than Reinsch). When the frequency is closer to pi/2, the sample size is much smaller where rounding error accumulation is negligible.
By the way, I noticed some saving opportunities in the Goertzel filter: https://dsp.stackexchange.com/questions/26971/special-case-peak-notch-filter-for-exactly-1-32-sampling-rate
I want to understand what you're saying here. Can you tell me, what does the phrase "resonator's coefficient for the previous output" mean. I'm trying to figure out what that phrase means relative to my above Figure 3(a).
Regards,
[-Rick-]
Ah, OK. You're bringing up both a good and an important implementation topic here. Yes, multiplication of a 10 bit number by a 10 bit number can produce a 20-bit number. But after the 10th Goertzel iteration the accumulated v(n) result may not require a full 100-bit word depth. Accumulator register overflow is not guaranteed. The required v(n) accumulator word depth depends on the magnitudes of the numerical values being summed. When implementing the Goertzel algorithm using fixed-point numbers, the user must do one of two things. (1), make sure the v(n) accumulator can accommodate the number of iterations, N, and the magnitudes of the input signal samples. Or (2), truncate the v(n) accumulator output after each accumulation. The following web page discusses the truncation scheme.
http://www.embedded.com/design/configurable-systems/4008873/2/PRODUCT-HOW-TO-Efficient-Fixed-Point-Implementation-of-the-Goertzel-Algorithm-on-a-Blackfin-DSP
[-Rick-]
I will paste my code below. It can be compiled with G++, a C++ compiler. If you direct the program's output to a file output.txt, you can plot it in Octave using x = csvread("output.txt"); plot(x(:, 1), x(:, 2));
#define _USE_MATH_DEFINES
#include
#include
#include
main() {
int k = 1;
int N = 50;
int numSamples = 3000000;
fesetround(FE_TONEAREST); // FE_TONEAREST, FE_UPWARD, FE_DOWNWARD, or FE_TOWARDZERO
float coef = 2*cos(2*M_PI*k/N);
float outHistory1 = 0, outHistory2 = 0;
float input = 1; // First input value
for (int count = 0; count < numSamples; count++) {
float out = input + coef*outHistory1 - outHistory2;
outHistory2 = outHistory1;
outHistory1 = out;
input = 0;
double re = outHistory1 - cos(2*M_PI*k/N)*outHistory2;
double im = sin(2*M_PI*k/N)*outHistory2;
printf("%d,%f\n", count, sqrt(re*re + im*im));
}
}
Perhaps, if you wish, we could continue our conversation using private e-mails.
I'm at R_dot_Lyons_at_ieee_dot_org. It's up to you.
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I am currently working on problem that I think could be expressed as an integer lattice problem.
Given $u \in \mathbb{R}^n$ and a
bounded integer lattice $L = \mathbb{Z}^n \cap [-M,M]^n$ I would like to find an integer vector $v \in L$ that minimizes the angle between $u$ and $v$. That is, I would like $$v \in \text{argmax}_{w \in L} \frac{u.w}{\|u\|\|w\|}$$
Here, the objective is maximizing the cosine of the angle between $u$ and $w$ (i.e. minimizing the angle between $u$ and $w$). The vectors $u$ and $w$ are said to be "similar" if this quantity is close to 1.
I am wondering:
Is this problem related to a well-known integer lattice problem (e.g. a closest vector problem)?
Could this problem be solved using existing lattice algorithms (e.g. the LLL algorithm?)
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Exponential Smoothing with a Wrinkle
Introduction
This is an article to hopefully give a better understanding to the Discrete Fourier Transform (DFT) by providing a set of preprocessing filters to improve the resolution of the DFT. Because of the exponential nature of sinusoidal functions, they have special mathematical properties when exponential smoothing is applied to them. These properties are derived and explained in this blog article.
Basic Exponential Smoothing
Exponential smoothing is also known as exponential averaging. The definition is straight forward. Suppose you have a sequence of values denoted as $ S_n $, then exponential smoothing produces another sequence using the simple formula:
$$ F_0 = S_0 \tag 1 $$
$$ F_n = b S_n + a F_{n-1} \tag 2 $$
Where $ a $ and $ b $ are scalar parameters having values between 0 and 1, and $ a + b = 1 $.
This definition is true for all values of $ n $, so $ n-1 $, $ n-2 $, etc., can be substituted in for n.
$$ F_{n-1} = b S_{n-1} + a F_{n-2} \tag 3 $$
Plugging (3) into (2) gives an expanded version of the definition.
$$ F_n = b S_n + a \left( b S_{n-1} + a F_{n-2} \right) \tag 4 $$
Using the definiton for $ n- 2 $, the expanded version can be expanded further.
$$ F_n = b S_n + a \left( b S_{n-1} + a \left( b S_{n-2} + a F_{n-3} \right) \right) \tag 5 $$
Now a pattern starts to emerge. All the summation terms except the last one have a factor of $ b $, which can be factored out.
$$ F_n = b \left( S_n + a S_{n-1} + a^2 S_{n-2} \right) + a^3 F_{n-3} \tag 6 $$
The series formed by the pattern can be put in summation form.
$$ F_n = b \sum_{k=0}^{2} { a^k S_{n-k} } + a^3 F_{n-3} \tag 7 $$
Repeating the pattern, assuming that $ |a| < 1 $ and all the $ S_n $ values are bounded, the series will converge when it is extended to infinity.
$$ F_n = b \sum_{k=0}^{\infty} { a^k S_{n-k} } \tag 8 $$
The exponential value of $ a^k $ is what gives the average its name.
Backward Exponential Smoothing
If the sequence represents time series values, then ascending subscript values represent moving forward in time. If the data has already been collected, as in a audio signal buffer, or any other after the fact collection, then the basic definition can be reversed.
$$ B_{N-1} = S_{N-1} \tag 9 $$
$$ B_n = b S_n + a F_{n+1} \tag {10} $$
The backward definition can be expanded in a manner similar to the forward case with the only difference being the sign of the iteration variable $(k)$ in the subscript of the sequence value.
$$ B_n = b \sum_{k=0}^{\infty} { a^k S_{n+k} } \tag {11} $$
Cancelling the Lags
One of the properties of exponential averages is that the average value represents the values in the near past. Thus, the average looks like it lags the signal. By symmetry, in the backward case, the lag is reversed and actually leads the signal when viewed in an ascending subscript manner. These effects can be seen in Figure 1. The black line is a sample signal, the green line is the forward exponential average, and the red line is the backward exponential average.
When the leading and trailing exponential averages are combined with an arithmetic average, the lags cancel themselves out on symmetric features of the signal. This can be seen in the blue line in Figure 2.
$$ A_n = ( B_n + F_n ) / 2 \tag {12} $$
First Derivative Like Behavior
Here is the wrinkle. When the difference is taken, instead of the sum, in the combination, the result acts a lot like a first derivative of the averaged values. This can be understood conceptually be realizing that the leading lag represents the expected value of the signal from the near future and the trailing lag represents the expected value of the signal from the near past. If the future value is higher than the past value, the signal is on an upward trend. If the future value is lower than the past value, the signal is on a downward trend. Just like with a derivative, if the future value is the same as the past value, the signal is horizontal. In a varying signal, this means either at a peak, a trough, or a horizontal inflection point. This can be seen in the cyan line in Figure 2.
$$ D_n = ( B_n - F_n ) / 2 \tag {13} $$
Sinusoidal Signal
A sinusoidal signal is a very special case of a time varying signal. Because it represents a pure tone when the signal is an audio signal, the sinusoidal case is very important in Digital Signal Processing (DSP). A pure tone signal can be modelled mathematically as:
$$ S_n = M cos ( f \cdot { 2\pi \over N } n + \phi ) \tag {14} $$
Where $ N $ is the number of points in a sample frame, $ f $ is the frequency in cycles per frame, $ M $ is the amplitude, and $ \phi $ is the phase shift.
For convenience, a variable ($\alpha$) can be defined to simplify the definition.
$$ \alpha = f \cdot { 2\pi \over N } \tag {15} $$
$$ S_n = M cos ( \alpha n + \phi ) \tag {16} $$
In practical applications, $ \alpha $ can be assumed to range from 0 to $ \pi $. A value of zero is the so called DC (direct current) case where the value of the signal remains constant. A value of $ \pi $ occurs at the Nyquist frequency, which is the limit of resolution for a DFT.
Exponential Smoothing of a Sinusoidal Signal
The definition of the signal (16) can be plugged into the forward exponential average series (8).
$$ F_n = b \sum_{k=0}^{\infty} { a^k M cos \left( \alpha ( n - k ) + \phi \right) } \tag {17} $$
With a little algebra, the argument of the cosine function can be rearranged to separate the $ n $ and the $ k $ values.
$$ F_n = M \cdot b \sum_{k=0}^{\infty} { a^k cos ( \alpha n + \phi - \alpha k ) } \tag {18} $$
The cosine expression can then be expanded by using the cosine angle addition formula.
$$ F_n = \\ M \cdot b \sum_{k=0}^{\infty} { a^k \left[ cos( \alpha n + \phi )cos( \alpha k ) + sin( \alpha n + \phi )sin( \alpha k ) \right] } \tag {19} $$
A similar treatment can be applied in the backward case.
$$ B_n = \\ M \cdot b \sum_{k=0}^{\infty} { a^k \left[ cos( \alpha n + \phi )cos( \alpha k ) - sin( \alpha n + \phi )sin( \alpha k ) \right] } \tag {20} $$
The forward and backward exponential averages of a sinusoidal signal are shown in Figure 3.
When the averaged sequence is calculated, the sine terms cancel each other out.
$$ A_n = M \cdot b \sum_{k=0}^{\infty} { a^k \left[ cos( \alpha n + \phi )cos( \alpha k ) \right] } \tag {21} $$
Since the $ n $ based cosine expression is independent of the summation iterator, it can be factored out of the summation.
$$ A_n = M cos( \alpha n + \phi ) \cdot b \sum_{k=0}^{\infty} { a^k cos( \alpha k ) } \tag {22} $$
The expression on the left of the dot is the definition of the original signal. The expression on the right of the dot is only dependent on the smoothing factor and the frequency. Notably, it is independent of the amplitude or phase of the signal, and it is independent of the sequence subscript. For a given frequency, the expression on the right will be the same for all subscript values. This means it is a multiplier of the signal. Since exponential averaging smooths down bumps, the multiplier is positive and has a value less than one. Therefore it can be called a dampening factor.
This proves that for a sinusoidal signal, the forward and backward lags do indeed cancel each other out exactly.
The same treatment can be applied to the difference case, the distinction being the cosine terms get cancelled and the sine terms remain.
$$ D_n = -M \cdot b \sum_{k=0}^{\infty} { a^k \left[ sin( \alpha n + \phi )sin( \alpha k ) \right] } \tag {23} $$
$$ D_n = -M sin( \alpha n + \phi ) \cdot b \sum_{k=0}^{\infty} { a^k sin( \alpha k ) } \tag {24} $$
Just like the averaged case, the expression on the left of the dot is a sinusoidal, and the expression on the right is a dampening factor.
The averaged and difference cases are shown in Figure 4.
Closed Form Equation for the Averaged Case
The summation in the dampening factors are a bit unwieldy. With a little bit of work the summation can be eliminated and replaced with a closed form expression.
Here is where the exponential nature of the sinusoidal functions comes into play.
$$ cos( \theta ) = { e^{ i \theta } + e^{ -i \theta } \over 2 } \tag {25} $$
The definition in (25) can be plugged into the summation expression of (22).
$$ \sum_{k=0}^{\infty} { a^k cos( \alpha k ) } = { 1 \over 2 } \sum_{k=0}^{\infty} { a^k ( e^{ i \alpha k } + e^{ -i \alpha k } ) } \tag {26} $$
The sum inside the summation allows the summation to be broken out as two geometric series summations.
$$ \sum_{k=0}^{\infty} { a^k cos( \alpha k ) } = { 1 \over 2 } \left[ \sum_{k=0}^{\infty} { ( a e^{ i \alpha } )^k } + \sum_{k=0}^{\infty} { ( a e^{ -i \alpha } )^k } \right]\tag {27} $$
The summation formula for an infinite geometric series can now be applied.
$$ \sum_{k=0}^{\infty} {x^k} = { 1 \over { 1 - x } } \tag {28} $$
$$ \sum_{k=0}^{\infty} { a^k cos( \alpha k ) } = { 1 \over 2 } \left[ { 1 \over { 1 - a e^{ i \alpha } } } + { 1 \over { 1 - a e^{ -i \alpha } } } \right] \tag {29} $$
The summations have now been eliminated. The fractions inside the brackets can be added to make a single fraction.
$$ \sum_{k=0}^{\infty} { a^k cos( \alpha k ) } = { 1 \over 2 } \left[ { 2 - a e^{ i \alpha } - a e^{ -i \alpha } } \over { 1 - a e^{ i \alpha } - a e^{ -i \alpha } + a^2 } \right] \tag {30} $$
The exponential definition of a cosine (25) can now be used in reverse, twice. The result is a simplified summation expression.
$$ \sum_{k=0}^{\infty} { a^k cos( \alpha k ) } = { { 1 - a cos \alpha } \over { 1 - 2 a cos \alpha + a^2 } } \tag {31} $$
Plugging the summation expression back into (22) and replacing $ b $ with $ ( 1 - a ) $ gives a final closed form expression for the averaged smoothing of a sinusoidal signal.
$$ A_n = M cos( \alpha n + \phi ) \cdot { { ( 1 - a ) ( 1 - a cos \alpha ) } \over { 1 - 2 a cos \alpha + a^2 } } \tag {32} $$
Interestingly, the denominator in the dampening expression is in the form of the Law of Cosines.
Closed Form Equation for the Difference Case
A closed form expression for the difference case can be derived in a similar manner to the average case. Since it is sine based, the exponential definition of the sine function is needed.
$$ sin( \theta ) = { e^{ i \theta } - e^{ -i \theta } \over 2 i } \tag {33} $$
The sine definition can then be substituted into the summation expression.
$$ \sum_{k=0}^{\infty} { a^k sin( \alpha k ) } = { 1 \over {2i} } \sum_{k=0}^{\infty} { a^k ( e^{ i \alpha k } - e^{ -i \alpha k } ) } \tag {34} $$
The summation can be split into two geometric series summations.
$$ \sum_{k=0}^{\infty} { a^k sin( \alpha k ) } = { 1 \over {2i} } \left[ \sum_{k=0}^{\infty} { ( a e^{ i \alpha } )^k } - \sum_{k=0}^{\infty} { ( a e^{ -i \alpha } )^k } \right]\tag {35} $$
The infinite geometric series summation formula can be applied to each summation.
$$ \sum_{k=0}^{\infty} { a^k sin( \alpha k ) } = { 1 \over {2i} } \left[ { 1 \over { 1 - a e^{ i \alpha } } } - { 1 \over { 1 - a e^{ -i \alpha } } } \right] \tag {36} $$
The difference of the two fractions in the brackets can be calculated.
$$ \sum_{k=0}^{\infty} { a^k sin( \alpha k ) } = { 1 \over {2i} } \left[ { a e^{ i \alpha } - a e^{ -i \alpha } } \over { 1 - a e^{ i \alpha } - a e^{ -i \alpha } + a^2 } \right] \tag {37} $$
The sine and cosine exponential definitions can be used to simplify the fraction.
$$ \sum_{k=0}^{\infty} { a^k sin( \alpha k ) } = { { a sin \alpha } \over { 1 - 2 a cos \alpha + a^2 } } \tag {38} $$
Finally, the summation expression can be replaced in the difference case (24).
$$ D_n = -M sin( \alpha n + \phi ) \cdot { { ( 1 - a ) a sin \alpha } \over { 1 - 2 a cos \alpha + a^2 } } \tag {39} $$
Comparison with the Derivative
If the signal (16) was defined as a function instead of a sequence, then the derivative could be taken.
$$ { dS \over dn } = -M sin( \alpha n + \phi ) \cdot \alpha \tag {40} $$
The similarity to (39) is apparent. When $ \alpha $ is small, which corresponds to many samples per cycle, the similarity is almost a proportional equivalence. When $ \alpha $ is small, $ sin( \alpha ) $ is roughly equal to $ \alpha $, and $ cos( \alpha ) $ is roughly equal to one.
$$ { { ( 1 - a ) a sin \alpha } \over { 1 - 2 a cos \alpha + a^2 } } \approx { { ( 1 - a ) a } \over { 1 - 2 a + a^2 } } \alpha = { { a } \over { 1 - a } } \alpha \tag {41} $$
When $ a $ has a value of 1/2, and $ \alpha $ is small, the difference case (39) becomes nearly equivalent to the derivative (40).
Amplitude Dampening of Smoothed Sinusoidal Signals
How much dampening occurs is a function of the smoothing parameter $ a $ and the frequency of the sinusoidal, represented by $ \alpha $. Figure 5 shows the dampening effect as a function of $ \alpha $ for five sample values of $ a $. The values are .1, .3, .5, .7, and .9. The higher the $ a $ value, the more dampening there is and the lower the curve on the graph. No matter what the smoothing parameter is, DC values (the far left axis) remain unscathed. As expected, higher frequency values are more impacted by smoothing than lower ones at any level of smoothing.
Figure 6 shows the dampening effect in the difference case. DC values are completely obliterated. The proportionality to $ \alpha $ when $ \alpha $ is small can be clearly seen in the graph. The steeper the line the higher the smoothing parameter. Note that the proportionality extends much further along the $ \alpha $ axis for the lower values of $ a $. Unlike the average case, high frequencies near the Nyquist limit (the far right axis) are also zeroed out.
In either case, the smoothing effect is local in the time domain, so noise gets dampened similarly to high frequency components.
Conclusion
The combination of exponential smoothing and the exponential nature of sinusoidal functions allows improved performance of the DFT by reducing the effects of noise and allowing the selection of frequency ranges to be emphasized. The frequency is preserved in both the averaged and difference cases. The phase is preserved in the average case and shifted by a quarter cycle ( $ \pi / 2 $ ) in the difference case. The original magnitude can be found by dividing the recovered magnitude by the dampening factor for the known frequency and smoothing factor. This allows the DFT to be used in a much more targeted, noise resistant way.
Previous post by Cedron Dawg:
Phase and Amplitude Calculation for a Pure Real Tone in a DFT: Method 1
Next post by Cedron Dawg:
DFT Bin Value Formulas for Pure Complex Tones
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Event detail Analysis and PDE Seminar: Fredholm theory and the resolvent of the Laplacian near zero energy on asymptotically conic spaces
Seminar | February 5 | 4-5 p.m. | 740 Evans Hall
András Vasy, Stanford University
We consider geometric generalizations of Euclidean low energy resolvent estimates, such as estimates for the resolvent of the Euclidean Laplacian plus a decaying potential, in a Fredholm framework. More precisely, the setting is that of perturbations \(P(\sigma )\) of the spectral family of the Laplacian \(\Delta _g-\sigma ^2\) on asymptotically conic spaces \((X,g)\) of dimension at least \(3\), and the main result is uniform estimates for \(P(\sigma )^{-1}\) as \(\sigma \to 0\) on microlocal variable order spaces under an assumption on the nullspace of \(P(0)\) on the appropriate function space (which in the Euclidean case translates to \(0\) not being an \(L^2\)-eigenvalue or having a half-bound state). These spaces capture the limiting absorption principle for \(\sigma \neq 0\) in a lossless, in terms of decay, manner.
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I would like to generate correlated data draws across two unrelated distributions (or across two empirical datasets that are from unknown distributions). The correct way to do this is to use a copula. I'm not a copula expert.
Mechanically, my steps are as follows:
Do multivariate normal draws with a fixed correlated structure. Map the resulting draws to their quantiles. Use these quantiles on the CDF (or eCDF) of the desired target distribution Take a rank correlation of the resulting data to verify the correlation structure is preserved. (Simulate this many times to ensure that finite sample bias is zero)
I wrote a bunch of custom code to do this for my use case, only to discover that I had a slight (3-5%) attenuation bias. Suppose I was drawing for 2 variables with correlation rho=0.6, I'd get data correlated at 0.58. Suppose rho=0.2, I'd get 0.19. Suppose rho=-0.2, I'd get -0.19. The magnitude of the bias is multiplicative and setting rho=1 or rho=0 worked the way you'd expect (1/0 observed correlation)
Figuring there was something wonky with my code, I tried using the
copula package in R to do it parametrically:
# Load copula librarylibrary(copula)# Normal copula, rho=0.6copula = normalCopula(0.6, dim = 2, dispstr="ex")# Output distributions: uniform, normal, parameterizeddecl = mvdc(copula = copula, margins = c("unif", "norm"), paramMargins = list(list(min=1, max=1000), list(mean=75, sd=8)))# Run 1,000 simulations of 1,000 observationsresults = replicate(1000, rMvdc(1000, decl))# Get the average correlation coefficient across all 1,000 simssummary(t(apply(results, 3, cor, method="spearman")))
The result is 0.58 correlation.
Before writing this question, I did a search on whether this is a known property of copula. This article suggests it is the case -- under "Using Rank Correlation Coefficients":
The correlation parameter, $\rho$, of the underlying bivariate normal determines the dependence between X1 and X2 in this construction. However, the linear correlation of X1 and X2 is not $\rho$. For example, in the original lognormal case, a closed form for that correlation is: $cor(X1, X2) = \frac{e^{\rho\sigma^2} - 1}{e^{\sigma^2} - 1}$ which is strictly less than $\rho$, unless $\rho$ is exactly 1. In more general cases such as the Gamma/t construction, the linear correlation between X1 and X2 is difficult or impossible to express in terms of $\rho$, but simulations show that the same effect happens.
Well, sugar. I'm in the "more general case" and I would like a closed form for the correlation so I can develop a correction to fix this. My end goal is writing a method in a package for simulating particular data in R (think
wakefield but a little different). It's not the end of the world if this is a known property that I cannot work around, but I would at least like to understand a little bit better what's going on to be able to document the behaviour.
Can anyone help intuit what's going on here or point me to a discussion of this phenomenon or, even better, let me know what kind of correction I can apply to resolve this given arbitrary marginal distributions and/or empirical distributions?
Thanks in advance.
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Attempted answers:
MD 1. \\(\Delta\\) is monotonic, because \\(a \leq b \to (a,a) \leq (b,b)\\), by definition of \\(\leq_{A\times A}\\)
MD 2. By the method in lecture 6, r(y,z) = least upper bound of X = \\(\\{x : \Delta (x) \leq (y,z) \\}\\). Since X is the set of elements of A less than min(y,z), r(y,z) is min(y,z).
MD 3. By duality, l(y,z) = max(y,z)
MD 4.
r: least upper bound of X = \\(\\{x : \Delta (x) \leq (y,z) \\}\\): least common multiple
l: greatest lower bound of X = \\(\\{x : \Delta (x) \geq (y,z) \\}\\) : greatest common divisor
I think I've been sloppy and got some of this flipped - to be fixed later.
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For a continuous signal (function), we have Bernstein inequality : $$ |{df(t)}/dt| \le 2AB\pi $$ where $A=\sup|f(t)|$ and $B$ is the bandwidth of $f(t)$. The question is: is there a relationship for a discrete function $x[n]$ like this? $$ |x[n] -x[n-1] | \le\ \mu\ W $$ where $$ X[k] = \sum\limits_{n = 0}^{N - 1} {x[n]{e^{ - j\frac{{2\pi }}{N}nk}}} $$ is the DFT for $x[n]$, $X[k]=0$ for $k> W$.
As shown in the answers to this question, for continuous-time signals, the bound predicted by Bernstein's inequality is achieved with equality by a sinusoidal signal with a frequency equal to the upper frequency limit.
In analogy to this, in this answer I'll show a bound on $\big|x[n]-x[n-1]\big|$ for a discrete-time sinusoidal signal $x[n]$ at angular frequency $W$:
$$x[n]=A\sin(nW+\phi)\tag{1}$$
For the signal $x[n]$ given by $(1)$, the largest value of $\big|x[n]-x[n-1]\big|$ for all $n\in\mathbb{Z}$ occurs for two values $x[k]$ and $x[k-1]$ symmetrical to the point of the largest derivative of a (continuous) sinusoid, i.e., for $x[k]=A\sin(W/2)$ and $x[k-1]=A\sin(-W/2)$. Consequently, the bound for a sinusoid with frequency $W$ is given by
$$\big|x[n]-x[n-1]\big|\le 2A\sin\left(\frac{W}{2}\right)\tag{2}$$
This bound is achieved with equality for the signals
$$x[n]=A\sin\left(nW+\frac{(2l+1)W}{2}\right),\qquad l\in\mathbb{Z}\tag{3}$$
Note that the constant $A$ does not generally equal the maximum amplitude $B=\max |x[n]|$ of $x[n]$. Depending on the sampling phase, the maximum amplitude $B$ lies in the following interval:
$$A\cos\left(\frac{W}{2}\right)\le B\le A\tag{4}$$
Consequently, we have
$$A\le\frac{B}{\cos\left(\frac{W}{2}\right)},\qquad 0<W<\pi\tag{5}$$
Combining $(2)$ and $(5)$ we get
$$\big|x[n]-x[n-1]\big|\le 2B\tan\left(\frac{W}{2}\right),\qquad B=\max_n\big|x[n]\big|\tag{6}$$
Note, however, that $(6)$ is only useful for $\tan(W/2)<1$, i.e., for $W<\pi/2$ because for any $x[n]$ with $\max|x[n]|=B$ we must have
$$\big|x[n]-x[n-1]\big|\le 2B\tag{7}$$
I believe that the bound $(6)$ holds for all discrete-time band-limited signals with a maximum frequency $W$ (i.e., with $X(e^{j\omega})=0$ for $|\omega|\in (W,\pi]$), but I don't know how to show it.
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There's a subtle issue here that is not mentioned in the question regarding estimation of the standard deviation of $\overline{x}$'s sampling distribution.
Suppose you have a sample from a population with mean $\mu$ and variance $\sigma^2$. When $\sigma^2$ is known, $${\rm SE}(\overline{x}) = \sigma/\sqrt{n}$$ is exactly the standard deviation of the sample mean. In practice, you usually don't know $\sigma^2$, so you instead plug in the sample variance $\hat\sigma$ to use $${\rm SE}(\overline{x}) = \hat\sigma/\sqrt{n}$$ to estimate the standard deviation of $\overline{x}$. This distinction is actually important - when the variance is unknown,
this additional uncertainty must be incorporated into the hypothesis test. This is why, even when the sample is normally distributed, the test statistic has a $t$-distribution (which has longer tails) instead of a normal distribution when $\sigma$ is unknown.
Is it correct to say that $t=(\overline{x}−μ)/SE(\overline{x})$ follows a normal distribution for ANY population (not just normally distributed), as long as the samples sizes are significant in size (by means of the central limit theorem)
This is almost correct. The population must have finite variance (i.e. not have tails that are "too long") for this to be the case. Even when the population does have a finite variance, the population distribution can have a large effect on how long until the CLT "kicks in". For shorter tailed distributions this convergence is faster. For long-tailed distributions it can take quite a while (e.g. see my example here).
Note that since we're talking about a "large sample" result here, this is true regardless of whether or not you know $\sigma$ since $\hat \sigma$ gets closer to the true $\sigma$ as the sample size increases.
And, is it correct that t follows a t-distribution when the sample size is small, but then the population only if the population is normally distributed, because the central limit theorem does not apply?
Again, assuming we're in the "$\sigma$ is unknown" world, $t$ only follows a $t$-distribution when the sample is normally distributed, which I think is what you're saying here. Related to what I said in the beginning, if $\sigma$ is known, then $t$ will have a (exact) normal distribution if the sample is normally distributed.
To summarize:
If $\sigma$ is known, and the population is normally distributed: $t$ has a normal distribution. If $\sigma$ is unknown, and the population is normally distributed: $t$ has a $t$-distribution. If the population is not normally distributed but meets the regularity requirements of the CLT: $t$ has an approximate normal distribution whether or not $\sigma$ is known. That is, the distribution of $t$ converges to a normal distribution as the sample size increases.
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With a similar reasoning as here, I might be able to give an answer to your question under certain conditions.
Let $x_{i}$ be your true value for the $i^{th}$ data point and $\hat{x}_{i}$ the estimated value. If we assume that the differences between the estimated and true values have
mean zero (i.e. the $\hat{x}_{i}$ are distributed around $x_{i}$)
follow a Normal distribution
and all have the same standard deviation $\sigma$
in short:
$$\hat{x}_{i}-x_{i} \sim \mathcal{N}\left(0,\sigma^{2}\right),$$
then you really want a confidence interval for $\sigma$.
If the above assumptions hold true $$\frac{n\mbox{RMSE}^{2}}{\sigma^{2}} = \frac{n\frac{1}{n}\sum_{i}\left(\hat{x_{i}}-x_{i}\right)^{2}}{\sigma^{2}}$$ follows a $\chi_{n}^{2}$ distribution with $n$ (not $n-1$) degrees of freedom. This means
\begin{align}P\left(\chi_{\frac{\alpha}{2},n}^{2}\le\frac{n\mbox{RMSE}^{2}}{\sigma^{2}}\le\chi_{1-\frac{\alpha}{2},n}^{2}\right) = 1-\alpha\\\Leftrightarrow P\left(\frac{n\mbox{RMSE}^{2}}{\chi_{1-\frac{\alpha}{2},n}^{2}}\le\sigma^{2}\le\frac{n\mbox{RMSE}^{2}}{\chi_{\frac{\alpha}{2},n}^{2}}\right) = 1-\alpha\\\Leftrightarrow P\left(\sqrt{\frac{n}{\chi_{1-\frac{\alpha}{2},n}^{2}}}\mbox{RMSE}\le\sigma\le\sqrt{\frac{n}{\chi_{\frac{\alpha}{2},n}^{2}}}\mbox{RMSE}\right) = 1-\alpha. \end{align}
Therefore, $$\left[\sqrt{\frac{n}{\chi_{1-\frac{\alpha}{2},n}^{2}}}\mbox{RMSE},\sqrt{\frac{n}{\chi_{\frac{\alpha}{2},n}^{2}}}\mbox{RMSE}\right]$$is your confidence interval.
Here is a python program that simulates your situation
from scipy import stats
from numpy import *
s = 3
n=10
c1,c2 = stats.chi2.ppf([0.025,1-0.025],n)
y = zeros(50000)
for i in range(len(y)):
y[i] =sqrt( mean((random.randn(n)*s)**2))
print "1-alpha=%.2f" % (mean( (sqrt(n/c2)*y < s) & (sqrt(n/c1)*y > s)),)
Hope that helps.
If you are not sure whether the assumptions apply or if you want to compare what I wrote to a different method, you could always try bootstrapping.
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It looks like the pivot is slightly displaced from the beam's center of mass. If the center of mass of the beam is displaced slightly below the pivot by some distance $D$, then just pick the pivot as the center of mass and calculate the torque from that small displacement as the angle of the beam changes. The force $g\, M$ of the center of mass is downward, and it moves to the right by $D \sin \theta$ when the beam tilts at angle $\theta$, so the torque due to that is just $g\, M\, D \sin \theta$.
The torque from the 5kg mass must be equal and opposite. If the smaller mass $m$ is a distance $d$ from the center of the beam, then the torque is approximately $g\, m\, d \cos\theta$ -- or more precisely $g\, m\, \sqrt{d^2 + D^2} \cos [\theta + \arccos(D/\sqrt{d^2 + D^2})]$. To get the second form here, you just use the magnitude of the force ($g\, m$) times the distance between the pivot and where that force is applied ($\sqrt{d^2 + D^2}$, which is the hypotenuse of the little triangle with legs $d$ and $D$) times the sine of the angle between $\vec{r}$ and $\vec{F}$ -- or equivalently the cosine of $90^\circ$ minus that angle. Now, when the beam is level that angle is $90^\circ - \arccos(D/\sqrt{d^2 + D^2})$, which you can see by looking at that triangle again. And as you tilt the beam, you just increase the angle (decreasing the torque).
I also want to add a more general comment about torque. Calculations with torque are really easy (especially in equilibrium), but students frequently forget the most important thing to do in such calculations: pick an origin and stick with it. You can pick any origin at all, and the calculation should come out right. Just be sure to use the same origin for all the torques you use.
Even though any origin should work, there is one little trick to choosing the origin that can make things easier. Depending on where you choose your origin, certain hard-to-calculate things may cancel out. Now, remembering that torque is $\vec{r} \times \vec{F}$, you'll want to choose an origin where $\vec{r}=0$ for some force, or the force is perpendicular to $\vec{r}$, for as many of the forces as possible.
In your example, choosing the origin somewhere along the beam makes it at least easier to calculate the angles. One natural choice would be at the pivot point, because $\vec{r}=0$ for that force, so there would be no force from the pivot. But you could also choose the origin at the location of the mass, in which case you would need the force of the pivot.
Now, you may be concerned that you have no information about the force exerted by the pivot. But you can make another equation that will tell you it. The sum of the torques must be zero in equilibrium, but the sum of the
forces must also be zero in equilibrium. This looks like a harder way to solve the problem in this particular case, but there are other cases where it can be useful.
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I am trying to Plot an integral equation that involves exponential function. My code is as follow,
L[\[Alpha]_] := NIntegrate[ 1/(k + I*0.1) ( Exp[I*k*x] (Exp[Sqrt[k^2 + \[Alpha]/w^2]*w] - 1) (Exp[k*w] - 1 + I*0.1) Sqrt[ k^2 + \[Alpha]/ w^2])/((Sqrt[k^2 + \[Alpha]/w^2] + k) (Exp[ Sqrt[k^2 + \[Alpha]/w^2]*w - Exp[k*w]]) + (Sqrt[ k^2 + \[Alpha]/w^2] - k) (Exp[(k + Sqrt[k^2 + \[Alpha]/w^2]) w] - 1)), {k, -100, 100}]; Plot[{Re[L[10]], Re[L[100]], Re[L[500]]}, {x, -0.45, 0.45}, PlotRange -> Full].
But this integral gives a lot of oscillations which it should not. This is fig 2 in this article “https://arxiv.org/pdf/1508.00836.pdf” that I am trying to plot. Any help will be highly appreciated.
So this is a question pertaining to the proof for $ PSPACE-COMPLETE$ (for TQBF for example). The idea is to first prove the $ L$ $ is$ $ PSPACE$ (easy part) and next is to prove $ PSPACE-COMPLETE$ . The latter requires demonstrating an algorithm which computes the L in polynomial space. This is usually achieved by having recursive calls such that is re-used.
In TBQF proof, the equation $ \phi_{i+1}(A,B)$ = $ \exists Z [\phi_{i+1}(A,Z) \land \phi_{i+1}(Z,B) ]$ ($ Z$ is mid-point )is default recursive relation for computing TBQF truth. In any standard proof, it is said that $ \phi_{i+1}(A,B)$ is computed two times and for $ m$ nodes, this formula explodes hence, other recursive-relation should be used to bound.
However in Savitch’s proof, the recursive relation is $ Path(a,b,t)$ = $ Path(a,mid,t-1)$ AND $ Path(mid,b,t-1)$ accepts then ACCEPT. In proof, it is stated that this relation reuses-spaces.
My Question is
Why in TBQF relation space explodes while in Path, it is reused? Both of these relations looks more or less same to me because both refers to i-1 instances and will need space to store them?.
Consider the class of problems that can be computed when you have access to exponentially many processors working in parallel.
How does one capture that in a proper formalism? Is there some literature on it already? Would it be true that such a class would be a superset of BQP?
I am implementing Kalman filtering in R. Part of the problem involves generating a really huge error covariance block-diagonal matrix (dim: 18000 rows x 18000 columns = 324,000,000 entries). We denote this matrix Q. This Q matrix is multiplied by another huge rectangular matrix called the linear operator, denoted by H.
I am able to construct these matrices but it takes a lot of memory and hangs my computer. I am looking at ways to make my code efficient or do the matrix multiplications without actually creating the matrices exclusively.
library(lattice) library(Matrix) library(ggplot2) nrows <- 125 ncols <- 172 p <- ncols*nrows #--------------------------------------------------------------# # Compute Qf.OSI, the "constant" model error covariance matrix # #--------------------------------------------------------------# Qvariance <- 1 Qrho <- 0.8 Q <- matrix(0, p, p) for (alpha in 1:p) { JJ <- (alpha - 1) %% nrows + 1 II <- ((alpha - JJ)/ncols) + 1 #print(paste(II, JJ)) for (beta in alpha:p) { LL <- (beta - 1) %% nrows + 1 KK <- ((beta - LL)/ncols) + 1 d <- sqrt((LL - JJ)^2 + (KK - II)^2) #print(paste(II, JJ, KK, LL, "d = ", d)) Q[alpha, beta] <- Q[beta, alpha] <- Qvariance*(Qrho^d) } } # dn <- (det(Q))^(1/p) # print(dn) # Determinant of Q is 0 # Sum of the eigen values of Q is equal to p #-------------------------------------------# # Create a block-diagonal covariance matrix # #-------------------------------------------# Qf.OSI <- as.matrix(bdiag(Q,Q)) print(paste("Dimension of the forecast error covariance matrix, Qf.OSI:")); print(dim(Qf.OSI))
It takes a long time to create the matrix Qf.OSI at the first place. Then I am looking at pre- and post-multiplying Qf.OSI with a linear operator matrix, H, which is of dimension 48 x 18000. The resulting H
Qf.OSIHt is finally a 48×48 matrix. What is an efficient way to generate the Q matrix? The above form for Q matrix is one of many in the literature. In the below image you will see yet another form for Q (called the Balgovind form) which I haven’t implemented but I assume is equally time consuming to generate the matrix in R.
There is an interesting observation using the first derivative to deconvolve an exponentially modified Gaussian:
The animation is here, https://terpconnect.umd.edu/~toh/spectrum/SymmetricalizationAnimation.gif
The main idea is that if we have an Exponentially Modified Gaussian (EMG) function, and we add a small fraction of first derivative to the original EMG, it results in recovering the original Gaussian while preserving the original area. The constant multiplier is the 1/time constant of the EMG. This is a very useful property.
Has anyone seen this deconvoluting property of the first derivative mentioned elsewhere in mathematical literature? An early reference from the 1960s from a Chemistry paper shows a picture a similar picture. This observation was just by chance, I am looking for a fundamental connection and if the first derivative can be used to deconvolute other types of convolutions besides the exponential ones.
Thanks.
Ref: J. W., and Charles N. Reilley. “De-tailing and sharpening of response peaks in gas chromatography.” Analytical Chemistry 37, (1965), 626-630.
While reading ‘Computing the Volume, Counting Integral points, and Exponential Sums’ by A. Barvinok (1993), I came across the following:
“Moreover, let $ K$ be the conic hull of linearly independent vectors $ u_{1}, …, u_{n}$ so $ K = co(u_{1}, …, u_{n})$ . Denote by $ K^{*} = \{x\in \mathbb{R}^{n}: <x, y> \le 0 \text{ for all } y\in K\}$ the polar of K. Then for all $ c \in \text{Int} K^{*}$ we have
\begin{equation} \int_{K}exp(<c, x>)dx = |u_{1} \land … \land u_{n}|\prod_{i=1}^{n}<-c_{i}, u_{i}>^{-1} \end{equation}
To obtain the last formula we have to apply a suitable linear transformation to the previous integral. “
I have tried proving this but I can’t find relevant links to help me. Also, I’m unsure what $ |u_{1} \land … \land u_{n}|$ is supposed to mean. Would greatly appreciate if someone could point me to a relevant resource or provide proof. Thanks!
Suppose $ (X,\omega,J)$ is a closed symplectic manifold with a compatible almost complex structure. The fact below follows from McDuff-Salamon’s book on $ J$ -holomorphic curves (specifically, Lemma 4.7.3).
Given $ 0<\mu< 1$ , there exist constants $ 0<C<\infty$ and $ \hbar>0$ such that the following property holds. Given a $ J$ -holomorphic map $ u:(-R-1,R+1)\times S^1\to X$ with energy $ E(u)<\hbar$ defined on an annulus (with $ R>0$ ), we have the exponential decay estimates
(1) $ E(u|_{[-R+T,R-T]\times S^1})\le C^2e^{-2\mu T}E(u)$
(2) $ \sup_{[-R+T,R-T]\times S^1}\|du\|\le Ce^{-\mu T}\sqrt{E(u)}$
for all $ 0\le T\le R$ . Here, we take $ S^1 = \mathbb R/2\pi\mathbb Z$ and use the standard flat metric on the cylinder $ \mathbb R\times S^1$ and the metric on $ X$ to measure the norm $ \|du\|$ .
Now, if $ J$ were integrable, we can actually improve this estimate further in the following way: at the expense of decreasing $ \hbar$ and increasing $ C$ , we can actually take $ \mu=1$ in (1) and (2) above. The idea would be to use (2) to deduce that $ u|_{[-R,R]\times S^1}$ actually maps into a complex coordinate neighborhood on $ X$ where we can use the Fourier expansion of $ u$ along the cylinder $ [-R,R]\times S^1$ to obtain the desired estimate.
I would like to know: is it possible to improve the estimate to $ \mu=1$ also in the case when $ J$ is not integrable? If so, a proof with some details or a reference would be appreciated. If not, what is the reason and is it possible to come up with a (counter-)example to illustrate this?
I read about NPC and its relationship to PSPACE and I wish to know whether NPC problems can be deterministicly solved using an algorithm with worst case polynomial space requirement, but potentially taking exponential time (2^P(n) where P is polynomial).
Moreover, can it be generalised to EXPTIME in general?
The reason I am asking this is that I wrote some programs to solve degenerate cases of an NPC problem, and they can consume very large amounts of RAM for hard instances, and I wonder if there is a better way. For reference see https://fc-solve.shlomifish.org/faq.html .
I’m learning about Exponential search and I keep reading that the worst-case performance is $ log_2(i)$ where $ i$ is the searched index.
I tried with an array containing $ 1073741824$ elements $ (1024*1024*1024)$ and I searched for the $ 1073741823$ th element (which is just $ n-1$ ).
Therefore, I used the exponential search to get the range containing $ i$ (which was $ [536870912 ; 1073741824]$ and it took $ log_2(n)$ to find. Then it took another $ log_2(n/2)$ to find the right index using Binary search in this range only.
Am I doing something wrong? Or is the complexity of the worst-case $ 2*log_2(n)-1$ ?
Suppose I have a real number $ $ x=\sum_{i=1}^n a_i e^{\lambda_i} $ $ where $ a_i,\lambda_i$ s are complex algebraic numbers.
Is there an algorithm to determine whether it is greater than 0 or less than 0?
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This is a quite simple question: quarks do mix (through the CKM matrix), neutrinos do mix (through the PMNS matrix).
Then
why do charged leptons not mix?
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Short answer: only the down-type quarks mix by the CKM matrix, by convention and without loss of generality. The lepton sector is completely analogous to the quark sector where the up-type quarks play the role of the neutrinos and the down-type quarks play the role of the charged leptons.
Long answer: Let's recall some facts about the standard model. You have three quark weak-doublets $Q_i = (u_L, d_L), (c_L, s_L), (t_L, b_L)$ for $i=1,2,3$, and six quark singlets $u_{Ri}=(u_R, c_R, t_R)$ and similarly for the down-type quarks $d_{Ri}$. The Yukawa couplings for these quarks, which are what end up giving them mass after Higgsing, look like this (I have made no attempt to match this to some standard notation, this should just serve to give the idea):
$$ \mathcal{L} \supset y^U_{ij} Q_i \Phi u_{Rj}+ y^D_{ij} Q_j \tilde{\Phi} d_{Rj} + h.c. $$
If it weren't for these couplings the standard model would have an $(SU(3))^3$ global symmetry under three independent unitaries acting on the generation indices of the doublet, up- and down-type singlets respectively. In other words we could make field redefinitions
$$Q_{i} \to U_{ij}^{Q}Q_{j}\\ u_{Ri} \to U_{ij}^{U}u_{Rj}\\ d_{Ri} \to U_{ij}^{D}d_{Rj} $$
with three independent unitaries $U^{Q,U,D}$ without changing anything else in the Lagrangian apart from the Yukawa terms. That means we could choose $U^Q$ and $U^U$ to diagonalise the up-type quark term (A general fact of matrix algebra: you can diagonalise a matrix if you are free to multiply on the right and left by independent unitaries). So you can diagonalise $y^U$ at the expense of choosing $U^Q$ and $U^U$. Now we try to diagonalise $y^D$, but there's a hitch: we're no longer free to choose $U^Q$. We can only choose $U^D$, so we can't completely diagonalise $y^D$. After symmetry breaking there will inevitably be off-diagonal terms (unless there is some really incredible unexpected flavour symmetry!) which will cause oscillations between the down-type quarks (d,s,b). The CKM matrix, which is related in some way to $y^D, U^Q$ and $U^D$, measures this.
So how are leptons different? You just replace $u\to e, d\to\nu_e,$ etc. In the pure standard model there are no right-handed neutrino singlets analogous to $d_{Ri}$. So you can diagonalise the lepton Yukawas and be done with it. But if you bring in an extension of the standard model to explain neutrino oscillations (which requires at least two right handed neutrinos - I'll assume for simplicity there are three, one for each flavour), then you will have a Yukawa matrix for the neutrinos like there was for the down-type quarks, either appearing directly in your theory or as an effective operator after integrating out some more complicated hidden sector. Now the same logic applies as before. You can always diagonalise the charged lepton Yukawas, no worries, but you don't have enough freedom to completely diagonalize the neutrino Yukawas. There is an analog of the CKM matrix called the PMNS matrix in this context.
This is not to say that neutrino oscillations have no (potentially) observable consequences on the charged leptons. Consider this diagram:
This is clearly forbidden as a real process by energy conservation, but if you radiate a photon then you can get the real lepton flavour violating decay $\mu^- \to e^- \gamma$ which people have looked for. (Check out the Particle Data Group for the current limits on the rate.) Neuneck points out that the standard model rate for this process is very small, so, while it is technically nonzero in SM, any observation of this process will be strong evidence for beyond standard model physics.
It is as @dmckee says, the weak eigenstates are the same as the mass eigenstates for the leptons, therefore the corresponding Cabibo angle is zero.
Without going into the mathematics this plot says all:
The Cabibbo angle represents the rotation of the mass eigenstate vector space formed by the mass eigenstates . θC = 13.04°.
The CKM matrix is an experimental fact:
Note, however, that the specific values of the angles are not a prediction of the standard model: they are open, unfixed parameters. At this time, there is no generally accepted theory that explains why the measured values are what they are.
Thus that the mass eigenstates and the weak eigenstates for leptons have zero angle is an experimental observation too.
Generally physics theories answer how one state implies/transforms into another. "Why" questions can be answered by "how" answers until one reaches the "that is what the experiment says". This is one of those "why" questions.
Somewhat reluctantly, I decided to add to the otherwise excellent technical answers above, since none confronted the fundamental false premise of your question, "why do charged leptons not mix?". Of course they do.
. Charged weak currents merely link generations
Let me review its antecedents as you seem to be aware of the phenomenon, when you have all quarks, ups and downs, mix, and not just the downs, as popular convention has it. Consider the early picture of quarks with only 2 generations, some time in the 70s, but without K-M. Schematically, skipping chiral structures and Lorentz indices, the charged current vertex in the weak effective hamiltonian is $$ \overline{\begin{bmatrix} u \\ c \end{bmatrix} }\cdot \begin{bmatrix} d^\prime \\ s^\prime \end{bmatrix} ~W^+= \overline{\begin{bmatrix} u \\ c \end{bmatrix} }\cdot\begin{bmatrix} \cos{\theta_\mathrm{c}} & \sin{\theta_\mathrm{c}} \\ -\sin{\theta_\mathrm{c}} & \cos{\theta_\mathrm{c}}\\ \end{bmatrix}\begin{bmatrix} d \\ s \end{bmatrix} W^+,$$plus the hermitean conjugate,$$ \overline{\begin{bmatrix} d' \\ s' \end{bmatrix} }\cdot \begin{bmatrix} u \\ c \end{bmatrix} ~W^-= \overline{\begin{bmatrix} d \\ s \end{bmatrix} }\begin{bmatrix} \cos{\theta_\mathrm{c}} & -\sin{\theta_\mathrm{c}} \\ \sin{\theta_\mathrm{c}} & \cos{\theta_\mathrm{c}}\\ \end{bmatrix}\cdot\begin{bmatrix} u \\ c \end{bmatrix} W^- . $$You did not ask about any up vs down asymmetries because there aren't any:if you chose, you could have left the down quarks alone and "mixed" the ups instead,$$ \begin{bmatrix} u' \\ c' \end{bmatrix}= \begin{bmatrix} \cos{\theta_\mathrm{c}} & -\sin{\theta_\mathrm{c}} \\ \sin{\theta_\mathrm{c}} & \cos{\theta_\mathrm{c}}\\ \end{bmatrix}\begin{bmatrix} u \\ c \end{bmatrix}. $$Since the weak vertices are specified by the same expression, it never occurs to you to ask who mixes: the up or the down quarks, because they all do. Historically, in 1960, when Gell-Mann and Levy introduced this mixing, only the
u, d, s quarks were known, so of course, they were looking only at the upper row of the Cabbibo matrix and naturally they had d and s mix into d' and s'. But, in full fairness, an s weak-couples to a "mixture" of u and c.
So, barring Majorana masses, and making the (monstrous?) assumption the PMNS matrix is as parameter-poor as the CKM, the analog vertex for just 2 generations (think MNS, 1962) is $$ \overline{\begin{bmatrix} e \\ \mu \end{bmatrix} }\cdot \begin{bmatrix} \nu_e \\ \nu_\mu \end{bmatrix} ~W^-= \overline{\begin{bmatrix} e \\ \mu \end{bmatrix} }\begin{bmatrix} \cos{\theta_\mathrm{p}} & -\sin{\theta_\mathrm{p}} \\ \sin{\theta_\mathrm{p}} & \cos{\theta_\mathrm{p}}\\ \end{bmatrix} \begin{bmatrix} \nu_1 \\ \nu_2 \end{bmatrix} W^- . $$This is a hypothetical world, since the PMNS mixing is
compared to the quark mixing, and no realistic split like this could be made, so $\nu_1,\nu_2$ are not the actual physical mass eigenstates $\nu_1,\nu_2$s of today's neutrino experiments: this is just to illustrate the point. huge
But you see the point: the r.h.side expression, written completely in terms of mass eigenstates of leptons, does not much care who is charged and who is neutral. Since we know so little about the $\nu_i$s it would be perverse to pretend the charged leptons mix into $l_1, ~l_2$, and some of the answers justify the logic of sticking to the present convention.
But, deep down, the formal structures have never failed to be the same. I urge my colleagues to give the $\nu_i$s memorable names (Huey, Dewey, and Ratatouille,
anything!) so they are thought of as the real, physical MCoys, and not weird formal entities of convenience, like $\nu_e$, $\nu_\mu$, $\nu_\tau$, with which we are stuck today, an effective confusion machine; but they are reluctant to go there before pinning down the hierarchy of masses...
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When we talk about the present-day number density of photons ($n_\gamma\approx 10^8 \:\mathrm m^{-3}$) in the universe do we mean the number density of CMB photons? I mean there are other sources of photons, for example, stellar or galactic sources. Are these photons included in $n_\gamma$ ?
Is it the same $n_\gamma$ which is used to quantify the baryon asymmetry in the universe?
In principle, the number density of photons include all photons, both of cosmic origin (e.g. the cosmic microwave background; CMB) and of astrophysical origin (starlight, gamma rays from gamma-ray bursts, radio waves from quasars, etc.).
However, CMB photons outnumber all other types of photons by more than 200:1.
The cosmic background radiation
The figure below, from Hill et al. (2018), shows the brightness of the sky over the full electromagnetic spectrum, from radio to gamma rays:
More specifically, the $y$ axis shows the
specific intensity $I_\nu$, multiplies by the frequency $\nu$. This is a convenient measure because it gives the contribution per logarithmic scale, so when plotted on a logarithmic scale, if two peaks are equally wide, the one with the highest value of $\nu I_\nu$ has a larger energy density.
Thus you see that, by far, the largest contribution comes from the CMB. The second largest contribution to the energy density are the cosmic infrared and optical background (CIB and COB), which come from galaxies. At extreme frequencies, you have the even lower X-ray and gamma-ray background (CXB and CGB), which come from active galactic nuclii (quasars etc.). See also model fit to these observation by Inoue 2014, fig. 1.
Photon number densities
However, since photons have different energies, a larger amount of photons is needed to produce a given energy for low-energy photons than for high-energy photons. Dividing by $\nu$ to get $I_\nu$ and by Planck's constant $h$ to get the number flux, and multiplying by $4\pi$ gives the photon flux from all directions, i.e. number of photons per second. Further dividing by the speed of light $c$ gives the number density. That is, $$ n = \nu I_\nu \times \frac{4\pi}{h \nu} \frac{1}{c} $$ In the plot below, I took the data from the plot above, interpolated a bit, and calculated the number density:
For each "family" of photons, I integrated the number densities across the frequency bands, writing the numbers in black. The CMB photons have a total number density of $411\,\mathrm{cm}^{-3}$, which is seen to be more than two orders of magnitude more than all of the other photons combined!
Note that the UV background is quite uncertain, both because UV observations from ground is very difficult so you'll have to go to space, and because interstellar hydrogen is very efficient at absorbing UV radiation.
Analytical expression for the number density
Because the CMB is described by a near-perfect blackbody of temperature $T = 2.7255\,\mathrm{K}$, their number density $n_\mathrm{CMB}$ can be calculated analytically as $$ \begin{array}{rcl} n_\mathrm{CMB} & = & 16\pi \left( \frac{kT}{hc} \right)^3 \zeta(3) \\ & \simeq & 411\,\mathrm{cm}^{-3}. \end{array} $$ Here, $k$, $h$, $c$, and $\zeta$ are Boltzmann's constant, Planck's constant, the speed of light, and the Riemann zeta function, respectively.
Baryon asymmetry
As for your second question, when particles and antiparticles annihiliate, they emit gamma-rays, which become a part of the CGB, but in the
very high frequency end, like $\nu>10^{20}$ Hz. Thus, the CGB can, as you suggest, be used to constrain the baryon asymmetry in the Universe (see e.g. Ballmoos 2014). But if you define $n_\gamma$ as the number density of all photons, these photons contribute negligibly to $n_\gamma$.
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Even before quantization, charged bosonic fields exhibit a certain "self-interaction". The body of this post demonstrates this fact, and the last paragraph asks the question.
Notation/ Lagrangians
Let me first provide the respective Lagrangians and elucidate the notation.
I am talking about complex scalar QED with the Lagrangian $$\mathcal{L} = \frac{1}{2} D_\mu \phi^* D^\mu \phi - \frac{1}{2} m^2 \phi^* \phi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$ Where $D_\mu \phi = (\partial_\mu + ie A_\mu) \phi$, $D_\mu \phi^* = (\partial_\mu - ie A_\mu) \phi^*$ and $F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$. I am also mentioning usual QED with the Lagrangian $$\mathcal{L} = \bar{\psi}(iD_\mu \gamma^\mu-m) \psi - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$ and "vector QED" (U(1) coupling to the Proca field) $$\mathcal{L} = - \frac{1}{4} (D^\mu B^{* \nu} - D^\nu B^{* \mu})(D_\mu B_\nu-D_\nu B_\mu) + \frac{1}{2} m^2 B^{* \nu}B_\nu - \frac{1}{4} F^{\mu \nu} F_{\mu \nu}$$
The four-currents are obtained from Noether's theorem. Natural units $c=\hbar=1$ are used. $\Im$ means imaginary part.
Noether currents of particles
Consider the Noether current of the complex scalar $\phi$ $$j^\mu = \frac{e}{m} \Im(\phi^* \partial^\mu\phi)$$ Introducing local $U(1)$ gauge we have $\partial_\mu \to D_\mu=\partial_\mu + ie A_\mu$ (with $-ieA_\mu$ for the complex conjugate). The new Noether current is $$\mathcal{J}^\mu = \frac{e}{m} \Im(\phi^* D^\mu\phi) = \frac{e}{m} \Im(\phi^* \partial^\mu\phi) + \frac{e^2}{m} |\phi|^2 A^\mu$$ Similarly for a Proca field $B^\mu$ (massive spin 1 boson) we have $$j^\mu = \frac{e}{m} \Im(B^*_\mu(\partial^\mu B^\nu-\partial^\nu B^\mu))$$ Which by the same procedure leads to $$\mathcal{J}^\mu = \frac{e}{m} \Im(B^*_\mu(\partial^\mu B^\nu-\partial^\nu B^\mu))+ \frac{e^2}{m} |B|^2 A^\mu$$
Similar $e^2$ terms also appear in the Lagrangian itself as $e^2 A^2 |\phi|^2$. On the other hand, for a bispinor $\psi$ (spin 1/2 massive fermion) we have the current $$j^\mu = \mathcal{J}^\mu = e \bar{\psi} \gamma^\mu \psi$$ Since it does not have any $\partial_\mu$ included.
"Self-charge"
Now consider very slowly moving or even static particles, we have $\partial_0 \phi, \partial_0 B \to \pm im\phi, \pm im B$ and the current is essentially $(\rho,0,0,0)$. For $\phi$ we have thus approximately $$\rho = e (|\phi^+|^2-|\phi^-|^2) + \frac{e^2}{m} (|\phi^+|^2 + |\phi^-|^2) \Phi$$ Where $A^0 = \Phi$ is the electrostatic potential and $\phi^\pm$ are the "positive and negative frequency parts" of $\phi$ defined by $\partial_0 \phi^\pm = \pm im \phi^\pm$. A similar term appears for the Proca field.
For the interpretation let us pass back to SI units, in this case we only get a $1/c^2$ factor. The "extra density" is $$\Delta \rho = e\cdot \frac{e \Phi}{mc^2}\cdot |\phi|^2$$ That is, there is an extra density proportional to the ratio of the energy of the electrostatic field $e \Phi$ and the rest mass of the particle $mc^2$. The sign of this extra density is dependent only on the sign of the electrostatic potential and both frequency parts contribute with the same sign (which is superweird). This would mean that
classicaly, the "bare" charge of bosons in strong electromagnetic fields is not conserved, only this generalized charge is.
After all, it seems a bad convention to call $\mathcal{J}^\mu$ the electric charge current. By multiplying it by $m(c^2)/e$ it becomes a matter density current with the extra term corresponding to mass gained by electrostatic energy. However, that does not change the fact that the "bare charge density" $j^0$ seems not to be conserved for bosons.
Now to the questions:
On a theoretical level, is charge conservation at least temporarily or virtually violated for bosons in strong electromagnetic fields? (Charge conservation will quite obviously not be violated in the final S-matrix, and as an $\mathcal{O}(e^2)$ effect it will probably not be reflected in first order processes.) Is there an intuitive physical reason why such a violation is not true for fermions even on a classical level? Charged bosons do not have a high abundance in fundamental theories, but they do often appear in effective field theories. Is this "bare charge" non-conservation anyhow reflected in them and does it have associated experimental phenomena? Extra clarifying question: Say we have $10^{23}$ bosons with charge $e$ so that their charge is $e 10^{23}$. Now let us bring these bosons from very far away to very close to each other. As a consequence, they will be in a much stronger field $\Phi$. Does their measured charge change from $e 10^{23}$? If not, how do the bosons compensate in terms of $\phi, B, e, m$? If this is different for bosons rather than fermions, is there an intuitive argument why?
This post imported from StackExchange Physics at 2015-06-09 14:50 (UTC), posted by SE-user Void
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Please assume that this graph is a highly magnified section of the derivative of some function, say $F(x)$. Let's denote the derivative by $f(x)$.Let's denote the width of a sample by $h$ where $$h\rightarrow0$$Now, for finding the area under the curve between the bounds $a ~\& ~b $ we can a...
@Ultradark You can try doing a finite difference to get rid of the sum and then compare term by term. Otherwise I am terrible at anything to do with primes that I don't know the identities of $\pi (n)$ well
@Silent No, take for example the prime 3. 2 is not a residue mod 3, so there is no $x\in\mathbb{Z}$ such that $x^2-2\equiv 0$ mod $3$.
However, you have two cases to consider. The first where $\binom{2}{p}=-1$ and $\binom{3}{p}=-1$ (In which case what does $\binom{6}{p}$ equal?) and the case where one or the other of $\binom{2}{p}$ and $\binom{3}{p}$ equals 1.
Also, probably something useful for congruence, if you didn't already know: If $a_1\equiv b_1\text{mod}(p)$ and $a_2\equiv b_2\text{mod}(p)$, then $a_1a_2\equiv b_1b_2\text{mod}(p)$
Is there any book or article that explains the motivations of the definitions of group, ring , field, ideal etc. of abstract algebra and/or gives a geometric or visual representation to Galois theory ?
Jacques Charles François Sturm ForMemRS (29 September 1803 – 15 December 1855) was a French mathematician.== Life and work ==Sturm was born in Geneva (then part of France) in 1803. The family of his father, Jean-Henri Sturm, had emigrated from Strasbourg around 1760 - about 50 years before Charles-François's birth. His mother's name was Jeanne-Louise-Henriette Gremay. In 1818, he started to follow the lectures of the academy of Geneva. In 1819, the death of his father forced Sturm to give lessons to children of the rich in order to support his own family. In 1823, he became tutor to the son...
I spent my career working with tensors. You have to be careful about defining multilinearity, domain, range, etc. Typically, tensors of type $(k,\ell)$ involve a fixed vector space, not so many letters varying.
UGA definitely grants a number of masters to people wanting only that (and sometimes admitted only for that). You people at fancy places think that every university is like Chicago, MIT, and Princeton.
hi there, I need to linearize nonlinear system about a fixed point. I've computed the jacobain matrix but one of the elements of this matrix is undefined at the fixed point. What is a better approach to solve this issue? The element is (24*x_2 + 5cos(x_1)*x_2)/abs(x_2). The fixed point is x_1=0, x_2=0
Consider the following integral: $\int 1/4*(1/(1+(u/2)^2)))dx$ Why does it matter if we put the constant 1/4 behind the integral versus keeping it inside? The solution is $1/2*\arctan{(u/2)}$. Or am I overseeing something?
*it should be du instead of dx in the integral
**and the solution is missing a constant C of course
Is there a standard way to divide radicals by polynomials? Stuff like $\frac{\sqrt a}{1 + b^2}$?
My expression happens to be in a form I can normalize to that, just the radicand happens to be a lot more complicated. In my case, I'm trying to figure out how to best simplify $\frac{x}{\sqrt{1 + x^2}}$, and so far, I've gotten to $\frac{x \sqrt{1+x^2}}{1+x^2}$, and it's pretty obvious you can move the $x$ inside the radical.
My hope is that I can somehow remove the polynomial from the bottom entirely, so I can then multiply the whole thing by a square root of another algebraic fraction.
Complicated, I know, but this is me trying to see if I can skip calculating Euclidean distance twice going from atan2 to something in terms of asin for a thing I'm working on.
"... and it's pretty obvious you can move the $x$ inside the radical" To clarify this in advance, I didn't mean literally move it verbatim, but via $x \sqrt{y} = \text{sgn}(x) \sqrt{x^2 y}$. (Hopefully, this was obvious, but I don't want to confuse people on what I meant.)
Ignore my question. I'm coming of the realization it's just not working how I would've hoped, so I'll just go with what I had before.
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abstract
Session 58 - Future of Antarctic Astrophysics.
Display session, Wednesday, June 10
Atlas Ballroom,
We report observations of the ^3P_1\rightarrow^3P_0 fine structure transition of neutral carbon ([C I], \lambda=609\ \mum) and the J=4\rightarrow3 rotational transition of CO (\lambda=652\ \mum) on the Large Magellanic Cloud.
These measurements were performed using the Antarctic Submillimeter Telescope and Remote Observatory (AST/RO) during the 1995, 1996 and 1997 seasons. Neutral carbon originates in the warm interface regions of molecular clouds known as Photodissociation Regions (PDRs), where CO is split into its atomic components by UV radiation. [C I] emission is thus thought to trace this transition occurring at a visual extinction of a few (A_v\sim1-3). Middle rotational transitions of CO trace relatively dense (n_H\approx10^5 cm^-3) and warm (T>20 K) molecular gas. PDR regions are enlarged in low metallicity environments, where the UV penetration is enhanced by the lower dust-to-gas ratios.
We present a semianalytical model for the dependence of the [C II]/CO and [C I]/CO intensity ratios with metallicity and compare its predictions with the available observations of PDRs in metal-poor environments. The study of PDRs in the low metallicity ISM will allow us to better understand the observational diagnostics for the first cosmic epochs of star formation.
This research was supported in part by the National Science Foundation under a cooperative agreement with the Center for Astrophysical Research in Antarctica (CARA), grant number NSF OPP 89-20223. CARA is a National Science Foundation Science and Technology Center.
Program
listing for Wednesday
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How many ways are there to fill a bag (maximum of weight $N$) with watermelons, apples, and grapes, where the number of apples has to be at least the number of watermelons? The weights are given as follows:
$\omega(watermelon) = 10$
$\omega(apple) = 2$
$\omega(grapes) = 1$
So far, I've tried creating the series for each fruit and using the product lemma to multiply them. The answer would be:
the coefficient of $x^N$ of $(\sum_{j=0}^{\inf} x^j)\ (\sum_{j=0}^{inf} x^{2j})\ (\sum_{j=0}^{inf} x^{10j})$
However, this does not account for the restriction that the number of apples has to be at least the number of watermelons. I don't really know where to go from here.
Edit: Thank you Nicholas for the answer. I found another way to get the same answer. I'd say this is more of an algebraic solution where as Nicholas' solution is more combinatoric. Here is my approach: let a = #apples, w = #watermelons, g = #grapes
we know that $a = w +y$, where $y \in Z, y\ge 0$. and $g + 2a + 10w = N$, substituting we get $g + 2y + 12w$ = N.
We can now write this in terms of generating functions as $(\sum_{j=0}^{\inf} x^j)\ (\sum_{j=0}^{inf} x^{2j})\ (\sum_{j=0}^{inf} x^{12j})$, which gives the same solution.
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What is the ratio of $\ce{MbO2/Mb}$ if the solution is saturated with air?
Details of the question: $$ \begin{align} \ce{Mb + O2 (aq) &-> MbO2}\\ \Delta G' &= -30.3~\mathrm{kJ/mol}\\ T &= 298~\mathrm{K}\\ \ce{pH} &= 7\\ \text{Henry's Constant for }\ce{O2} &= 42 \cdot 10^3~\mathrm{bar}\\ \text{Content of }\ce{O2}\text{ in dry air} &= 21\%\\ \end{align} $$ Attempt at Solution:
So, basically this is a two part question, as I see it. First I need to figure out the concentration of dissolved $\ce{O2}$ using Henry's Law. Then I can simply plug this value into the $$\Delta G' = -RT\ln K$$ equation where $$K = \frac{\ce{[MbO2]}}{\ce{[O2][Mb]}}.$$
Therefore: $$ \begin{align} P &= K\cdot m\\ 0.21~\mathrm{bar} &= (42 \cdot 10^3)\cdot m\\ m &= 5 \cdot 10^{-6} \pu{mol/kg}\\ -30.300~\mathrm{J/mol} &= -8.314 \cdot 298 \cdot \ln\left(\frac{\ce{[MbO2]}}{[5\cdot10^{-6}]\ce{[Mb]}}\right) \end{align} $$ After dividing and raising both sides to $e^x$ (to get rid of $\ln$) I get $$\frac{\ce{MbO2}}{\ce{Mb}} = 1.02.$$
However the answer sheet says it is $21.6$. I cannot figure out why this is so.
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Part of your question is to find the threshold price of a crafted item that covers exactly its cost of production, especially when production is random. Once you know this threshold, the price you will ask to a buyer has to be greater than this threshold. Otherwise, you make losses.
Consider the craft of a precise item.First, you need to determine a value for each item that enters as inputs. The sum will be denoted $V_I$. If you do not have any idea about the values, you can set them as average market prices, so that $V_I$ is the average market price for all the inputs. In other words, $V_I$ is what it costs you to produce the item if you have to buy the inputs on the market.
If there were no randomness in production, you would ask at least a price $p\geq V_I$ to any buyer who does not provide you with any input. Conversely, if the buyer comes with all the inputs, you have zero cost and your constraint is simply $p\geq 0$. If the buyer comes with some inputs, the price should be higher than the value of the remaining inputs.
It becomes more difficult with randomness in production. You need to determine two elements: i) the probability $q$ to fail in producing the item, ii) the value of non-desirable outcomes $V_N$.You can estimate $q$ by dividing the number of times you failed by the number of times you tried.If the non-desirable outcomes are not always the same, $V_N$ can be an average. Again, if you do not know which value to associate, you can take average market prices. In that case, $V_N$ would be on average the money you get from selling the non-desirable outcomes on the market.
There are two possibilities. Either you make the buyer bears the risk, or you cover the risk.In the first case, imagine for instance that you manage to produce the desirable item after 3 trials. The price you will ask to the buyer should be at least $3V_I-2V_N$ because you needed 3 times the inputs but you will keep twice non-desirable outputs.Whether a buyer is lucky or unlucky, it is not your problem. However, you do not have an invariant price of the crafted item.
In the second case, you can ask a fixed price to the buyers (higher for non-members), irrespective of their luck.If you make enough transactions, the lucky buyers will pay for the unlucky ones, and you can make positive profits at the same time. We need some maths to find the expected (or average) cost of producing the item. The average cost is the sum of the values of inputs minus the values of non-desirable outcomes, weighted by the probability of each event.
$$AvCost = \underset{1\ trial}{\underbrace{(1-q)V_I}} + \underset{2\ trials}{\underbrace{q(1-q)(2V_I-V_N)}}+\underset{3\ trials}{\underbrace{q^2(1-q)(3V_I-2V_N)}}+...$$This writes as$$AvCost = (1-q)\sum_{k=0}^\infty q^k[(k+1)V_I-kV_N].$$Using $\sum_{k=0}^\infty q^k(k+1)=\frac{1}{(1-q)^2}$, we obtain$$AvCost=\frac{V_I-qV_N}{1-q}.$$
The condition to make positive profits on average is to fix a price $p\geq \frac{V_I-qV_N}{1-q}$. In other words, if you want to be the most generous with members, they have to pay at least $\frac{V_I-qV_N}{1-q}$ so that you make no losses on average. If a buyer comes with some inputs, you reduce this cost by the value of her inputs.
Members vs non-membersYou can also decide to make losses on average for your members, $p_{member}< \frac{V_I-qV_N}{1-q}$. These losses would be compensated by gains from non-members, $p_{non-member}> \frac{V_I-qV_N}{1-q}$. You have to make sure that the gains compensate the losses.
About being competitive on the marketWe have established the conditions to make positive profits, but there is no guarantee that you will be competitive on the market. It is possible that the minimum price you ask to a buyer ($\frac{V_I-qV_N}{1-q}$) will still be higher than the price of your competitors. They can produce the inputs more easily for instance, or they can value the fact that the skill to produce will level up.
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Geometry is basically the study of different properties of point, line and plane.
Angles: Right angle: An angle of 90° is called a right angle. Acute angle: An angle of less than 90° is called an acute angle. Obtuse angle: An angle of greater than 90° is called an obtuse angle. Reflex angle: An angle of greater than 180° is called reflex angle. Complementary angles: If the sum of the two angles is 90°, then they are called complementary angles. Supplementary angles: If the sum of the two angles is 180°, then they are called supplementary angles. Angle Relations: $\angle $a = $\angle $c and $\angle $b = $\angle $d a, c and b, d are vertically opposite angles. Angles in parallel lines: If a transversal intersects two parallel lines: a. each pair of consecutive interior angles are supplementary. $\angle $4 + $\angle $5 = 180° $\angle $3 + $\angle $6 = 180° b. each pair of alternate interior angles are equal. $\angle $5 =$\angle $3, $\angle $6 = $\angle $4 $\angle $5 = $\angle $3 =$\angle $1 =$\angle $7 $\angle $6 = $\angle $4 = $\angle $2 = $\angle $8 Collinear: Three or more than three points are said to be collinear, if there is a line which contains them all. Concurrent: Three or more than three lines are said to be concurrent, if there is a point which lies on them all. In the above diagram, $\displaystyle\frac{{AB}}{{BC}} = \frac{{DE}}{{EF}}$ or $\displaystyle\frac{{AC}}{{BC}} = \frac{{DF}}{{EF}}$ 1. In the adjoining figure, find $\angle $ C. Sol: $\angle $ E = 180 - 120 = 60 $\angle $ E + $\angle $ X + $\angle $ F = 180 $ \Rightarrow $ 60 + 50 + F = 180 $ \Rightarrow $ F = 70 Now $\angle $ G = 180 - F = 110. As AB // CD, $\angle $G = $\angle $C So $\angle $C = 110 Sol: As C // E, $\angle $B = $\angle $Y
Angles:
Right angle: An angle of 90° is called a right angle.
Acute angle: An angle of less than 90° is called an acute angle.
Obtuse angle: An angle of greater than 90° is called an obtuse angle.
Reflex angle: An angle of greater than 180° is called reflex angle.
Complementary angles: If the sum of the two angles is 90°, then they are called complementary angles.
Supplementary angles: If the sum of the two angles is 180°, then they are called supplementary angles.
Angle Relations:
Vertically opposite angles:If two lines intersect each other, then the vertically opposite angles are equal.
$\angle $a = $\angle $c and $\angle $b = $\angle $d
a, c and b, d are vertically opposite angles.
Angles in parallel lines: If a transversal intersects two parallel lines:
a. each pair of consecutive interior angles are supplementary.
$\angle $4 + $\angle $5 = 180°
$\angle $3 + $\angle $6 = 180°
b. each pair of alternate interior angles are equal.
$\angle $5 =$\angle $3, $\angle $6 = $\angle $4
$\angle $5 = $\angle $3 =$\angle $1 =$\angle $7
$\angle $6 = $\angle $4 = $\angle $2 = $\angle $8
Collinear: Three or more than three points are said to be collinear, if there is a line which contains them
all.
Concurrent: Three or more than three lines are said to be concurrent, if there is a point which lies on them
all.
Proportionality Theorem:
In the above diagram, $\displaystyle\frac{{AB}}{{BC}} = \frac{{DE}}{{EF}}$ or $\displaystyle\frac{{AC}}{{BC}} = \frac{{DF}}{{EF}}$
Solved Examples:
1. In the adjoining figure, find $\angle $ C.
If AB // CD
Now $\angle $ G = 180 - F = 110.
As AB // CD, $\angle $G = $\angle $C
So $\angle $C = 110
2. Find the value of x in following figure.
Sol: As C // E, $\angle $B = $\angle $Y
Also $\angle $Y + $\angle $Z = 180 $ \Rightarrow $ $\angle $Z = 180 - 130 = 50
Now in $\Delta $ AGF, $\angle $A + $\angle $Z + $\angle $X = 180 $ \Rightarrow $ $\angle $X = 180 - 20 - 50 = 110
3. Find $\angle $OBD in the given figure. It is given that EC || AO.
Sol: From the diagram, in $\Delta $OBD, 110 + 20 + $\angle $B = 180 $ \Rightarrow $ $\angle $B = 50
4. If the two lines n and m are parallel, then find the value of $\phi $ in the given figure.
Sol: Draw a parellel to n. Now from the diagram $\phi $ = 50 + 30 = 80 (Alternate interior angles are equal)
Alternate method: you can also solve this problem by calculating interior angles of the triangle.
You may Like Triangles Click Here
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Assuming that you have a infinite long conducting cylindrical wire: Is it possible to let the magnetic field be constant inside the cavity and why?
Using Ampere's law
$$\nabla \times \mathbf H = \mathbf{J} + \frac{\partial \mathbf D}{\partial t}$$
and assuming to have steady-state condition ($\partial \over \partial t$ = 0)
yields to
$$\nabla \times \mathbf H = \mathbf{J} \quad .$$
It is common to make use of the
vector potential $\mathbf{A}$, defined by $\mathbf{B} = \nabla \times \mathbf{A}$ :
$$ \nabla \times \mathbf H = \nabla \times \Bigg( \cfrac{1}{\mu} \nabla \times \mathbf A \Bigg) = \cfrac{1}{\mu} \nabla \big( \nabla \cdot \mathbf A \big) - \cfrac{1}{\mu} \big( \nabla \cdot \nabla \big) \mathbf A =\mathbf J $$
Using the Coloumb gauge $\nabla \cdot \mathbf A = 0$ and $\mathbf B = \mu \mathbf H$ yields
$$\Delta \mathbf A = -\mu \mathbf J \quad .$$
Since you have an cylindrical wire with a cavity inside it make sense to use cylindrical coordinates to solve this equation. If you assume that you have an infinite long wire and symmetry the derivatives $\frac{\partial}{\partial \phi} = \frac{\partial}{\partial z} = 0 $ and the Poison's equation reduces to a very simple differential equation:
$$\Delta_{\mathrm{cyl}} \mathbf{A(r,\phi,z)} = \cfrac{1}{r} \Bigg[ \cfrac{\partial}{\partial r} \Bigg( r \cfrac{\partial \mathbf{A_z}}{\partial r} \Bigg) \Bigg] \mathbf{e_z} = - \mu \mathbf{J}\quad .$$
You have only to solve for the
z component:
$$ \cfrac{1}{r} \Bigg[ \cfrac{\partial}{\partial r} \Bigg( r \cfrac{\partial A_z}{\partial r} \Bigg) \Bigg] = -\mu J_z \qquad \qquad \mathrm (1). $$
Case-to-case Analysis
Now you have to distinguish between the inner non-conducting area, and the outer conducting area!
1. Inner (cavity) case
If we integrate this differential equation to get a solution for the cavity we can set $J_z = 0$. Now by integrating once by $\mathrm{d}r$ you get
$$\cfrac{\partial A_z}{\partial r} = \cfrac{c}{r} \quad ,$$
where $c$ is an integration constant. The solution can be easily found to be
$$A_r = c_1, \quad A_{\phi}= c_2, \quad A_z = c_2 \mathrm{ln}(r) + c_3 \quad , $$
where $c_2$ and $c_3$ are integration constants.
To get the magnetic field you have to take the curl of $\mathbf A$, which is in our case with $\frac{\partial}{\partial \phi} = \frac{\partial}{\partial z} = 0 $:
$$\mathbf B = \nabla_{\mathrm{cyl}} \times \mathbf A = - \cfrac{\partial A_z}{\partial r} \mathbf e_{\phi} \qquad \qquad \mathrm{(2)} .$$
Hence the solution including the integration constant determined by $\mathbf B_0$, and setting $c_4 = -c_2$, is
$$ \mathbf B_i = \mathbf B_{0} + \cfrac{c_4}{r} \mathbf e_{\phi}$$ or since you need it later using $\mathbf H = \frac{1}{\mu} \mathbf B$ $$ \mathbf H_i = \mathbf H_{i0} + \cfrac{c_5}{r} \mathbf e_{\phi} \qquad \qquad \mathrm{(3)} .$$ The index indicates that it is the inner (cavity) area.
2. Conducting Ring
Now you have to find a solution for the conducting ring of the cylinder. So using equation (1) and integrating once yields
$$r \cfrac{\partial A_z}{\partial r} = -\mu \int r \cdot J_z\; \mathrm{d}r + c_6 \quad , $$ or
$$\cfrac{\partial A_z}{\partial r} = -\cfrac{\mu}{r} \int r \cdot J_z\; \mathrm{d}r + \cfrac{c_6}{r} \quad . $$
Using equation (2) again and taking the last equation
$$\mathbf B_o = \nabla_{\mathrm{cyl}} \times \mathbf A = - \cfrac{\partial A_z}{\partial r} \mathbf e_{\phi} \\ \quad \quad = \cfrac{\mu}{r} \int r \cdot J_z\; \mathrm{d}r \; \mathbf e_{\phi} - \cfrac{c_6}{r} \; \mathbf e_{\phi} + \mathbf B_{o0} \quad ,$$
and since you will need $\mathbf H_o$ later: $$\mathbf H_o = \cfrac{1}{r} \int r \cdot J_z\; \mathrm{d}r \; \mathbf e_{\phi} + \cfrac{c_7}{r} \; \mathbf e_{\phi} + \mathbf H_{o0} \qquad \qquad \mathrm{(4)}.$$
The index $o$ denotes the outer (conducting) area.
3. Solving for an inner magnetic field to be constant
Now you can solve your problem. The figure below shows the tangential components of the magnetic field at a conducting boundary surface, as in your problem. (Figure taken from Griffith's
Introduction to Electrodynamics 4e, pp.250)
Therefore the boundary condition at the surface is:
$$\mathbf{H_{o\ \mathrm{tangential}}} - \mathbf{H_{i\ \mathrm{tangential}}} = \mathbf K \times \mathbf{\hat{n}}$$
where $\mathbf K$ denotes the surface current.
In other words, at the radius $R$, where the cavity and the conducting medium touch each other, the condition above has to be meet.
In cylindrical coordinates and for a cylinder as in your case that means:
$$H_{o}(R) \; \mathbf e_{\phi} = H_{i}(R) \; \mathbf e_{\phi} \quad .$$
You can take now equation (4) and (3) and set it equal as described above:
$$ \cfrac{1}{r} \int r \cdot J_z\; \mathrm{d}r \; \mathbf e_{\phi} + \cfrac{c_7}{r} \; \mathbf e_{\phi} + \mathbf H_{o0} \; \mathbf e_{\phi} = \mathbf H_{i0} \; \mathbf e_{\phi}+ \cfrac{c_5}{r} \mathbf e_{\phi}\qquad.$$
Multiplying this equation with $r$ on both sides and differentiating both sides by $\frac{\mathrm{d}}{\mathrm{d}r}$ yields to
$$r \cdot J_z = \mathbf H_{i0} - \mathbf H_{o0} \cdot \mathbf e_{\phi}\\ r \cdot J_z = K(R) \quad, $$
therefore the current density has to be
$$\bbox[5px,border:2px solid red]{ J_z = \cfrac{1}{r}J_{R} }$$
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Well, let's try it, and see how hard it is to forge a message.Let's say for illustrative purposes that each character is a block, and that numbers represent the length indicator section. And let's start by putting the length indicator at the end. So,XXXXXXX7represents a 7-block message, with the '7' indicator at the end. Let's also say that,$_{MAC}...
The problem with CBC-MAC for variable-length messages is that CBC-MAC applied to a one-block message essentially amounts to an oracle for evaluating the block cipher at values of the adversary's choice. And that oracle allows an adversary to break the scheme.Consider first CMAC restricted to messages that consist of a whole number of blocks. Then the ...
So, I'll answer the theoretical part of your question, since we need a key to address the practical part.Why is padding used in CBC?Blockcipher such as AES are encrypting blocks of a fixed given size only, we call it the "blocksize". So, what if your data is smaller than the blocksize ? An easy solution is to add what we call "padding" to your plaintext ...
I'll assume All ciphered blocks means the same as ciphertext for CBC-Encryption with implicit zero IV, while CBC-MAC is the last block of that.All ciphered blocks is unsafe as a message authenticator for messages longer than one block, for it succumbs to a trivial attack (here with two blocks):Eve intercepts message $M=M_0||M_1$ and its authenticator $A=...
You'll find there's a lot of splitting hairs regarding this topic, especially key derivation. But yes, your pseudocode is fine, although you may want to revise (0, 128) => (0, 127) and (129, 256) => (128, 255) ... (correct me if I'm wrong?). Also, you might want to implement a constant-time comparison function for verifying the mac.Have I derived the ...
This scheme is not worth the name MAC; it is horribly weak.First and foremost, the tag/MAC is unchanged when two blocks of plaintext are exchanged (because of the commutativity and associativity of the $\oplus$ operation). If follows that from any message with at least two different blocks, we can make a different message for which we know the tag/MAC. ...
Yes, this is exactly what a message authentication code is for. Its job is to prevent an attacker from tampering with your message, or from forging completely bogus messages. For a secure MAC, it should not matter what these messages contain.(And no, a secure MAC cannot compromise your key; if it did, it would by definition not be secure, since an ...
Can anyone explain why CBC-MAC is not secure for variable length message?For the previous question I'll quote Matthew Green's post from 2013:A quick reminder. CBC-MAC is very similar to the classic CBC mode forencryption, with a few major differences. First, the InitializationVector (IV) is a fixed value, usually zero. Second, CBC-MAC only...
When can I consider a ciphersuite an Authenticated Encryption?To cite from Wikipedia: Authenticated Encryption:Authenticated encryption ... is a form of encryption which simultaneously provides confidentiality, integrity, and authenticity assurances on the data.Note the focus on simultaneously.Is AES-CBC in TLS 1.2 an Authenticated Encryption (...
Also, there is CBC-MAC for providing integrity and confidentiality. Which one is better CBC-MAC or CBC with HMAC?Generally, asking which one is better results in opinionated answers.However, since CBC-MAC cannot be used for dynamically sized messages and may lead to compromise when the same key is used, HMAC definitely is less prone to abuse. Differently ...
Well, yes, it does matter; however the terminology 'CBC-MAC' does not specify which.CBC-MAC is a generic construction that takes an arbitrary block cipher, and turns it into an object that acts like a MAC for fixed length messages (much like CBC mode is a generic construction that takes an arbitrary block cipher, and turns it into a object that encrypts ...
IMO, you code looks pretty solid. A few things I might suggest taking a closer look at are:You haven't specified what iteration count you're using for PBKDF2. You should make the iteration count as high as practical.PKCS #5 suggests a minimum of 1000 iterations, but that recommendation comes from nearly two decades ago. IMO, nowadays there's very ...
It is certainly wrong to state that "MAC can only be produced with AES in CBC and CFB mode", but there seems to be a simple reason that people were inspired by these modes when thinking up possible MAC constructions: They carry along some state that incorporates information from the message while traversing the input blocks. In both modes, encrypting a block ...
The CBC-MAC construction indeed can use a PRF instead of PRP. It is now based on PRP due to historical reasons: the blockciphers used for CBC-MAC were based on permutations.From the security point of view there will be no difference: the security proof for the CBC-MAC first converts PRP to PRF (which is indistinguishable up to $2^{n/2}$ queries) and then ...
Because CBC-MAC with inputs that are not prefix free is weak against existential forgery, meaning it is not a "secure" MAC. More precisely, CBC-MAC is easily distinguishable from a random function (i.e. not a PRF) when the input domain is not prefix-free. This is because an adversary can request the CBC-MAC of messages $M_0$ and $M_1$, and then xor the MAC ...
updated per comments;Currently Netflix Uses AES-GCMI am now studying the AES encryption for real-time video stream. It seems that Netflix uses the AES-GCM (or CBC + MAC) mode for real-time video encryption and authentication.With MAC authentication, client can only get the MAC message after the> whole video is encrypted and authenticated. After ...
Why over-complicate it like that,D1 and D2 generates random 64-bit P (half the block size of AES)they send it to each otherboth generate AES(key,P_own||P_other) and again send to each other (note that these are different for each)then both can verify that what they received is equal to AES(key,P_other||P_own)Upside here is that it is a fully ...
1) The adversary queries the oracle (with some randomly chosen message $m$) and gets as a result a message $m=m_1|m_2|...$ and its tag $t=(t_0,F_{k_2}(t_r))$. She then draws $\rho$ uniformly at random in $\{0,1\}^n$ and outputs the message $m=\rho\oplus m_1|m_2|...$ and its (valid) tag $t=(\rho\oplus t_0,F_{k_2}(t_r))$.2) The adversary queries the oracle (...
The quoted sentences means: if there is a collision among the MACs of the $2^{(n+1)/2}$ messages submitted, the attacker playing the distinguishing game announces that the oracle is a random function; else announces that the oracle is CBC-MAC.This works because the messages submitted differ only in their first block, thus will never collide under CBC-MAC, ...
Yes. Assume that the attacker knows the ciphertext $c = c_1 \mathbin\| c_2$, the initialization vector $v$ and the plaintext $m = m_1 \mathbin\| m_2$. This tells them that$D_k(c_1) = m_1 \oplus v$ and$D_k(c_2) = m_2 \oplus c_1$,where $D_k(\cdot)$ denotes block cipher decryption under the (unknown) key $k$.In particular, this implies that, if the ...
If key 2 and key 3 has a nonnegligible chance to be the same, then the attacker has a nonnegligible chance of being able to generate a valid (Message, MAC) pair.Here's how it works, if the message is not a multiple of 16, then XCBC pads the message out to the next multiple of 16; if it already is, the message remains the same.Then, XCBC logically does a ...
Yes, this is secure.(one of the few cases where I'm pretty confident about this).Here are the arguments:Combining a secure (e.g. SUF-CMA) MAC with a secure (e.g. CPA-secure) encryption method in encrypt-then-authenticate is generally proven secure. This was shown in "Authenticated Encryption: Relations among notions and analysis of the generic ...
An attacker can trivial forge the MAC of any message, given one valid MAC of a known message, in either CBC mode or CTR mode.Let us assume that the attacker knows a message $m$ and its MAC $E(k, H(m)) = IV, E(k, IV, H(m))$; he has a message $m'$ he wants to form the message to. He computes $\delta = H(m) \oplus H(m')$, then:For CTR mode, he computes $IV,...
It sounds like you have one big misconception in your question:Also, I hear about CBC-MAC which can provide integrity and confidentiality. Which one is better CBC-MAC or CBC with HMAC?The first sentence reads like you misunderstand CBC-MAC to be an algorithm that provides two properties, integrity and confidentiality. But in reality, CBC-MAC can only ...
From the sound of your questions, it almost appears that you have some confusion between the CBC-MAC key and the CBC-MAC tag.The CBC-MAC algorithm takes the message (in this case, most likely the ciphertext) and a secret key; it outputs a tag (which can be public). The security property of CBC-MAC is that someone who does not know the key cannot generate ...
An answer surfaced from careful reading of appropriate documentation.The MAC in the question is also defined in ANSI X9.19, and is supported by some PKCS#11 tokens as the mechanism CKM_DES3_X919_MAC_GENERAL.Other than that, this MAC can be simulated using CKM_DES_MAC_GENERAL (or CKM_DES_CBC or CKM_DES3_CBC) for all but the last block, then CKM_DES3_CBC; ...
What you think of is called an extension attack and it turns out that this is the way to go if you would like to break the general CBC-MAC when the message length is not fixed.All that an adversary needs to do is to mount a chosen message attack.Suppose he asks for the tag on the message $m=m_1||m_2||...||m_l$. The resulting CBC MAC would be $MAC_k(m)=t$....
Which MAC algorithm is faster - CBC based MAC's or HMAC - depends completely on which ciphers and hashes are used. Furthermore, it depends on the runtime environment that contains the hash and cipher implementations.With regard to the leading CPU architecture for PC's, there are the Intel whitepapers. Both AES and SHA-2 performance can be enhanced using ...
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Article
Keywords: Cesàro mean; Abel mean; growth order; uniformly continuous operator semi-group and cosine function
Summary: It will be proved that if $N$ is a bounded nilpotent operator on a Banach space $X$ of order $k+1$, where $k\geq 1$ is an integer, then the $\gamma$-th order Cesàro mean $C_{t}^{\gamma}:=\gamma t^{-\gamma}\int_{0}^{t}(t-s)^{\gamma-1}T(s)\,ds$ and Abel mean $A_{\lambda}:=\lambda\int_{0}^{\infty}e^{-\lambda s}T(s)\,ds$ of the uniformly continuous semigroup $(T(t))_{t\geq 0}$ of bounded linear operators on $X$ generated by $iaI+N$, where $0\neq a\in \mathbb{R}$, satisfy that (a) $\|C_{t}^{\gamma}\|\sim t^{k-\gamma}\;(t\to\infty)$ for all $0< \gamma\leq k+1$; (b) $\|C_{t}^{\gamma}\|\sim t^{-1}\;(t\to\infty)$ for all $\gamma\geq k+1$; (c) $\|A_{\lambda}\|\sim \lambda\;(\lambda\downarrow 0)$. A similar result will be also proved for the uniformly continuous cosine function $(C(t))_{t\geq 0}$ of bounded linear operators on $X$ generated by $(iaI+N)^{2}$.
References:
[1] Chen J.-C., Sato R., Shaw S.-Y.:
Growth orders of Cesàro and Abel means of functions in Banach spaces
. Taiwanese J. Math.(to appear). MR 2674604
[3] Sato R.:
On ergodic averages and the range of a closed operator
. Taiwanese J. Math. 10 (2006), 1193–1223. MR 2253374
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I have an equation like this
$$-K(y)-\frac{\pi J}{2 \sqrt{\lambda }}+\frac{E(y)}{1-y}=0$$
1/(1 - y) EllipticE[y] - EllipticK[y] - (J π /(2 Sqrt[λ])) == 0
Where $E(y)$ and $K(y)$ are the Elliptic integrals in the Mathematica notation and $y$ is Real. I want to solve the equation numerically for $y$ for different values of the variables $(\lambda, J)$. That is $(\lambda, J)$ acts as a $(x,y)$ grid where $y$ gives the values along $z$ axis. Now I want to take these values of $y$ for different values of $(\lambda, J)$ and evaluate the function
$$\Delta= \frac{2 \sqrt{\lambda} }{\pi} \left(K(y)+\frac{\left(\sqrt{y}-1\right) E(y)}{1-y}\right)$$
(2 Sqrt[λ])/π ((Sqrt[y] - 1)/(1 - y) EllipticE[y] + EllipticK[y])
and plot it in 3d along the $\Delta, \lambda, J$ axes. So the values on the $(\lambda, J)$ plane remains the same, but instead of $y$, we want to plot $\Delta(y)$ at each $(\lambda, J)$ value.
Can someone help me write the code? Thanks!
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Power Series Expansion for Exponential Function Theorem
Let $\exp x$ be the exponential function.
Then:
\(\displaystyle \forall x \in \R: \ \ \) \(\displaystyle \exp x\) \(=\) \(\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\) \(\displaystyle \) \(=\) \(\displaystyle 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots\) Proof
From Higher Derivatives of Exponential Function, we have:
$\forall n \in \N: \map {f^{\paren n} } {\exp x} = \exp x$ Since $\exp 0 = 1$, the Taylor series expansion for $\exp x$ about $0$ is given by: $\displaystyle \exp x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ From Radius of Convergence of Power Series over Factorial, we know that this power series expansion converges for all $x \in \R$.
From Taylor's Theorem, we know that
$\displaystyle \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$
where $0 \le \eta \le x$.
Hence:
\(\displaystyle \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }\) \(=\) \(\displaystyle \size {\frac {x^n} {n!} \map \exp \eta}\) \(\displaystyle \) \(\le\) \(\displaystyle \frac {\size {x^n} } {n!} \map \exp {\size x}\) Exponential is Strictly Increasing \(\displaystyle \) \(\to\) \(\displaystyle 0\) \(\displaystyle \text { as } n \to \infty\) Series of Power over Factorial Converges So the partial sums of the power series converge to $\exp x$.
The result follows.
$\blacksquare$
Sources 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables... (previous) ... (next): $\S 20$: Series for Exponential and Logarithmic Functions: $20.15$ 1971: George E. Andrews: Number Theory... (previous) ... (next): $\text {3-4}$ Generating Functions: Example $\text {3-7}$ 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach... (previous) ... (next): $\S 15.5$ 1992: Larry C. Andrews: Special Functions of Mathematics for Engineers... (previous) ... (next): $\S 1.3.2$: Power series: $(1.45)$ 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms(3rd ed.) ... (previous) ... (next): $\S 1.2.9$: Generating Functions: $(22)$
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Here is one solution:
Clearly Double-SAT belongs to ${\sf NP}$, since a NTM can decide Double-SAT as follows: On a Boolean input formula $\phi(x_1,\ldots,x_n)$, nondeterministically guess 2 assignments and verify whether both satisfy $\phi$.
To show that Double-SAT is ${\sf NP}$-Complete, we give a reduction from SAT to Double-SAT, as follows:
On input $\phi(x_1,\ldots,x_n)$:
Introduce a new variable $y$. Output formula $\phi'(x_1,\ldots,x_n, y) = \phi(x_1,\ldots,x_n) \wedge (y \vee \bar y)$.
If $\phi (x_1,\ldots,x_n)$ belongs to SAT, then $\phi$ has at least 1 satisfying assignment, and therefore $\phi'(x_1,\ldots,x_n, y)$ has at least 2 satisfying assignments as we can satisfy the new clause ($y \vee \bar y$) by assigning either $y = 1$ or $y = 0$ to the new variable $y$, so $\phi'$($x_1$, ... ,$x_n$, $y$) $\in$ Double-SAT.
On the other hand, if $\phi(x_1,\ldots,x_n)\notin \text{SAT}$, then clearly $\phi' (x_1,\ldots,x_n, y) = \phi (x_1,\ldots,x_n) \wedge (y \vee \bar y)$ has no satisfying assignment either, so $\phi'(x_1,\ldots,x_n,y) \notin \text{Double-SAT}$.
Therefore, $\text{SAT} \leq_p \text{Double-SAT}$, and hence Double-SAT is ${\sf NP}$-Complete.
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Production of light nuclei and anti-nuclei in $pp$ and Pb-Pb collisions at energies available at the CERN Large Hadron Collider
(American Physical Society, 2016-02)
The production of (anti-)deuteron and (anti-)$^{3}$He nuclei in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been studied using the ALICE detector at the LHC. The spectra exhibit a significant hardening with ...
Forward-central two-particle correlations in p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(Elsevier, 2016-02)
Two-particle angular correlations between trigger particles in the forward pseudorapidity range ($2.5 < |\eta| < 4.0$) and associated particles in the central range ($|\eta| < 1.0$) are measured with the ALICE detector in ...
Measurement of D-meson production versus multiplicity in p-Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV
(Springer, 2016-08)
The measurement of prompt D-meson production as a function of multiplicity in p–Pb collisions at $\sqrt{s_{\rm NN}}=5.02$ TeV with the ALICE detector at the LHC is reported. D$^0$, D$^+$ and D$^{*+}$ mesons are reconstructed ...
Measurement of electrons from heavy-flavour hadron decays in p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2016-03)
The production of electrons from heavy-flavour hadron decays was measured as a function of transverse momentum ($p_{\rm T}$) in minimum-bias p–Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with ALICE at the LHC for $0.5 ...
Direct photon production in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2016-03)
Direct photon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 2.76$ TeV was studied in the transverse momentum range $0.9 < p_{\rm T} < 14$ GeV/$c$. Photons were detected via conversions in the ALICE ...
Multi-strange baryon production in p-Pb collisions at $\sqrt{s_\mathbf{NN}}=5.02$ TeV
(Elsevier, 2016-07)
The multi-strange baryon yields in Pb--Pb collisions have been shown to exhibit an enhancement relative to pp reactions. In this work, $\Xi$ and $\Omega$ production rates have been measured with the ALICE experiment as a ...
$^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ production in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2016-03)
The production of the hypertriton nuclei $^{3}_{\Lambda}\mathrm H$ and $^{3}_{\bar{\Lambda}} \overline{\mathrm H}$ has been measured for the first time in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE ...
Multiplicity dependence of charged pion, kaon, and (anti)proton production at large transverse momentum in p-Pb collisions at $\sqrt{s_{\rm NN}}$= 5.02 TeV
(Elsevier, 2016-09)
The production of charged pions, kaons and (anti)protons has been measured at mid-rapidity ($-0.5 < y < 0$) in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV using the ALICE detector at the LHC. Exploiting particle ...
Jet-like correlations with neutral pion triggers in pp and central Pb–Pb collisions at 2.76 TeV
(Elsevier, 2016-12)
We present measurements of two-particle correlations with neutral pion trigger particles of transverse momenta $8 < p_{\mathrm{T}}^{\rm trig} < 16 \mathrm{GeV}/c$ and associated charged particles of $0.5 < p_{\mathrm{T}}^{\rm ...
Centrality dependence of charged jet production in p-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 5.02 TeV
(Springer, 2016-05)
Measurements of charged jet production as a function of centrality are presented for p-Pb collisions recorded at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector. Centrality classes are determined via the energy ...
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Babbys first HoTT Note Curry-Howard correspondence and geometric semantics
syntax computation logic geometry $P$ type proposition space $x$ with $x:P$ term $x$ of type $P$ argument $x$ for proposition $P$ point $x$ of $P$ $\bf{0}$ empty type false proposition empty set ${\bf 1}$ type which does have one unique term true proposition singleton ${\bf 2}$ Bool true proposition$^\star$ set with two points
$^\star$Some type theories make any type into a proposition and any proposition into a type and a proposition is true iff we can construct some term of the corresponding type. HoTT is a type theory where only some types are propositions, namely those who have
up to one unique term, see below.
For $x:P$, consider a well formed expression $Q(x)$. This can be viewed as an $x$-parametrized family of types, a predicate with an open variable from, or a collection of fibres over a base.
syntax computation logic geometry ${\large{\prod}}_{x:P} Q(x)$ type of functions $s$ mapping an $x$ into $Q(x)$ $\forall (x\in P).\,Q(x)$ sections ${\large{\sum}}_{x:P} Q(x)$ type of pairs $\langle x,y\rangle$ with $x:P$ and $y:Q(x)$ $\exists (x\in P).\,Q(x)$ total space
Constructing a term $s:{\large{\prod}}_{x:P} Q(x)$ is finding an argument $s$ for $\forall (x\in P).\,Q(x)$. The latter means being able to translate any argument $x$ for $P$ into an argument $s(x)$ of $Q(x)$. This is a functional assignment corresponding to a section of a fibration. The sum type works analogously.
If $Q$ is free of $x$, we can define implication $\implies$ and conjunction $\land$ as special cases of the above:
$P\to Q \equiv {\large{\prod}}_{x:P} Q$
$P\times Q \equiv {\large{\sum}}_{x:P} Q$
If we tag the terms of a type $P$ resp. $Q$ with the value of Bool, we can identify disjunction $\lor$ as $P+Q\equiv{\large{\sum}}_{x:{\bf 2}}\dots$ where the dots represent a suitable $if$-construction. Negation $\neg P$ is read as $P$ leading to absurum, i.e. $P\to{\bf 0}$. And lastly
syntax computation logic geometry $P'$ with $P'<:P$ subtype implied propositions subspace
We have, for all $P$, a poset of types/propositions/spaces, with ${\bf 1}$ the maximum (by definition, we know the term of ${\bf 1}$ and hence, given a term of ${\bf 1}\to P$, we obtain one of $P$).
Identity type
We saw how a predicate logic can be expressed as type theory. When this is extended to equalitarian logics, we deal with
Intentional type theory alla Per Martin-Löf. Here, for any type $P$, we get an identity type $=_P:P\to P\to\mathrm{Type}$. It must be defined in a way which respects symmetry and substitution, which implies transitivity, reflexivity etc. Example 1
For example, we can specify the theory of semigroup as
$Semigroup\equiv{\large{\sum}}_{A:\mathrm{Type}}{\large{\sum}}_{m:A\to A\to A}{\large{\prod}}_{x:A}{\large{\prod}}_{y:A}{\large{\prod}}_{z:A}\ m(x,m(y,z))=_A m(m(x,y),z)$
A term of $SemiGroupStr$ is a pair $\langle A,\langle m,p\rangle\rangle$, where the first entry is a type $A$ and the second entry is pair, where the first entry is a function $m:A\to A\to A$ and the second entry is a proof $p$ that this $m$ fulfills the usual axiom. We do constructive mathematics here, so $p$ is a function which takes any $x$ to a function which takes any $y$ to a … to a function which takes any $m(m(x,y),z)$ to a term which demonstrates the required equality.
Example 2
As another example (which incidentally is also a semigroup, but that's not the point here) we can inductively define the natural numbers $\mathbb N$ as type by postulating
$0:\mathbb N$
and
$S:\mathbb N\to \mathbb N$.
Now
$0,S0,SS0,SSS0,\dots$
are all terms of $\mathbb N$. The picture on the right shows arithmetic alla Peano in first order logic
Using the 3rd and 4th axioms we can define an addition function $+:\mathbb N\to\mathbb N\to\mathbb N$, which we denote as $plus$ here, as
$plus(n,0):=n$
$plus(n,Sm):=S(plus(n,m))$
Using the 1st and 2nd we can define an identity type $=_\mathbb N:\mathbb N\to\mathbb N\to\mathrm{Type}$, which denote as $Code$ here, as
$Code(Sn,0)\ :=\ {\bf 0}$
$Code(0,Sm)\ :=\ {\bf 0}$
$Code(Sn,Sm)\ :=\ Code(n,m)$
$Code(0,0)\ :=\ {\bf 1}$
For any $n,m$, the above identity type $n=_\mathbb{N}m$ is defined in a way so that its either ${\bf 0}$ or ${\bf 1}$. So it either has a term or not, and under Curry-Howard the former means it's false and the latter means it's true. The dependent type $=_\mathbb{N}$ is equivalent to a binary predicate or a function of type $\mathbb N\to\mathbb N\to\mathbb {\bf 2}$ but to consider it a map into $\mathrm{Type}$ becomes important in the following.
There are a million ways to define and work with equality. For example, depending on the theory you treats functions intentional or extenional, the function mapping $n$ to $2(n+5)$ and the function mapping $n$ to $2n+10$ might be judged to be different or the same. For example, the former theory might be suited to study the difference between sorting algorithms (which do the same job, but possibly need different number of computing steps), while the latter may not. Intensional theories also have better decidablity properties regarding type checking but, as one can imagine, they are also much more complicated to work with.
We now turn to a variation of intentional type theory with an additional axiom which implies function extensionality.
Homotopy type theory
The following statements relating to the above construction are fairly clear: The type $Code(n,m)$ reduces to the same type as $Code(m,n)$. If $Code(n,k)$ and $Code(k,m)$ reduces to ${\bf 1}$, then so does $Code(n,m)$. That $Code(0,0)$ has a known term follows from the reflexivity property.
Abstractly, the rules for identity constructively say that if $p:x=_Py$, then there is a term $i(p):y=_Px$, if further $q:y=_Pz$, then there is a term $t(p,q):x=_Pz$, and for any $x$, there is a trivial term $refl_x:x=_Px$. Note that nothing holds us from defining identity types with more than one term.
Motivated by this, homotopy type theory generally defines identity types $x=_Py$ as path spaces where the above functions are given by inversion $p^{-1}$, concatenation $p\circ q$ and constant path. Path spaces also behave like hom-classes $\mathrm{Hom}_P[x,y]$ of groupoids (categoris where everything is invertible) and in fact, since terms $p,q:x=_Py$ of an identity type can also be compared for identity via $p=_{x=_Py}q$, we deal with homotopies or higher categories.
Example
We can inductively define the circle $\mathbb S$ as type by postulating $base:\mathbb S$ and $loop:base=_{\mathbb S}base$. Now $\dots,loop^{-2},loop^{-1},refl_{base},loop,loop^2,loop^3,\dots$ are all terms of $base=_{\mathbb S}base$. This type can be shown to be equivalent (in the formal sense discussed below) to the integers. Indeed, recall that the first homotopy group of the circle is isomorphic to $\mathbb Z$.
The above construction is synthetic in the sense Euclid writes down geometry. This is in contrast to a Cartesian approach, where we define the circle analytically as a set of points.
Propositions and sets
In this framework, not all types are sets, proposition or either of those. The following identifications are made:
We say a type is a proposition iff its boolean, i.e. if it either has no term, or one unique term. This can be characterized as saying that all terms must be the same
$isProp(A):={\large{\prod}}_{x:A}{\large{\prod}}_{y:A}\,x=y$
Recall that for any $n,m:\mathbb N$, we could code up $n=_\mathbb{N}m$ in a way so that it's either ${\bf 0}$ or ${\bf 1}$.
We say a type is a set iff $x=y$ is a proposition.
$isSet(A):={\large{\prod}}_{x:A}{\large{\prod}}_{y:A}\,isProp(x=y)$
The above says that $x$ connects to $y$ via at most one invertible path and hence this characterizes as set as a discrete category. Since sets, characterized like here as types with lack of other structure, are generally different from sets characterized in a theory like ZFC.
Homotopy theory
The general point here is that when path spaces already come together with the simple logical notion of equality, we can do a lot of geometry in the native language of the theory. We can define geometric quite directly. E.g. for any $\pi:A\to B$ and $b:B$, the type ${\large{\sum}}_{x:A}\, \pi(x)=b$ is a collection of $x's$ together with demonstrations that $\pi(x)=b$, i.e. the $x$'s are values in a fibre over $A$. You can express that a space is contractible (shrinkable to one point) by using $isProp$ and you can in turn use this to express homotopy for functions and then homotopy equivalence $\simeq$ of spaces. The
univalence axiom of homotopy type theory finally says that you should generally view a type as homotopy type of a space (that naming overlap is a happy coincidence) and accordingly guarantees a term of the type
$(A\simeq B)\to (A=B)$
converting any homotopy to an argument for equality.
Some more curious identifications: If $f$ is a function, then $x=_Ay$ implies $f(x)=f(y)$. Constructively, this means we can lift a path $p:x=_Ay$ to a path $p_*:f(x)=f(y)$. This makes any function into a functor with $\mathrm{fmap}\ p:=p_*$. Moreover, the condition for $H$ to be a homotopy between two functions is exactly the one of being a natural transformation.
Internal logic of a cateogry and models
Now if we read $P:\mathrm{Ob}_{\bf C}$ as types, then an arrow $\pi:{\bf C}[Q,P]$, viewed as a projection map for a fibre bundle, defines fibres and hence a dependent type ${\large{\prod}}_{x:P} Q(x)$. If ${\bf C}$ is a topos, then we have subobjects, i.e. subspaces.
In this way a) each category itself provides a syntactic framework b) categories provide models for type theories. For example, simply typed lambda calculus corresponds to Cartesian closed categories (it has function spaces as objects etc.) and homotopy type theory corresponds to some wild higher topos.
Here's how to semantically caputure identity types/path spaces:
Generally, if $\pi:E\to B$ is a map and $b:B$, then $F(b):=\pi^{-1}(b)$ is a fibre.
As a special case, let $I$ by some sort of interval so that $B^I\equiv I\to B$ is the space of paths. Then $\pi:B^I\to (B\times B)$ gives us, for each $\langle b_{start},b_{end}\rangle:B\times B$ a space of paths $Path(b_{start},b_{end}):=\pi^{-1}(\langle b_{start},b_{end}\rangle)$.
(Another note: Roughly, in topos theory the dependent product “$S=\prod_{x:B}p^{-1}(x)$” w.r.t. an arrow $p:E\to B$ can be captured (I put the expression in quotes because we don't really have arguments $x$ when we don't deal with sets/types), as the object so that $p\circ eval=\pi$, were $eval:S\times B\to B$. See the hard to read article dependent product.)
Reference
Wikipedia: Homotopy type theory
homotopytypetheory.org: HoTT Book
todo
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Does anyone here understand why he set the Velocity of Center Mass = 0 here? He keeps setting the Velocity of center mass , and acceleration of center mass(on other questions) to zero which i dont comprehend why?
@amanuel2 Yes, this is a conservation of momentum question. The initial momentum is zero, and since there are no external forces, after she throws the 1st wrench the sum of her momentum plus the momentum of the thrown wrench is zero, and the centre of mass is still at the origin.
I was just reading a sci-fi novel where physics "breaks down". While of course fiction is fiction and I don't expect this to happen in real life, when I tired to contemplate the concept I find that I cannot even imagine what it would mean for physics to break down. Is my imagination too limited o...
The phase-space formulation of quantum mechanics places the position and momentum variables on equal footing, in phase space. In contrast, the Schrödinger picture uses the position or momentum representations (see also position and momentum space). The two key features of the phase-space formulation are that the quantum state is described by a quasiprobability distribution (instead of a wave function, state vector, or density matrix) and operator multiplication is replaced by a star product.The theory was fully developed by Hilbrand Groenewold in 1946 in his PhD thesis, and independently by Joe...
not exactly identical however
Also typo: Wavefunction does not really have an energy, it is the quantum state that has a spectrum of energy eigenvalues
Since Hamilton's equation of motion in classical physics is $$\frac{d}{dt} \begin{pmatrix} x \\ p \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \nabla H(x,p) \, ,$$ why does everyone make a big deal about Schrodinger's equation, which is $$\frac{d}{dt} \begin{pmatrix} \text{Re}\Psi \\ \text{Im}\Psi \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \hat H \begin{pmatrix} \text{Re}\Psi \\ \text{Im}\Psi \end{pmatrix} \, ?$$
Oh by the way, the Hamiltonian is a stupid quantity. We should always work with $H / \hbar$, which has dimensions of frequency.
@DanielSank I think you should post that question. I don't recall many looked at the two Hamilton equations together in this matrix form before, which really highlight the similarities between them (even though technically speaking the schroedinger equation is based on quantising Hamiltonian mechanics)
and yes you are correct about the $\nabla^2$ thing. I got too used to the position basis
@DanielSank The big deal is not the equation itself, but the meaning of the variables. The form of the equation itself just says "the Hamiltonian is the generator of time translation", but surely you'll agree that classical position and momentum evolving in time are a rather different notion than the wavefunction of QM evolving in time.
If you want to make the similarity really obvious, just write the evolution equations for the observables. The classical equation is literally Heisenberg's evolution equation with the Poisson bracket instead of the commutator, no pesky additional $\nabla$ or what not
The big deal many introductory quantum texts make about the Schrödinger equation is due to the fact that their target audience are usually people who are not expected to be trained in classical Hamiltonian mechanics.
No time remotely soon, as far as things seem. Just the amount of material required for an undertaking like that would be exceptional. It doesn't even seem like we're remotely near the advancement required to take advantage of such a project, let alone organize one.
I'd be honestly skeptical of humans ever reaching that point. It's cool to think about, but so much would have to change that trying to estimate it would be pointless currently
(lol) talk about raping the planet(s)... re dyson sphere, solar energy is a simplified version right? which is advancing. what about orbiting solar energy harvesting? maybe not as far away. kurzgesagt also has a video on a space elevator, its very hard but expect that to be built decades earlier, and if it doesnt show up, maybe no hope for a dyson sphere... o_O
BTW @DanielSank Do you know where I can go to wash off my karma? I just wrote a rather negative (though well-deserved, and as thorough and impartial as I could make it) referee report. And I'd rather it not come back to bite me on my next go-round as an author o.o
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The MathJax Processing Model¶
The purpose of MathJax is to bring the ability to include mathematics easily in web pages to as wide a range of browsers as possible. Authors can specify mathematics in a variety of formats (e.g., MathML or LaTeX), and MathJax provides high-quality mathematical typesetting even in those browsers that do not have native MathML support. This all happens without the need for special downloads or plugins, but rendering will be enhanced if high-quality math fonts (e.g., STIX) are available to the browser.
MathJax is broken into several different kinds of components: page preprocessors, input processors, output processors, and the MathJax Hub that organizes and connects the others. The input and output processors are called jax, and are described in more detail below.
When MathJax runs, it looks through the page for special tags that hold mathematics; for each such tag, it locates an appropriate input jax which it uses to convert the mathematics into an internal form (called an element jax), and then calls an output jax to transform the internal format into HTML content that displays the mathematics within the page. The page author configures MathJax by indicating which input and output jax are to be used.
Often, and especially with pages that are authored by hand, themathematics is not stored (initially) within the special tags neededby MathJax, as that would require more notation than the average pageauthor is willing to type. Instead, it is entered in a form that ismore natural to the page author, for example, using the standard TeXmath delimiters
$...$ and
$$...$$ to indicate what part of thedocument is to be typeset as mathematics. In this case, MathJax canrun a preprocessor to locate the math delimiters and replace them bythe special tags that it uses to mark the formulas. There arepreprocessors for TeX notation, MathMLnotation, and the jsMath notation that uses span and div tags.
For pages that are constructed programatically, such as HTMLpages that result from running a processor on text in some otherformat (e.g., pages produced from Markdown documents, or via programslike tex4ht), it would be best to use MathJax’s special tagsdirectly, as described below, rather than having MathJax runanother preprocessor. This will speed up the final display of themathematics, since the extra preprocessing step would not be needed,and it also avoids the conflict between the use of the less-than sign,
<, in mathematics and asn an HTML special character (that startsan HTML tag).
How mathematics is stored in the page¶
In order to identify mathematics in the page, MathJax uses special
<script> tags to enclose the mathematics. This is done becausesuch tags can be located easily, and because their content is notfurther processed by the browser; for example, less-than signs can beused as they are in mathematics, without worrying about them beingmistaken for the beginnings of HTML tags. One may also consider themath notation as a form of “script” for the mathematics, so a
<script> tag makes at least some sense for storing the math.
Each
<script> tag has a
type attribute that identifies thekind of script that the tag contains. The usual (and default) valueis
type="text/javascript", and when a script has this type, thebrowser executes the script as a javascript program. MathJax,however, uses the type math/tex to identify mathematics in the TeXand LaTeX notation, and math/mml for mathematics in MathMLnotation. When the tex2jax or mml2jax preprocessors run, theycreate
<script> tags with these types so that MathJax can processthem when it runs its main typesetting pass.
For example,
<script type="math/tex">x+\sqrt{1-x^2}</script>
represents an in-line equation in TeX notation, and
<script type="math/tex; mode=display"> \sum_{n=1}^\infty {1\over n^2} = {\pi^2\over 6}</script>
is a displayed TeX equation.
Alternatively, using MathML notation, you could use
<script type="math/mml"> <math> <mi>x</mi> <mo>+</mo> <msqrt> <mn>1</mn> <mo>−<!-- − --></mo> <msup> <mi>x</mi> <mn>2</mn> </msup> </msqrt> </math></script>
for in-line math, or
<script type="math/mml"> <math display="block"> <mrow> <munderover> <mo>∑<!-- ∑ --></mo> <mrow> <mi>n</mi> <mo>=</mo> <mn>1</mn> </mrow> <mi mathvariant="normal">∞<!-- ∞ --></mi> </munderover> </mrow> <mrow> <mfrac> <mn>1</mn> <msup> <mi>n</mi> <mn>2</mn> </msup> </mfrac> </mrow> <mo>=</mo> <mrow> <mfrac> <msup> <mi>π<!-- π --></mi> <mn>2</mn> </msup> <mn>6</mn> </mfrac> </mrow> </math></script>
for displayed equations in MathML notation. As other input jax are created, they will use other types to identify the mathematics they can process.
Page authors can use one of MathJax’s preprocessors to convert frommath delimiters that are more natural for the author to type (e.g.,TeX math delimiters like
$$...$$) to MathJax’s
<script>format. Blog and wiki software could extend from their own markuplanguages to include math delimiters, which they could convert toMathJax’s
<script> format automatically.
Note, however, that Internet Explorer has a bug that causes it toremove the space before a
<script> tag if there is also a spaceafter it, which can cause serious spacing problems with in-line mathin Internet Explorer. There are three possible solutions to this inMathJax. The recommended way is to use a math preview (an elementwith class
MathJax_Preview) that is non-empty and comes rightbefore the
<script> tag. Its contents can be just the word
[math], so it does not have to be specific to the mathematicsscript that follows; it just has to be non-empty (though it could haveits style set to
display:none). See also the
preJax and
postJax options in the Core Configuration Options document for another approach.
The components of MathJax¶
The main components of MathJax are its preprocessors, its input and output jax, and the MathJax Hub, which coordinates the actions of the other components.
Input jax are associated with the different script types (like math/tex or math/mml) and the mapping of aparticular type to a particular jax is made when the various jaxregister their abilities with the MathJax Hub at configuration time.For example, the MathML input jax registers the math/mmltype, so MathJax will know to call the MathML input jax when it seesmath elements of that type. The role of the input jax is to convertthe math notation entered by the author into the internal format usedby MathJax (called an element jax). This internal format isessentially MathML (represented as JavaScript objects), so an inputjax acts as a translator into MathML. Output jax convert that internal element jax format into a specificoutput format. For example, the NativeMML output jax inserts MathMLtags into the page to represent the mathematics, while the HTML-CSSoutput jax uses HTML with CSS styling to lay out the mathematics sothat it can be displayed even in browsers that dont understandMathML. Output jax could be produced that render the mathematicsusing SVG, for example, or that speak an equation for the blindusers. The MathJax contextual menu can be used to switch between theoutput jax that are available.
Each input and output jax has a small configuration file that is loaded when that input jax is included in the jax array in the MathJax configuration, and a larger file that implements the core functionality of that particular jax. The latter file is loaded when the first time the jax is needed by MathJax to process some mathematics.
The
MathJax Hub keeps track of the internal representations of thevarious mathematical equations on the page, and can be queried toobtain information about those equations. For example, one can obtaina list of all the math elements on the page, or look up a particularone, or find all the elements with a given input format, and so on.In a dynamically generated web page, an equation where the sourcemathematics has changed can be asked to re-render itself, or if a newparagraph is generated that might include mathematics, MathJax can beasked to process the equations it contains.
The Hub also manages issues concerning mouse events and other user interaction with the equation itself. Parts of equations can be made active so that mouse clicks cause event handlers to run, or activate hyperlinks to other pages, and so on, making the mathematics as dynamic as the rest of the page.
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Given a forward rate, for example:$ F(t, T, T+\delta)$The instantaneous forward rate $f(t,T)$ fixed in $t$ is the limit when $\delta \rightarrow 0$ of your forward rate.If the relation between forward rate and zero coupon bond is:$F(t,T,T+\delta) = \frac{p(t,T) - p(t,T+\delta)}{\delta p(t,T+\delta)}$We have,\begin{equation}f(t,T) = \lim_{\...
1. Observable instruments, spot rates, and forward ratesFirst remember that something observable means that you can observe/find the rate in the market by looking at traded rate instruments or fixings.1.1. Observed spot ratesFor simplicity, assume Zero Coupon Bonds (ZCBs) are traded with time left to maturity of 10Y, 15Y and 20Y. Hence, by observing ...
A forward rate agreement is an agreement to exchange a fixed for a floating rate over one period, with the payment being made at the start of the period.A zero coupon swap (with both legs paid at maturity) is an agreement to exchange a fixed for floating rate over one or more periods, with the payments being made at the end of the final period.So the two ...
You do not need zero rates to estimate a parametricmodel of the yield curve, such as Nelson-Siegel.Suppose for instance that you have a cross-section ofbond prices. Then:For given parameters for your yield-curve model, compute yield curve;with this yield curve, calculate theoretical bond prices;compute discrepancy between theoretical bondprices and ...
In general futures contracts are leverage instruments. They never require the investment of principal. They do however require margin: you need to fund your account at a futures exchange so that they have insurance against any losses you incur, as an example this might be 2 days standard volatility. On 1 ED contract for 5bps a day thats probably 10bps margin ...
By definition of the $T$-forward measure $P_T$, the process $\Big\{\frac{P(t,S)}{P(t,T)} \mid t\geq 0\Big\}$ is a martingale under the measure $P_T$, without assuming any specific models of the short rate $r_t$.That is, this martingale property is model independent.However, as a good exercise, you can also do the following:Given the CIR interest rate ...
In part (a) use discount rate $e^.07 -1 = .072508181$ to get the right answer.For part (b) I am just giving you hint:Calculate bond price at the end of 1st year and 2nd year in the same way as you did in part (a).Use the above calculated price to buy bond from the dividend at the end of first and second year. You may assume bond can be purchased in ...
Who knows what the 5 year zero coupon rate is in that case, there could be an event 4.5 years out that will have serious interest rate implications that we don't know about. The only thing you can do with these three numbers is extrapolate and say the rate should 9%. You should be aware of what assumptions you're making when you do something like that, but I'...
Here is a general proof for all parameters in an open domain.$$dr = adt+bdW:=r\big(k(\theta-x)+\frac12\sigma^2\big)dt+\sigma rdW.$$Let$$u(r(s),s):=e^{-\int_t^sr}B(r(s),s,T)=:\phi(s) B.$$Then$$u(r(t),t)=\mathbf E\big[u(r(s),s)\big|r(t)\big],\, \forall t<s. \tag{1}$$So, by Ito's Lemma,\begin{align}du(r(s),s) &= Bd\phi +\phi dB \\&= \phi \...
The CMT yields published by the Fed/US Treasury are par yields calculated using a cubic spline model. In other words, these are the yields to maturity as well as coupon rates on bonds whose theoretic prices are 100. With this information in mind, you can linearly interpolate between these yields, or use a cubic spline to fill in rates at other tenors, ...
Under the Hull-White interest rate model, the short rate $r_t$ satisfies a risk-neutral SDE of the form\begin{align*}dr_t = (\theta(t)-a r_t)dt+ \sigma dW_t.\end{align*}The price at time $t$ of a zero-coupon bond with maturity $T$ and unit face value is then given by\begin{align*}P(t, T) &= A(t, T) e^{-B(t, T) r_t},\end{align*}where\begin{...
I think one way to approach the answer is thinking what are these two rates used for. Starting with zero coupon rates, it's aiming for getting the par value back at maturity (similar to a bank's loan, where in the end payments are all up). For forward rates however, is calculated under the risk neutral measure and is mostly used for option pricing in fixed ...
The website below shows how to price bonds from curves, currently it only supports fixed rate and zero-coupon bonds, but it might give you an idea how to price a floater using similar concept:Goto: https://www.opencminc.comSwitch to Yield Curves under the Market Data sectionClick on any curve point. For example: click on a rate under 10Yhttps://www....
You think you make a mistake where you actually don´t make one.The exercise is just like it is. Resulting in $$r_{6m}>r_{12m}$$The difference in your both answers, based on the same rounding, lays in the different basis for the logarithm.$$r_{6m} = - 2 \log_e \left( \frac{99.8-102.5 e^{-r_{1y}} }{2.5} \right) = \textbf{6.118%}$$$$r_{6m} = - 2 \log_{...
There is only one cashflow for the zero-coupon bond. At maturity, it pays the par value.They are always issued below par, as the buyer is paying the NPV for the bond that matures in the future.Here is a brief reference at Investopedia.A zero-coupon bond, also known as an "accrual bond," is a debt security that doesn't pay interest (a coupon) but is ...
You have to be very careful with terminology here. In particular "yield" is being thrown around carelessly by both of you.The textbook is correct if the (meaningless) phrase "at a 6% yield (rate)" is crossed out and replaced by "at a 6% discount rate". And this is how Tbill's are handled when they are issued (the press release by the US Treasury speaks of ...
We shall prove this by contradiction. Let $\theta=0$ and $\sigma=0$. $X_t=X_0e^{-kt}$ and$$B(0,t)=\exp\Big(-\int_0^te^{X_0e^{-ks}}ds\Big).$$Suppose the contrary that $B(0,t)$ is affine. We should have$$B(0,t)=\exp{\left(A(0,t)-C(0,t)e^{X_0}\right)}\;\;\ \forall (t,X_0), \tag{1}$$Differentiate the logarithm of Equation (1) with respect to $t$ side,$...
The short answer is - you need more data.If you want to build a full zero-rate swap curve, typically these curves go out to 30 years.In general, the front of the curve is made from LIBOR rates, which you have. Typically you don't see practitioners use anything past the 3M point but some will use up to the 6M point.For the 2nd part of the curve, from ...
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Typesetting and Converting Mathematics¶
There are two main uses for MathJax:
Typesetting all the mathematics within a web page, and Converting a string containing mathematics into another form.
In version 2, MathJax could perform the first function very well, but it was much harder to do the second. MathJax version 3 makes both easy to do. Both these tasks are described below.
Typesetting Math in a Web Page¶
MathJax makes it easy to typeset all the math in a web page, and in fact it will do this automatically when it is first loaded unless you configure it not to. So this is one of the easiest actions to perform in MathJax; if your page is static, there is nothing to do but load MathJax.
If your page is dynamic, and you may be adding math after the page isloaded, then you will need to tell MathJax to typeset the mathematicsonce it has been inserted into the page. There are two methods fordoing that:
MathJax.typeset() and
MathJax.typesetPromise().
The first of these,
MathJax.typeset(), typesets the page, anddoes so immediately and synchronously, so when the call finishes, thepage will have been typeset. Note, however, that if the math includesactions that require additional files to be loaded (e.g., TeX inputthat uses require, or that includes autoloaded extensions), thenan error will be thrown. You can use the
try/catch command totrap this condition.
The second,
Mathjax.typesetPromise(), performs the typesettingasynchronously, and returns a promise that is resolved when thetypesetting is complete. This properly handles loading of externalfiles, so if you are expecting to process TeX input that can includerequire or autoloaded extensions, you should use this form oftypesetting. It can be used with
await as part of a larger
async function.
Both functions take an optional argument, which is an array of elements whose content should be processed. An element can be either an actual DOM element, or a CSS selector string for an element or collection of elements. Supplying an array of elements will restrict the typesetting to the contents of those elements only.
Handling Asynchronous Typesetting¶
It is generally a bad idea to try to perform multiple asynchronoustypesetting calls simultaneously, so if you are using
MathJax.typesetPromise() to make several typeset calls, youshould chain them using the promises they return. For example:
MathJax.typesetPromise().then(() => { // modify the DOM here MathJax.typesetPromise();}).catch((err) => console.log(err.message));
This approach can get complicated fast, however, so you may want to maintain a promise that can be used to chain the later typesetting calls. For example,
let promise = Promise.resolve(); // Used to hold chain of typesetting callsfunction typeset(code) { promise = promise.then(() => {code(); return MathJax.typesetPromise()}) .catch((err) => console.log('Typeset failed: ' + err.message)); return promise;}
Then you can use
typeset() to run code that changes the DOMand typesets the result. The
code() that you pass it does theDOM modifications and returns the array of elements to typeset, or
null to typeset the whole page. E.g.,
typeset(() => { const math = document.querySelector('#math'); math.innerHTML = '$$\\frac{a}{1-a^2}$$'; return math;});
would replace the contents of the element with
id="math" with thespecified fraction and have MathJax typeset it (asynchronously).Because the
then() call returns the result of
MathJax.typesetPromise(), which is itself a promise, the
then() will not resolve until that promise is resolved; i.e.,not until the typesetting is complete. Finally, since the
typeset() function returns the
promise, you can use
await in an
async function to wait for the typesetting tocomplete:
await typeset(...);
Note that this doesn’t take the initial typesetting that MathJaxperforms into account, so you might want to use
MathJax.startup.promise in place of
promise above.I.e., simply use
function typeset(code) { MathJax.startup.promise = MathJax.startup.promise .then(() => {code(); return MathJax.typesetPromise()}) .catch((err) => console.log('Typeset failed: ' + err.message)); return MathJax.startup.promise;}
This avoids the need for the global
promise variable, andmakes sure that your typesetting doesn’t occur until the initialtypesetting is complete.
Resetting Automatic Equation Numbering¶
The TeX input jax allows you to automatically number equations. When modifying a page, this can lead to problems as numbered equations may be removed and added; most commonly, duplicate labels lead to issues.
You can reset equation numbering using the command
MathJax.texReset([start])
where
start is the number at which to start equation numbering.
If you have inserted new content, that may require the entire page to be reprocessed in order to get the automatic numbering, labels, and references to be correct. In that case, you can do
MathJax.startup.document.state(0);MathJax.texReset();MathJax.typeset();
to force MathJax to reset the page to the state it was before MathJax processed it, reset the TeX automatic line numbering and labels, and then re-typeset the contents of the page from scratch.
Loading MathJax Only on Pages with Math¶
The MathJax combined configuration files are large, and so you maywish to include MathJax in your page only if it is necessary. If youare using a content-management system that puts headers and footersinto your pages automatically, you may not want to include MathJaxdirectly, unless most of your pages include math, as that would loadMathJax on
all your pages. Once MathJax has been loaded, it shouldbe in the browser’s cache and load quickly on subsequent pages, butthe first page a reader looks at will load more slowly. In order toavoid that, you can use a script like the following one that checks tosee if the content of the page seems to include math, and only loadsMathJax if it does. Note that this is not a very sophisticated test,and it may think there is math in some cases when there really isn’tbut it should reduce the number of pages on which MathJax will have tobe loaded.
Create a file called
check-for-tex.js containing the following:
(function () { var body = document.body.textContent; if (body.match(/(?:\$|\\\(|\\\[|\\begin\{.*?})/)) { if (!window.MathJax) { window.MathJax = { tex: { inlineMath: {'[+]': [['$', '$']]} } }; } var script = document.createElement('script'); script.src = 'https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js'; document.head.appendChild(script); }})();
and then use
<script src="check-for-tex.js" defer></script>
in order to load the script when the page content is ready. Notethat you will want to include the path to the location where youstored
check-mathjax.js, that you should change
tex-chtml.js to whatever component file you want to use, and thatthe
window.MathJax value should be set to whatever configurationyou want to use. In this case, it just adds dollar signs to thein-line math delimiters. Finally, adjust the
body.match() regularexpression to match whatever you are using for math delimiters.
This simply checks if there is something that looks like a TeX in-line or displayed math delimiter, and loads MathJax if there is. If you are using different delimiters, you will need to change the pattern to include those (and exclude any that you don’t use). If you are using AsciiMath instead of TeX, then change the pattern to look for the AsciiMath delimiters.
If you are using MathML, you may want to use
if (document.body.querySelector('math')) {...}
for the test instead (provided you aren’t using namespace prefixes, like <m:math>).
Converting a Math String to Other Formats¶
An important use case for MathJax is to convert a string containing mathematics (in one of the three forms that MathJax understands) and convert it into another form (either MathML, or one of the output formats that MathJax supports). This was difficult to do in MathJax version 2, but easy to do in version 3.
When MathJax startup up, it creates methods for converting from theinput format(s) to the output format(s) that you have loaded, and toMathML format. For example, if you have loaded the MathML input jaxand the SVG output jax (say by using the
mml-svg component), thenMathJax will create the following conversion methods for you:
MathJax.mathml2svg(math[,options])
MathJax.mathml2svgPromise(math[,options])
MathJax.mathml2mml(math[,options])
MathJax.mathml2mmlPromise(math[,options])
If you had loaded the TeX input jax as well, you would also get fourmore methods, with
tex in place of
mathml.
As the names imply, the
Promise functions perform the conversionasynchronously, and return promises, while the others operatesynchronously and return the converted form immediately. The firsttwo functions (and any others like them) produce DOM elements as theresults of the conversion, with the promise versions passing that totheir
then() functions as their argument (see the section onAsynchronous Conversion below), and the non-promise versions returningthem directly. You can insert these DOM elements into the documentdirectly, or you can use their
outerHTML property to obtaintheir serialized string form.
The functions that convert to MathML produce serialized MathML stringsautomatically, rather than DOM elements. (You can use the browser’s
DOMParser object to convert the string into a MathML DOM treeif you need one.)
Conversion Options¶
All four of these functions require an argument that is the mathstring to be converted (e.g., the serialized MathML string, or in thecase of
tex2chtml(), the TeX or LaTeX string). You can alsopass a second argument that is an object containing options thatcontrol the conversion process. The options that can be included are:
display, a boolean specifying whether the math is in display-mode or not (for TeX input). Default is
true.
em, a number giving the number of pixels in an
emfor the surrounding font. Default is
16.
ex, a number giving the number of pixels in an
exfor the surrounding font. Default is
8.
containerWidth, a number giving the width of the container, in pixels. Default is 80 times the
exvalue.
lineWidth', a number giving the line-breaking width in
emunits. Default is a very large number (100000), so effectively no line breaking.
scale, a number giving a scaling factor to apply to the resulting conversion. Default is 1.
For example,
let html = MathJax.tex2chtml('\\sqrt{x^2+1}', {em: 12, ex: 6, display: false});
would convert the TeX expression
\sqrt{x^2+1} to HTML as anin0line expression, with
em size being 12 pixels and
ex sizebeing 6 pixels. The result will be a DOM element containing the HTMLfor the expression. Similarly,
let html = MathJax.tex2chtml('\\sqrt{x^2+1}', {em: 12, ex: 6, display: false});let text = html.outerHTML;
sets
text to be the serialized HTML string for the expression.
Obtaining the Output Metrics¶
Since the
em,
ex, and
containerWidth alldepend on the location where the math will be placed in the document(they are values based on the surrounding text font and the containerelements width), MathJax provides a method for obtaining these valuesfrom a given DOM element. The method
MathJax.getMetricsFor(node, display)
takes a DOM element (
node) and a boolean (
display), indicatingif the math is in display mode or not, and returns an objectcontaining all six of the options listed above. You can pass thisobject directly to the conversion methods discussed above. So you cando something like
let node = document.querySelector('#math');let options = MathJax.getMetricsFor(node, true);let html = MathJax.tex2svg('\\sqrt{x^2+1}', options);node.appendChild(html);
in order to get get the correct metrics for the (eventual) location ofthe math that is being converted. Of course, it would be easier tosimply insert the TeX code into the page and use
MathJax.typeset() to typeset it, but this is just an exampleto show you how to obtain the metrics from a particular location inthe page.
Note that obtaining the metrics causes a page refresh, so it is expensive to do this. If you need to get the metrics from many different locations, there are more efficient ways, but these are advanced topics to be dealt with elsewhere.
Obtaining the Output Stylesheet¶
The output from the SVG and CommonHTML output jax both depend on CSS stylesheets in order to properly format their results. You can obtain the SVG stylesheet element by calling
MathJax.svgStylesheet();
and the HTML stylesheet from
MathJax.chtmlStylesheet();
The CommonHTML output jax CSS can be quite large, so the output jaxtries to minimize the stylesheet by including only the styles that areactually needed for the mathematics that has been processed by theoutput jax. That means you should request the stylesheet only
afteryou have typeset the mathematics itself.
Moreover, if you typeset several expressions, the stylesheet will include everything needed for all the expressions you have typeset. If you want to reset the stylesheet, then use
MathJax.startup.output.clearCache();
if the output jax is the CommonHTML output jax. So if you want toproduce the style sheet for a single expression, issue the
clearCache() command just before the
tex2chtml() call.
Asynchronous Conversion¶
If you are converting TeX or LaTeX that might use require to load extensions, or where extensions might be autoloaded, you will either need to use one of the “full” components that include all the extensions, or preload all the extensions you need if you plan to use the synchronous calls listed above. Otherwise, you can use the promise-based calls, which handle the loading of extensions transparently.
For example,
let node = document.querySelector('#math');let options = MathJax.getMetricsFor(node, true);MathJax.tex2chtmlPromise('\\require{bbox}\\bbox[red]{\\sqrt{x^2+1}}', options) .then((html) => { node.appendChild(html); let sheet = document.querySelector('#MJX-CHTML-styles'); if (sheet) sheet.parentNode.removeChild(sheet); document.head.appendChild(MathJax.chtmlStylesheet()); });
would get the metrics for the element with
id="math", convertthe TeX expression using those metrics (properly handling theasynchronous load needed for the
\require command); then when theexpression is typeset, it is added to the document and the CHTMLstylesheet is updated.
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Terms sourced from: http://iupac.org/publications/pac/62/11/2167/
"Glossary of atmospheric chemistry terms (Recommendations 1990)", Calvert, J.G.,
Pure and Applied Chemistry 1990, 62(11), 2167
abatement absorbance \(A\)absorber absorption absorption coefficient absorption cross-section \(\sigma\)absorption line absorption spectrum accommodation coefficient accretion accuracy accuracy acid acid deposition acid rain actinic flux \(S_{\lambda}\)activity coefficient \(f\),\(\gamma \)adiabatic lapse rate adsorbate adsorbent adsorber adsorption advection aeration aeromete aerometric measurement air contaminant air mass air monitoring station air pollutant air pollution air pollution index air pollution survey air quality characteristic air resource management air sampling network Aitken particles albedo aldehydes alert levels aliquot altocumulus cloud altostratus cloud ambient air ambient air quality Ames/salmonella test amperometric detection method analytical function analytical unit aneroid barometer anticyclone appearance energy arrester Arrhenius equation ash aspirator atmosphere atmosphere atomize
background concentration baffle chamber bag filter baghouse base baseline Beer–Lambert law biosphere bivane blank value blowdown bond-dissociation energy \(D\)boundary layer breeching breeze Brownian motion bubbler baseline concentration
calibration component calibration function calibration gas mixture calibration mixture capacitance hygrometer carbon black carrier gas cascade impactor catalyst catalytic cracking ceilometer ceramic filter chain reaction chemical ionization chemiluminescence analyser chemisorption chimney effect chromatogram chromatography cloud coagulation coefficient of haze \(\text{COH}\)collection efficiency collector collision number colorimeter combustion gas composition of pure air condensation condensation nuclei continuous analyser continuous indication analyser continuous measuring cell controlled atmosphere convection (as applied to air motion) correlation coefficient coulometric detection method cracking critical point critical temperature \(T_{\text{c}}\)cryogenic cryogenic sampling cumulative sample cumulonimbus cloud cumulus cloud cupola cut off cyclone (collector) channel black concentration
dark reaction dead time decay rate deliquescence demister denuder system (tube or assembly) deodorizer deposition deposition velocity desorption desulfurization dew point hygrometer (cooled surface condensation) dew point dichotomous sampler diffraction diffraction analysis diffuser diffusion diffusion battery diluent gas dioxin discomfort threshold discontinuous analyser discontinuous indication discontinuous measuring cell dispersion diurnal variation Dobson unit dosage downwash droplet dry bulb temperature drying agent dust collector dust fall dust dynamic range
eddy eddy dispersion effective chimney height efflorescence effluent einstein electrical hygrometer electrochemical detector electrochemical method of detection electrolytic hygrometer electron capture detector electron charge electrostatic filter electrostatic precipitator elute elutriation emission control equipment emission flux emission enthalpy \(H\)entrainment entropy \(S\)equilibrium constant equivalent diameter evaporation explosivity limits
fall time fallout fanning filter filtration flame ionization detector flame photometric detector flame photometry flammable limits flash point floccule flow analysis flue gas flue gas scrubber fluorimeter fly ash fog fog horizon fossil fuel fragment ion free energy free radical freezing out frequency distribution frost point hygrometer fume fumes fumigation fabric filter
gas analysis installation gas black gas chromatography gaseous diffusion separator Gaussian band shape Gibbs energy (function) \(G\)grab sampling gradient gravimetric method greenhouse effect grit ground level concentration ground level inversion gustiness
haze horizon haze helium ionization detector Henry's law humidity hydrocracking unit hydrometeor hydrosphere hygrometer hygrometry (moisture analysis) hysteresis
ideal gas illuminance \(E\),\(E_{\text{v}}\)immission dose immission flux immission immission rate impaction impingement impinger incinerator individual perception threshold inert gas inertial separator instability (with reference to instrumentation) instantaneous sampling intensity inversion height ion collector ion source isobar isokinetic line isokinetic sampling isotherm isotope pattern isotropic
lachrymator lapse rate LIDAR light scattering limiting condition of operation linear (decadic) absorption coefficient linear range lithometeor lithosphere log-normal distribution luminance luminescence
macrometeorology mass balance mass peak mass range mass spectrometer mass spectrometer focusing system (deflection system) mass spectrometric detector mass spectrum matrix isolation maximum allowable concentration maximum emission concentration maximum storage life mean free path \(\lambda\)measurement resolution measurement threshold mechanical hygrometer memory effect mesopause mesoscale mesosphere microclimatology micrometeorology middle atmosphere Mie scattering mist mixing height mixing ratio mobile phase mobility mode
particle concentration particle size distribution particle size particulate matter passive sampler peak concentration (trace atmospheric component) permeation tube pH photochemistry photoionization detector photolysis photometry photophoresis photostationary state photosynthesis phytotoxicant plume pollution (pollutant) potential temperature potentiometric detection method precipitation precision primary mixture primary pollutant psychrometric hygrometer psychrometry potentiometer
radical (free radical) radiometry radiosonde rain out range of measurement Rayleigh scattering reduced species reference material reference procedure relative density \(d\)relative humidity repeatability reproducibility residual fuel/oil residual spectrum/background spectrum resolution response time retention efficiency Ringelmann chart rise time rotometer residence time
sanitary land fill saturation vapour pressure scanning method scattering scattering angle \(\theta\)scattering cross-section \(\sigma_{\text{scat}}\)scavenging scrubber scrubbing secondary pollution (emissions) sedimentation sensitivity \(A\)sequential analyser sequential indication sequential measuring cell settling chamber settling velocity sink smog chamber smog smog index smoke soiling solar flare solar radiation solubility solution soot sorption specific gravity spectral irradiance \(E_{\lambda}\)spectral spheradiance stagnant inversion standard conditions for gases standard deviation \(s\)static pressure static stability steric factor stochastic sampling stoichiometric Stokes law Stokes number \(St\)STP stratocumulus cloud stratopause stratosphere stratus cloud suspended matter synoptic scale systematic error selectivity steady state supersaturation
temperature inversion temperature lapse rate thermal conductivity detector thermodynamic temperature \(T\)thermosphere titration tracer transfer line Troe expression tropopause troposphere turbidity \(\tau\)
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The two biggest challenges of ACT Math are the time crunch—the math test has 60 questions in 60 minutes!—and the fact that the test doesn’t provide you with any formulas.
All the formulas and math knowledge for the ACT comes from what you’ve learned and memorized.
In this complete list of critical formulas you'll need on the ACT, I'll lay out every formula you
must have memorized before test day, as well as explanations for how to use them and what they mean. I'll also show you which formulas you should prioritize memorizing (the ones that are needed for multiple questions) and which ones you should memorize only when you've got everything else nailed down tight.
Already Feeling Overwhelmed?
Does the prospect of memorizing a bunch of formulas make you want to run for the hills? We've all been there, but don't throw in the towel just yet! The good news about the ACT is that it is designed to give all test-takers a chance to succeed. Many of you will already be familiar with most of these formulas from your math classes.
The formulas that show up on the test the most will also be most familiar to you. Formulas that are only needed for one or two questions on the test will be least familiar to you. For example, the equation of a circle and logarithm formulas only ever show up as one question on most ACT math tests. If you’re going for every point, go ahead and memorize them. But if you feel overwhelmed with formula lists, don’t worry about it—it’s only one question.
So let’s look at all the formulas you absolutely must know before test day (as well as one or two that you can figure out yourself instead of memorizing yet another formula).
Algebra Linear Equations & Functions
There will be at least five to six questions on linear equations and functions on every ACT test, so this is a very important section to know.
Slope
Slope is the measure of how a line changes. It’s expressed as: the change along the y-axis/the change along the x-axis, or $\rise/\run$.
Given two points, $A(x_1,y_1)$, $B(x_2,y_2)$, find the slope of the line that connects them:
$$(y_2 - y_1)/(x_2 - x_1)$$
Slope-Intercept Form A linear equation is written as $y=mx+b$ mis the slope and bis the y-intercept (the point of the line that crosses the y-axis) A line that passes through the origin (y-axis at 0), is written as $y=mx$ If you get an equation that is NOT written this way (i.e. $mx−y=b$), re-write it into $y=mx+b$
Midpoint Formula Given two points, $A(x_1,y_1)$, $B(x_2,y_2)$, find the midpoint of the line that connects them:
$$((x_1 + x_2)/2, (y_1 + y_2)/2)$$
Good to Know Distance Formula Find the distance between the two points
$$√{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
You don’t actually need this formula,as you can simply graph your points and then create a right triangle from them. The distance will be the hypotenuse, which you can find via the pythagorean theorem
Logarithms There will usually only be one question on the test involving logarithms. If you’re worried about having to memorize too many formulas, don’t worry about logs unless you’re trying for a perfect score.
$log_bx$ asks “to what power does
b have to be raised to result in x?” Most of the time on the ACT, you’ll just need to know how to re-write logs
$$log_bx=y => b^y=x$$
$$log_bxy=log_bx+log_by$$
$$log_b{x/y} = log_bx - log_by$$
Statistics and Probability Averages
The average is the same thing as the mean
Find the average/mean of a set of terms (numbers)
$$\Mean = {\sum\of\the\terms}/{\the\number(\amount)\of\different\terms}$$
Find the average speed
$$\Speed = {\total\distance}/{\total\time}$$
May the odds be ever in your favor.
Probabilities
Probability is a representation of the odds of something happening. A probability of 1 is guaranteed to happen. A probability of 0 will never happen.
$${\Probability\of\an\outcome\happening}={\number\of\desired\outcomes}/{\total\number\of\possible\outcomes}$$
Probability of two independent outcomes bothhappening is
$$\Probability\of\event\A*\probability\of\event\B$$
e.g., Event A has a probability of $1/4$ and event B has a probability of $1/8$. The probability of both events happening is: $1/4 * 1/8 = 1/32$. There is a 1 in 32 chance of bothevents A and event B happening.
Combinations
The possible amount of different combinations of a number of different elements
A “combination” means the order of the elements doesn’t matter (i.e. a fish entree and a diet soda is the same thing as a diet soda and a fish entree) Possible combinations = number of element A * number of element B * number of element C…. e.g. In a cafeteria, there are 3 different dessert options, 2 different entree options, and 4 drink options. How many different lunch combinations are possible, using one drink, one, dessert, and one entree? The total combinations possible = 3 * 2 * 4 = 24
Percentages Find xpercent of a given number n
$$n(x/100)$$
Find out what percent a number nis of another number m
$$(100n)/m$$
Find out what number nis xpercent of
$$(100n)/x$$
The ACT is a marathon. Remember to take a break sometimes and enjoy the good things in life. Puppies make everything better.
Geometry Rectangles Area
$$\Area=lw$$
lis the length of the rectangle wis the width of the rectangle
Perimeter
$$\Perimeter=2l+2w$$
Rectangular Solid Volume
$$\Volume = lwh$$
his the height of the figure
Parallelogram
An easy way to get the area of a parallelogram is to drop down two right angles for heights and transform it into a rectangle.
Then solve for husing the pythagorean theorem Area
$$\Area=lh$$
(This is the same as a rectangle’s lw. In this case the height is the equivalent of the width)
Triangles Area
$$\Area = {1/2}bh$$
bis the length of the base of triangle (the edge of one side) his the height of the triangle The height is the same as a side of the 90 degree angle in a right triangle. For non-right triangles, the height will drop down through the interior of the triangle, as shown in the diagram.
Pythagorean Theorem
$$a^2 + b^2 = c^2$$
In a right triangle, the two smaller sides (a and b) are each squared. Their sum is the equal to the square of the hypotenuse (c, longest side of the triangle)
Properties of Special Right Triangle: Isosceles Triangle An isosceles triangle has two sides that are equal in length and two equal angles opposite those sides. An isosceles right triangle always has a 90 degree angle and two 45 degree angles. The side lengths are determined by the formula: x, x, x√2, with the hypotenuse (side opposite 90 degrees) having a length of one of the smaller sides * √2. E.g., An isosceles right triangle may have side lengths of 12, 12, and 12√2.
Properties of Special Right Triangle: 30, 60, 90 Degree Triangle A 30, 60, 90 triangle describes the degree measures of its three angles. The side lengths are determined by the formula: x, x√3, and 2 x. The side opposite 30 degrees is the smallest, with a measurement of x. The side opposite 60 degrees is the middle length, with a measurement of x√3. The side opposite 90 degree is the hypotenuse, with a length of 2 x. For example, a 30-60-90 triangle may have side lengths of 5, 5√3, and 10.
Trapezoids Area Take the average of the length of the parallel sides and multiply that by the height.
$$\Area = [(\parallel\side\a + \parallel\side\b)/2]h$$
Often, you are given enough information to drop down two 90 angles to make a rectangle and two right triangles. You’ll need this for the height anyway, so you can simply find the areas of each triangle and add it to the area of the rectangle, if you would rather not memorize the trapezoid formula. Trapezoids and the need for a trapezoid formula will be at most one question on the test. Keep this as a minimum priority if you're feeling overwhelmed.
Circles
Area
$$\Area=πr^2$$
πis a constant that can, for the purposes of the ACT, be written as 3.14 (or 3.14159) Especially useful to know if you don’t have a calculator that has a $π$ feature or if you're not using a calculator on the test. ris the radius of the circle (any line drawn from the center point straight to the edge of the circle).
Area of a Sector Given a radius and a degree measure of an arc from the center, find the area of that sector of the circle. Use the formula for the area multiplied by the angle of the arc divided by the total angle measure of the circle.
$$Area\of\an\arc = (πr^2)(\degree\measure\of\center\of\arc/360)$$
Circumference
$$\Circumference=2πr$$
or
$$\Circumference=πd$$
dis the diameter of the circle. It is a line that bisects the circle through the midpoint and touches two ends of the circle on opposite sides. It is twice the radius.
Length of an Arc Given a radius and a degree measure of an arc from the center, find the length of the arc. Use the formula for the circumference multiplied by the angle of the arc divided by the total angle measure of the circle (360).
$$\Circumference\of\an\arc = (2πr)(\degree\measure\center\of\arc/360)$$
Example: A 60 degree arc has $1/6$ of the total circle's circumference because $60/360 = 1/6$
An alternative to memorizing the “formulas” for arcs is to just stop and think about arc circumferences and arc areas logically. If you know the formulas for the area/circumference of a circle and you know how many degrees are in a circle, put the two together. If the arc spans 90 degrees of the circle, it must be $1/4$th the total area/circumference of the circle, because $360/90 = 4$. If the arc is at a 45 degree angle, then it is $1/8$th the circle, because $360/45 = 8$. The concept is exactly the same as the formula, but it may help you to think of it this way instead of as a “formula” to memorize.
Equation of a Circle Useful to get a quick point on the ACT, but don’t worry about memorizing it if you feel overwhelmed; it will only ever be worth one point. Given a radius and a center point of a circle $(h, k)$
$$(x - h)^2 + (y - k)^2 = r^2$$
Cylinder
$$\Volume=πr^2h$$
Trigonometry
Almost all the trigonometry on the ACT can be boiled down to a few basic concepts
SOH, CAH, TOA
Sine, cosine, and tangent are graph functions
The sine, cosine, or tangent of an angle (theta, written as Θ) is found using the sides of a triangle according to the mnemonic device SOH, CAH, TOA.
Sine - SOH
$$\Sine Θ = \opposite/\hypotenuse$$
Opposite = the side of the triangle directly opposite the angle Θ Hypotenuse = the longest side of the triangle
Sometimes the ACT will make you manipulate this equation by giving you the sine and the hypotenuse, but not the measure of the opposite side. Manipulate it as you would any algebraic equation:
$Sine Θ = \opposite/\hypotenuse$ => $\hypotenuse * \sine Θ = \opposite$
Cosine - CAH
$$\Cosine Θ = \adjacent/\hypotenuse$$
Adjacent = the side of the triangle nearest the angle Θ (that creates the angle) that is not the hypotenuse Hypotenuse = the longest side of the triangle
Tangent - TOA
$$\Tangent Θ = \opposite/\adjacent$$
Opposite = the side of the triangle directly opposite the angle Θ Adjacent = the side of the triangle nearest the angle Θ (that creates the angle) that is not the hypotenuse
Cosecant, Secant, Cotangent Cosecant is the reciprocal of sine $\Cosecant Θ = \hypotenuse/\opposite$ Secant is the reciprocal of cosine $\Secant Θ = \hypotenuse/\adjacent$ Cotangent is the reciprocal of tangent $\Cotangent Θ = \adjacent/\opposite$
Useful Formulas to Know $$\Sin^2Θ + \Cos^2Θ = 1$$
$${\Sin Θ}/{\Cos Θ} = \Tan Θ$$
Hurray! You've memorized your formulas. Now treat yo' self.
But Keep in Mind
Though these are all the
formulas you should memorize to do well on the ACT math section, this list by no means covers all aspects of the mathematical knowledge you’ll need on the exam. For example, you’ll also need to know your exponent rules, how to FOIL, and how to solve for absolute values. To learn more about the general mathematical topics covered by the test, see our article on what's actually tested on the ACT math section.
What's Next?
Now that you know the critical formulas for the ACT, it might be time to check out our article on How to an Perfect Score on the ACT Math by a 36 ACT-Scorer.
Don't know where to start? Look no further than our article on what is considered a good, bad, or excellent ACT score. Want to improve your score by 4+ points? Our completely online and customized prep program adapts to your strengths, weaknesses, and needs. And we guarantee your money back if you don't improve your score by 4 points or more. Sign up for your free trial today.
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Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time.
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In last week’s discussion of proofs by contradiction and nonconstructive proofs, we showed:
Theorem: There exist irrational numbers \(x\) and \(y\) with the property that \(x^y\) is rational.
However, our proof was
nonconstructive: it did not pinpoint explicit values for \(x\) and \(y\) that satisfy the condition, instead proving only that such numbers must exist. Would a more constructive proof be more satisfying? Let’s see! I claim \(x=\sqrt{2}\) and \(y=\log_2 9\) work, because \(\sqrt{2}\) we already know to be irrational, \(y=\log_2 9\) can be similarly proved to be irrational (try this!), and $$x^y = \sqrt{2}^{\log_2 9} = \sqrt{2}^{\log_{\sqrt{2}}3}=3,$$ which is rational.
Let’s further discuss why last week’s proof was less satisfying. The following rephrasing of this proof may help shed some light on the situation:
Proof: Assume the theorem were false, so that any time \(x\) and \(y\) were irrational, \(x^y\) would also be irrational. This would imply that \(\sqrt{2}^{\sqrt{2}}\) would be irrational, and by applying our assumption again, \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}\) would also be irrational. But this last number equals 2, which is rational. This contradiction disproves our assumption and thereby proves the theorem, QED.
So perhaps this argument seems less satisfactory simply because it is, at its core, a proof by contradiction. It does not give us evidence for the positive statement “\(x\) and \(y\) exist”, but instead only for the negative statement “\(x\) and \(y\)
don’t not exist.” (Note the double negative.) This distinction is subtle, but a similar phenomenon can be found in the English language: the double negative “not bad” does not mean “good” but instead occupies a hazy middle-ground between the two extremes. And even though we don’t usually think of such a middle-ground existing between logic’s “true” and “false”, proofs by contradiction fit naturally into this haze. In fact, these ideas motivate a whole branch of mathematical logic called Constructive logic that disallows double negatives and proofs by contradiction, instead requiring concrete, constructive justifications for all statements.
But wait; last week’s proof that \(\sqrt{2}\) is irrational used contradiction, and therefore is not acceptable in constructive logic. Can we prove this statement constructively? We must show that \(\sqrt{2}\) is not equal to any rational number; what does it even mean to do this constructively? First, we turn it into a positive statement: we must show that \(\sqrt{2}\)
is unequal to every rational number. And how do we constructively prove that two numbers are unequal? By showing that they are measurably far apart. So, here is a sketch of a constructive proof: \(\sqrt{2}\) is unequal to every rational number \(a/b\) because $$\left|\sqrt{2} – \frac{a}{b}\right| \ge \frac{1}{3b^2}.$$ See if you can verify this inequality! [1]
PS. In case you are still wondering whether \(\sqrt{2}^{\sqrt{2}}\) is rational or irrational: It is irrational (moreover, transcendental), but the only proof that I know uses a very difficult theorem of Gelfond and Schneider.
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Much in the fashion of a few problems already answered on Mathematica.SE
(1, 2), I am trying to solve the Laplace equation in velocity potential $\phi$ to simulate irrotational, inviscid flow over a cylinder (in Cartesian coordinates).
I find that my pressure field is flipped. i.e., at the front ($\theta$=0) the stagnation pressure must be a maximum while it is zero at the top and bottm ($\theta=\pi/2$ and $\theta=3\pi/2$). However, my stagnation pressure is maximum at the top and minimum in the front and back.
The Bernoulli equation was used to calculate the pressure field as follows: $$\frac{P_\text{stag}}{\rho g} = \frac{P_1}{\rho g} + \frac{\vec{V}_1^2}{2 g}$$ $$\vec{V}_1 = \frac{\partial \phi}{\partial y} + \frac{\partial \phi}{\partial x}$$ and $$\vec{V}_1^2 = \left(\frac{\partial \phi}{\partial y}\right)^2 + \left(\frac{\partial \phi}{\partial x}\right)^2$$
My code
Needs["DifferentialEquations`InterpolatingFunctionAnatomy`"]Needs["NDSolve`FEM`"]Lx = 40; Ly = 40; lx = 2; ly = 2; cx = Lx/2; cy = Ly/2; r = 5;Ω = RegionDifference[Rectangle[{0, 0}, {Lx, Ly}], Disk[{cx, cy}, r]];RegionPlot[Ω]cellmeasure = 1; iorder = 2;sol = NDSolveValue[{Laplacian[u[x, y], {x, y}] == NeumannValue[1., x == 0] + NeumannValue[-1, x == Lx], DirichletCondition[u[0, 0] == 0, x == Lx && y == 0]}, u, {x, y} ∈ Ω, Method -> {"PDEDiscretization" -> {"FiniteElement", "MeshOptions" -> {"MaxCellMeasure" -> cellmeasure}, "IntegrationOrder" -> iorder}}]{cylmesh} = InterpolatingFunctionCoordinates[sol];cylmesh["Wireframe"]
Velocity vectors seem fine
ClearAll[f];f[x_, y_] := Evaluate[Grad[sol[x, y], {x, y}, "Cartesian"]]StreamPlot[f[x, y], {x, 0, Lx}, {y, 0, Ly}, AspectRatio -> Automatic, Frame -> True, Epilog -> {Thickness[0.005], Line[{{0, lx}, {0, ly}}]}, StreamPoints -> 50]
Pressure contours are INCORRECT
ClearAll[p];P∞ = 0;(*U∞= Evaluate[D[sol[x,y],y] + D[sol[x,y],x]];*)ρ = 1; (*Air at 25 degree C*)(*p=P∞+0.5ρ Evaluate[D[sol[x,y],y]^2 + \D[sol[x,y],x]^2];*)p = P∞ + 0.5 ρ Evaluate[Norm[f[x, y], 2]];pplot = ContourPlot[p, {x, y} ∈ Ω, PlotLegends -> Automatic, Mesh -> True, ColorFunction -> "Temperature", Contours -> 25]
Velocity field around the cylinder is INCORRECT (non zero value at $\theta=0$)
I can only surmise that my "syntax" or manner of usage of some function is incorrect (assuming that the Bernoulli equation has been utilized correctly).
What went wrong?
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Let $E=\{A,B\}$ be a set and $X_{1,t}, X_{2,t}, X_{3,t}$ three independent Markov chains on the set $E$ with respective transition probability $P^{(1)}, P^{(2)}, P^{(3)}$ where $$P^{(i)}=\begin{bmatrix}p^{(i)} & 1-p^{(i)}\\ 1-p^{(i)} & p^{(i)} \end{bmatrix}$$
Let $P = P^{(1)}\otimes P^{(2)}\otimes P^{(3)}$ where $\otimes$ is the Kronecker product.
The process $Y_t = X_{1,t}\times X_{2,t}\times X_{3,t}$ (product of the three Markov chain) is a Markov chain with transition probability $P$ and state space $E_Y = E\times E\times E = \{AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB\}$ and we can use the standard Markov theory to compute anything we want. But, here are my questions :
1 - As $ ''AAB '' $ and $ ''ABA '' $ are the same value equal to $A^2B$, then $E_Y$ can be reduce to $\{A^3, A^2B, AB^2, B^3\}$. Is $Y_t$ still a Markov chain if we combine the transition probability in other to set it in that reduced space?
2 - How can I determine the time distribution probability of begin in a particular state. In other word, how can I compute $$\mathbb{P}[Y_{t+1} \neq A^2B, Y_t = A^2B, Y_{t-1} = A^2B, \dots, Y_1 = A^2B | Y_0=A^2B]$$ for each $t\in \mathbb{N}$.
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The following is adapted from an answer I gave to a similar but not identical question.
Let the position of the (unmoving) sun be $(S_x,S_y)$.
Drag this position to the origin. Let the current position of the planet be $(A_x,A_y)$. Our dragging procedure drags $(A_x,A_y)$ to $(A_x-S_x,A_y-S_y)$.
For simplicity of notation, let $u=A_x-S_x$, $v=A_y-S_y$. Now we determine the angle that the positive $x$-axis has to be rotated through (counterclockwise) to hit the line from the origin to $(u,v)$. Call this angle $\theta$.
Then $\theta$ is the angle, say in the interval $(-\pi,\pi]$, whose cosine is $u/r$, and whose sine is $v/r$, where $r$ is the radius of the circular orbit. So from now on we can take $\theta$ as a
known angle. It can be calculated once and for all from knowledge of $(S_x,S_y)$ and $(A_x,A_y)$. (Be careful to take the signs of $u$ and $v$ into account when calculating $\theta$.)
Assume that we are travelling counterclockwise around the sun, at an angular speed of $\alpha$ radians per time unit. Then after elapsed time $t$, we have travelled through an angle $\alpha t$. (If we are travelling clockwise, use $-\alpha t$.)
After the travel, our angle with the positive $x$-axis is $\theta+\alpha t$. This means that we are at the point with coordinates$$(r\cos(\theta+\alpha t), \: \: r\sin(\theta+\alpha t)).$$
Now transform back, by adding $(S_x,S_y)$ to the point. We obtain$$(S_x+r\cos(\theta+\alpha t),\:\: S_y+r\sin(\theta+\alpha t)).$$This is the position of the planet $t$ time units after it was in position $(A_x,A_y)$. All the components of this formula are known, so now we can compute.
Comment: By using the addition laws for cosine and for sine, we can obtain the following alternate (and for many purposes more useful) version of the answer.Let $(A_x(t), A_y(t)$ be the position of the planet at time $t$. So $(A_x, A_y)$ could also be written as $(A_x(0), A_y(0))$.We obtain$$A_x(t)=S_x+(A_x(0)-S_x)\cos(\alpha t)-(A_y(0)-S_y)\sin(\alpha t),$$$$A_y(t)=S_y+(A_x(0)-S_x)\sin(\alpha t)+(A_y(0)-S_y)\cos(\alpha t).$$
Alternately, we could look up how to rotate a point around the origin by multiplying by a suitable matrix. In the long run, that is a better way of viewing the matter. The calculation we have done can be viewed as a derivation of that matrix formula.
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The first cohomology of parabolic actions for some higher-rank abelian groups and representation theory
1.
Department of Mathematics, The Pennsylvania State University, University Park, PA 16802, United States
parabolicoperators. First, let $N$ be the upper-triangular group of $SL(2,\mathbb{C})$, $\Gamma$ any lattice and $\pi = L^2(SL(2,\mathbb{C})$/$\Gamma)$ the usual left-regular representation. We show that the first smooth almost-cohomology group $H_a^1(N, \pi)$ ≃ $H_a^1(SL(2,\mathbb{C}) , \pi)$. In addition, we show that the first smooth almost-cohomology of actions of certain higher-rank abelian groups $A$ acting by left translation on $(SL(2,\mathbb{R}) \times G)$/$\Gamma$ trivialize, where $G = SL(2,\mathbb{R})$ or $SL(2,\mathbb{C})$ and $\Gamma$ is any irreducible lattice. The abelian groups $A$ are generated by various mixtures of the diagonal and/or unipotent generators on each factor. As a consequence, for these examples we prove that the only smooth time changes for these actions are the trivial ones (up to an automorphism). Mathematics Subject Classification:28Dxx, 43A85, 22E27, 22E40, 58J42. . Citation:David Mieczkowski. The first cohomology of parabolic actions for some higher-rank abelian groups and representation theory. Journal of Modern Dynamics, 2007, 1 (1) : 61-92. doi: 10.3934/jmd.2007.1.61
[1] [2] [3]
Anatole Katok, Federico Rodriguez Hertz.
Measure and cocycle rigidity for certain nonuniformly hyperbolic actions of higher-rank abelian groups.
[4] [5] [6]
Danijela Damjanovic and Anatole Katok.
Local rigidity of actions of higher rank abelian groups and KAM method.
[7] [8] [9]
Changguang Dong.
Separated nets arising from certain higher rank $\mathbb{R}^k$ actions on homogeneous spaces.
[10] [11] [12]
Danijela Damjanović, James Tanis.
Cocycle rigidity and splitting for some discrete parabolic actions.
[13]
James Tanis, Zhenqi Jenny Wang.
Cohomological equation and cocycle rigidity of discrete parabolic actions.
[14] [15]
Danijela Damjanović.
Central extensions of simple Lie groups and rigidity of some abelian partially hyperbolic algebraic actions.
[16] [17]
Manfred Einsiedler, Elon Lindenstrauss.
On measures invariant under diagonalizable actions: the Rank-One case and the general Low-Entropy method.
[18]
Franz W. Kamber and Peter W. Michor.
Completing Lie algebra actions to Lie group actions.
[19] [20]
K. H. Kim, F. W. Roush and J. B. Wagoner.
Inert actions on periodic points.
2018 Impact Factor: 0.295
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1.Kathy buys a watch for Rs.500 and sells it to Jake for Rs.600. Find the profit percent.
a. 20%
b. 25%
c. 30%
d. 35%
Answer: A
Explanation:
Profit percent = $\dfrac{{sp - cp}}{{cp}} \times 100 = \dfrac{{600 - 500}}{{500}} \times 100$ = 20%
Profit percent = $\dfrac{{sp - cp}}{{cp}} \times 100
= \dfrac{{600 - 500}}{{500}} \times 100$ = 20%
2.John made a profit of 25% while selling a book for Rs.250. Find the cost price of the book.
a. Rs.160
b. Rs.170
c. Rs.180
d. Rs.200
Answer: D
Explanation:
From the above formulas, $cp \times \left( {\dfrac{{100 + 25}}{{100}}} \right) = 250$ $ \Rightarrow C.P = \dfrac{{100}}{{125}} \times 250 = 200$
From the above formulas, $cp \times \left( {\dfrac{{100 + 25}}{{100}}} \right) = 250$
$ \Rightarrow C.P = \dfrac{{100}}{{125}} \times 250 = 200$
3.Nivas sold a pen for Rs.900 thus making 10% loss. Find the cost price.
a. Rs.850
b. Rs.1000
c. Rs.1200
d. Rs.1300
Answer: B
Explanation:
From the above formulas, $cp \times \left( {\dfrac{{100 - 10}}{{100}}} \right) = 900$ $ \Rightarrow cp = \dfrac{{100}}{{90}} \times 900 = 1000$.
From the above formulas, $cp \times \left( {\dfrac{{100 - 10}}{{100}}} \right) = 900$
$ \Rightarrow cp = \dfrac{{100}}{{90}} \times 900 = 1000$.
4.A trader buys oranges at 7 for a rupee and sells them at 40% profit. How many oranges does he sell for a rupee?
a. 3
b. 4
c. 5
d. 6
Answer: C
Explanation:
Cost price of 1 apple = $\dfrac{1}{7}$ Selling price of 1 apple at 40% profit is equal to 140% of its cost price. Therefore, Selling price of 1 apple = $\dfrac{1}{7} \times 140\% = \dfrac{1}{7} \times \left( {\dfrac{{140}}{{100}}} \right) = \dfrac{1}{5}$ Therefore, 5 oranges are sold are 1 rupee.
Cost price of 1 apple = $\dfrac{1}{7}$
Selling price of 1 apple at 40% profit is equal to 140% of its cost price.
Therefore, Selling price of 1 apple = $\dfrac{1}{7} \times 140\% = \dfrac{1}{7} \times \left( {\dfrac{{140}}{{100}}} \right) = \dfrac{1}{5}$
Therefore, 5 oranges are sold are 1 rupee.
5.On selling mangoes at 36 for a rupee, a shopkeeper loses 10%. How many mangoes should he sell for a rupee in order to gain 8%?
a. 25
b. 30
c. 35
d. 40
Answer: B
Explanation:
Selling price of 1 apple = Rs.$\dfrac{1}{{36}}$ Cost price of 1 apple = $\dfrac{1}{{36}} \times \dfrac{1}{{90\% }}$ Therefore, New selling price = $\left( {\dfrac{1}{{36}} \times \dfrac{1}{{90\% }}} \right) \times 108\% $ = $\dfrac{{108}}{{36 \times 90}} = \dfrac{1}{{30}}$ Therefore, For one rupee, he should sell 30 mangoes.
Selling price of 1 apple = Rs.$\dfrac{1}{{36}}$
Cost price of 1 apple = $\dfrac{1}{{36}} \times \dfrac{1}{{90\% }}$
Therefore, New selling price = $\left( {\dfrac{1}{{36}} \times \dfrac{1}{{90\% }}} \right) \times 108\% $ = $\dfrac{{108}}{{36 \times 90}} = \dfrac{1}{{30}}$
Therefore, For one rupee, he should sell 30 mangoes.
6.A boy buys eggs at 10 for Rs.1.80 and sells them at 11 for Rs. 2. What is his gain or loss per cent?
a. 1.27%
b. 1.01%
c. 1.68%
d. 1.77%
Answer: B
Explanation:
To avoid fractions, let the number of eggs purchased be, LCM of 10 and 11 = 110 CP of 110 eggs = $\dfrac{{110 \times 1.80}}{{10}}$ = Rs. 19.80 SP of 110 eggs = $\dfrac{{110 \times 2.0}}{{11}}$ = Rs. 20.00 Profit per cent = $\dfrac{{20 - 19.80}}{{19.80}} \times 100$ = $\dfrac{1}{{99}} \times 100$ = 1.01%
To avoid fractions, let the number of eggs purchased be, LCM of 10 and 11 = 110
CP of 110 eggs = $\dfrac{{110 \times 1.80}}{{10}}$ = Rs. 19.80
SP of 110 eggs = $\dfrac{{110 \times 2.0}}{{11}}$ = Rs. 20.00
Profit per cent = $\dfrac{{20 - 19.80}}{{19.80}} \times 100$ = $\dfrac{1}{{99}} \times 100$ = 1.01%
7.A woman buys apples at 15 for a rupee and the same number at 20 a rupee. She mixes and sells them at 35 for 2 rupees. What is her gain per cent or loss per cent?
a. 2.04%
b. 3.5%
c. 4.4%
d. 5.4%
Answer: A
Explanation:
Suppose the woman buys two lots of (LCM of 15, 20 and 35) = 420 apples. One apple cost from the first lot = $\dfrac{1}{{15}}$. 420 apples cost = $\dfrac{1}{{15}} \times 420 = 28$ One apple cost from the second lot = $\dfrac{1}{{20}}$. 420 apples cost = $\dfrac{1}{{20}} \times 420 = 21$ Total cost for 840 apples = 28 + 21 = Rs. 49 Selling price of 1 apple = $\dfrac{2}{{35}}$. 840 apples selling price = $\dfrac{2}{{35}} \times 840 = 48$ Loss per cent = $\dfrac{1}{{49}} \times 100 = 2.04\% $
Suppose the woman buys two lots of (LCM of 15, 20 and 35) = 420 apples.
One apple cost from the first lot = $\dfrac{1}{{15}}$.
420 apples cost = $\dfrac{1}{{15}} \times 420 = 28$
One apple cost from the second lot = $\dfrac{1}{{20}}$.
420 apples cost = $\dfrac{1}{{20}} \times 420 = 21$
Total cost for 840 apples = 28 + 21 = Rs. 49
Selling price of 1 apple = $\dfrac{2}{{35}}$.
840 apples selling price = $\dfrac{2}{{35}} \times 840 = 48$
Loss per cent = $\dfrac{1}{{49}} \times 100 = 2.04\% $
8.Some quantity of coffee is sold at Rs. 22 per kg, making 10% profit. If total gain is Rs. 88, what is the quantity of coffee sold?
a. 44
b. 55
c. 60
d. 70
Answer: A
Explanation:
Cost price of the coffee = $\dfrac{{22}}{{110\% }} = 22 \times \dfrac{{100}}{{110}} = 20$ Profit on one kg of coffee = 22 - 20 = Rs.2. Let $x$ kgs of coffee sold. So total profit = $2x$. $ \Rightarrow 2x = 88$ $ \Rightarrow $ x = 44
Cost price of the coffee = $\dfrac{{22}}{{110\% }} = 22 \times \dfrac{{100}}{{110}} = 20$
Profit on one kg of coffee = 22 - 20 = Rs.2.
Let $x$ kgs of coffee sold. So total profit = $2x$.
$ \Rightarrow 2x = 88$
$ \Rightarrow $ x = 44
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Setting
The (sub-)circuit in question looks like this:
(Both OP amps can be considered as ideal.)
– Schematic created using CircuitLab
It's part of a larger question about an Ackerberg-Mossberg biquad (which looks like this) where one is asked to calculate
a) this subcircuits DC voltage gain \$A_{v,DC}=\frac{V_{out}}{V_{in}}\$, and later
b) its transfer function \$A_v(s)=\frac{V_{out}(s)}{V_{in}(s)}\$.
As I like to derive my answers for DC analysis from the transfer function \$A_v(s)\$, I tried the following:
1) Directly determine the transfer function \$A_v(s)\$, which answers b): $$A_v(s)=\frac{V_{out}(s)}{V_{in}(s)}=\frac{1}{sR_1C_1}$$
It should be the transfer function of a
non-inverting integrator amplifier.
2) Compute the DC gain by using \$\lim\limits_{s \rightarrow 0}{A_v(s)} \$. That is $$A_{v,DC}=\lim\limits_{s \rightarrow 0}{\left(A_v(s)\right)}=\lim\limits_{s \rightarrow 0}{\left(\frac{1}{sR_1C_1}\right)=\infty}$$
But here's the thing: the solution of a) simply says that \$A_{v,DC}=-\infty\$ which contradicts my answer.
Questions
So here's my questions for you:
i) Which answer to a) is correct?
ii) If my answer is wrong, one possible explanation would be that I can't simply use the limit for \$s\rightarrow 0\$ because in reality it's defined as \$s:= \sigma + i\omega\$. This means that I would have to assume \$\sigma=0\$ and do the limit for \$\omega\rightarrow 0\$ which brings me to $$A_{v,DC}=\lim\limits_{\omega \rightarrow 0}{\left(A_v(s=0+i\omega)\right)}=\lim\limits_{\omega \rightarrow 0}{\left|\frac{1}{(0+i\omega)R_1C_1}\right|}=\lim\limits_{\omega \rightarrow 0}{\left|\frac{-i}{\omega R_1C_1}\right|}=-\infty$$ Would this be a correct derivation?
iii) If the above doesn't work, why is it so? And is there a way to calculate a) from b)?
DC analysis can become very confusing when one has to deal with open loop amplifiers, especially in bigger circuits. So it would be very nice if one can derive the DC analysis results from the laplace domain transfer function.
iv) To go beyond the exercise question: I'm curious about the bode plot, because the single pole at \$s=0\$ (and no zeros) suggests that the magnitude goes from infinity to zero (with -20dB/decade) as the frequency goes upwards but what happens with the phase?
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Taken from Chapter 17 of Mas Colell "Microeconomic Theory"
Consider an exchange economy with two commodities and two consumers. Both consumers have homothetic preferences of the constant elasticity variety. Morover, the elasticity of substitution is the same for both consumers and is small (i.e. goods are close to perfect complements). Specifically
$u_1(x_{11},x_{21})=(2x_{11}^{\rho}+x_{21}^{\rho})^{1/\rho}$ and $u_1(x_{12},x_{22})=(x_{12}^{\rho}+2x_{22}^{\rho})^{1/\rho}$
And $\rho=-4$, The endowments are $w_1=(1,0)$ and $w_2=(0,1)$. Compute the excess demand function of the economy and verify that there are multiple equilibria.
My attempt
After applying fist-condition and normalizing prices as $\frac{P_1}{P2}=p$, I have that the demand functions are:
For the first consumer: $x_1=\frac{p}{p+(p/2)^{\frac{1}{1-\rho}}}$ and $x_2=\frac{p(p/2)^{\frac{1}{1-\rho}}}{p+(p/2)^{\frac{1}{1-\rho}}}$
For the second consumer: $x_1=\frac{1}{p+(2p)^{\frac{1}{1-\rho}}}$ and $x_2=\frac{(2p)^{\frac{1}{1-\rho}}}{p+(2p)^{\frac{1}{1-\rho}}}$
So the excess demand function would be:
$\begin{pmatrix} z_1\\z_2 \end{pmatrix}=\begin{pmatrix} \frac{p}{p+(p/2)^{\frac{1}{1-\rho}}}+\frac{1}{p+(2p)^{\frac{1}{1-\rho}}}-1 \\ \frac{p(p/2)^{\frac{1}{1-\rho}}}{p+(p/2)^{\frac{1}{1-\rho}}}+\frac{(2p)^{\frac{1}{1-\rho}}}{p+(2p)^{\frac{1}{1-\rho}}}-1 \end{pmatrix}$
The thing is that this is not the answer stated in Mas Colell's Solutions. Here it is:
I guess the elasticity of substitution has something to do here? But I don't get it anyway because if we had some corner solution, the answers would be $(x_1,x_2)=(1,0)$ for this consumer and $(x_1,x_2)=(0,1)$ for second consumer.
Any idea?
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April 2010
"To a factor près": Cayley’s Partial Anticipation of the Weierstrass P-Function By: Adrian Rice arice4@rmc.edu Although Arthur Cayley is perhaps best remembered for his contributions to matrix algebra and group theory, one of his prime interests was the subject of elliptic functions, among the most vibrant areas of mathematics in the nineteenth century. His publications on this topic included a little-known but interesting anticipation of an identity later made famous by Weierstrass. What is particularly pleasing about Cayley's derivation of this identity is that it relied totally on his use of invariant theory, a subject that seems, on the face of it, to have little connection to the theory of elliptic functions. In this paper, we compare Cayley's derivation to the standard Weierstrassian approach, and discuss reasons for the obscurity of the former compared to the relative fame of the latter.
Transitive Decompositions of Graphs and Their Links with Geometry and Origami By: Geoffrey Pearce Modular origami is a popular offshoot of the traditional Japanese art of paper folding; it involves building large and elaborate geometrical structures by fitting together a number of "modules" folded from separate squares of paper. We show how the idea of a "transitive decomposition" of a graph can be used to find highly symmetrical and decorative colorings of these structures.
Trigonometric Identities à la Hermite By: Warren P. Johnson wpjoh@conncoll.edu Hermite once observed that a certain product of cotangents can be integrated by breaking it into a sum of cotangents, where the coefficients are themselves products of cotangents. Why should such an identity exist? We give two derivations, one based on the partial fractions expansion of the cotangent. Hermite seems to have used a mixture of the two. We also discuss and extend a second theorem of Hermite, which leads to generalizations of his cotangent identity. The paper veers off into determinants at the end.
Quasi-Cauchy Sequences By: David Burton and John Coleman dburton@franciscan.edu, jcoleman@franciscan.edu A quasi-Cauchy sequence is one in which the distance between successive terms tends to zero. This is a far weaker property than that of being Cauchy, although students in undergraduate real analysis classes often struggle with the distinction between the two concepts. Nevertheless, quasi-Cauchy sequences have many interesting properties. In this paper we investigate such sequences in both the real number system and in general metric spaces.
p-Free l p Inequalities By: Grahame Bennett bennettg@indiana.edu We show how certain simple l p inequalities may be proved by "ignoring the p." This entails revisiting that weird and wonderful world of Hardy, Littlewood, and Pólya (wherein everything is more-or-less and nothing’s as it seems). We encounter many familiar characters along the way, among them: Cauchy’s Inequality, Theory of Means, Convexity, Rolle’s Theorem, and Descartes’ Rule of Signs. What makes our trip worthwhile is the realization that these old chestnuts still have something new to tell us: at any rate, they all appear here in novel, and sometimes surprising, ways.
Notes
A Short Proof of ζ (2) = π 2/6 By: T. H. Marshall tmarshall@aus.edu We show that the sum $\sum 1/\log 2( z)$, taken over all branches of the logarithm, is a rational function, and use this to give a short proof of \sum 1/ n 2=\π 2/6.
The Group of Symmetries of the Tower of Hanoi Graph By: So Eun Park soeun.park@berkeley.edu The Tower of Hanoi problem, one of the most famous mathematical puzzles, has many interesting aspects to study, such as the properties of its graph in the case of 3 pegs (the most widely known form of the puzzle) and the shortest paths (or geodesics) in generalized Tower of Hanoi problems.
In particular, Frame and Stewart, in response to
Monthly problem 3918, suggested a way to narrow down the possible cases of shortest paths of the generalized problems with more than 3 pegs (this Monthly {48} (1941) 216--219); the problem remains unsolved. In this note, we look at the generalized Tower of Hanoi problems, with the number of pegs greater than 3, from the perspective of graph theory. We prove that no matter how many pegs and disks we are playing with in the problem, the group of automorphisms of its graph is always isomorphic to the group of the peg permutations, i.e., the number of disks of the problem does not affect the automorphism group of the Tower of Hanoi graph.
Recurrent Proofs of the Irrationality of Certain Trigonometric Values By: Li Zhou and Lubomir Markov lzhou@polk.edu, lmarkov@mail.barry.edu Consider the set of transcendental functions ${F=\{ \cos , \sin , \tan , \cosh , \sinh , \tanh , \exp \}$. If x is a nonzero real number and $f\in F$, then x 2and ( f( x)) 2 cannot both be rational. This result is known and the proofs of some of its corollaries, such as the irrationality of π, are classical. It is the purpose of our paper to offer simple new proofs of these results, accessible to a calculus student.
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This question already has an answer here:
\begin{equation*} U(x, y) = (ax^{-c} + by^{-c})^{-\frac{1}{c}} \end{equation*}
I ask this mainly because after logging both sides of the Utility equation (the first step to proving the assertion, I assume), I am left with:
\begin{equation*} \lim_{c \rightarrow 0} \dfrac{-\ln(ax^{-c} + by^{-c})}{c} \end{equation*} I know that the bottom will go to 0, and I have a feeling that the top will go to 0 to. However, all I am left with on the top is essentially $a + b$, and for it to go to 0, $a + b = 1$.
How can $a + b = 1$? Is this the right direction? What does $a + b = 1$ mean? Why does $a + b = 1$?
Edit: And once proven, what does this whole "limit" thing say about the original function? What is so special about this particular equation such that its limit as $c \rightarrow 0$ is the Cobb Douglas function?
Edit 2: Upon further research, I have discovered a suspiciously similar function known as the CES. $a$ and $b$, however, are instead $a$ and $(1-a)$ !! Now I'm even more confused. How am I supposed to derive that complementary relationship from this equation? This is supposed to be consumer theory!
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I'm basically just going to copy Matthew Doty's puzzles but with lexicographical order:
**Definiton.** The **lexicographical order** of two posets \\((A,\leq_A)\\) and \\((B,\leq_B)\\) is \\((A \times B, \leq^{lex})\\) where
$$
(a_1,b_1) \leq^{lex} (a_2,b_2) \Longleftrightarrow a_1 \lt_A a_2 \text{ or } (a_1 =_A a_2 \text{ and } b_1 \leq_B b_2)
$$
---------------------------
Let \\(A\\) be a lattice.
**AV Puzzle 1**: Show that \\(\Delta\\) is monotonically increasing on \\(\leq^{lex}\\)
**AV Puzzle 2**: Find the *right adjoint* \\(r : A\times A \to A\\) to \\(\Delta\\) such that:
$$
\Delta(x) \leq^{lex} (y,z) \Longleftrightarrow x \leq_{A} r(y,z)
$$
**AV Puzzle 3**: Find the *left adjoint* \\(l : A\times A \to A\\) to \\(\Delta\\) such that:
$$
l(x,y) \leq_{A} z \Longleftrightarrow (x,y) \leq^{lex} \Delta(z)
$$
I think there are solutions but I could be wrong.
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I have a question about the contents of this paper*, which links a building energy model and a utility-maximization component. In it, the author tests several electricity prices using a Cobb-Douglas utility function. As I understand, the C-D function ($U=X^\alpha*Y^\beta$) stipulates that $X=\frac{\alpha}{\alpha+\beta}*\frac{Income}{P_x}$ at optimality. If Income and share parameters are fixed, that would mean $(X)(P_x)$ is always constant; if $P_x$ doubles, $X$ is halved, etc..
That is not what the paper finds in the linkage of the two components. P_electricity*ElectricityUse is not always constant. The expenditure equation implied by the C-D function is violated, but the paper applies an expenditure equation from the building energy model instead of it. They just maximize the utility function as the objective, subject to the budget constraint, and given the electricity use and expenditures from the building model, but do not factor in the implied relations.
Is this a violation that would nullify such linkage?
*Sorry if it is blocked behind a paywall, there isn't an alternate free copy.
Citation: Matar, W. "Households' response to changes in electricity pricing schemes: Bridging microeconomic and engineering principles." Energy Economics 75.
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Inhalt des Dokuments Berliner Kolloquium Wahrscheinlichkeitstheorie und Seminar RTG 1845 Sommerstemester 2015
Datum Sprecher GK-Seminar 17:15-18:00 Sprecher Kolloquium (18:00-19:00) Ort 22. 4. 2015 (keiner, Ausstellung) Herbert Spohn (München) 17:15-18:15 U Potsdam Haus 8, Raum 0.58 6.5.2015 Maite Wilke Berenguer (TUB) Patrick Cattiaux (Toulouse) TU Berlin MA 041 20. 5. 2015 (kein Kolloquium, Euler-Lecture am 22. 5.) U Potsdam 3.6.2015 Alberto Chiarini (TUB) Sabine Jansen (Bochum) TU Berlin MA 041 17.6.2015 Matti Leimbach (TUB) Jan Swart (Prag) TU Berlin MA 041 1.7.2015 Alexandros Saplaouras (TUB) Yuri Kifer (Jerusalem) TU Berlin MA 041 15.7.2015 Jennifer Krüger/Eva Lang (TUB) Dirk Blömker (Augsburg) U Potsdam
Das Kolloquium im Sommersemester findet an der TU Berlin (MA 041) und der U Potsdam statt. Organisation: S. Roelly (UP) und N. Kurt (TUB).
Abstracts 22. April
Kolloquium: Herbert Spohn (TU München) The KPZ (Kardar-Parisi-Zhang) equation and its universality class Abstract: The one-dimensional KPZ equation is a stochastic PDE which describes the dynamics of surface growth. It is one representative of a much larger universality class. I will discuss a few models in this class and explain how they are connected. They all share the common feature to be stochastic integrable.
6. Mai:
RTG-Seminar: Maite Wilke Berenguer (TUB)
Lipschitz Percolation
Abstract:
Kolloquium: Patrick Cattiaux (Toulouse) Central Limit Theorem for additive functionals of some Markov processes: anomalous results Abstract: In this talk we will consider an ergodic Markov process $X_t$ ($t \in \mathbb N$ or $t \in \mathbb R^+$) with unique invariant probability $\mu$, and some additive functional $S_t=\sum_{k=1}^t \, f(X_k)$ or $S_t=\int_0^t \, f(X_s)ds$ for some $\mu$ centered $f$. If $f \in \mathbb L 2(\mu)$ the expected appropriate normalization is $\sqrt{\Var(S_t)}$ (expected to be of order $\sqrt t$), and the expected limit is then a standard gaussian. If $f \in mathbb L^p(\mu)$ ($1<p<2$) one expects in some cases some stable limit after appropriate normalization. It turns out that the mixing rate of the process (equivalently the rate of convergence to equilibrium) is of particular importance for these results to hold true. We shall recall some of the main recent (and less recent) results in this direction and explain how the mixing rate enters into the game. We shall also discuss a particular class of examples for which depending on whether the convergence to equilibrium is quick enough or not, anomalous limit (with some variance breaking) or anomalous normalization appear. At the level of the invariance principle instead of the simple CLT theorem, the expected limiting process becomes a fractional Brownian motion instead of the usual one. These examples correspond to a special class of kinetic P.D.E.'s with heavy tails equilibria.
3. Juni:
RTG-Seminar: Alberto Chiarini (TUB) Extremes of the supercritical Gaussian Free FieldAbstract: We show that the rescaled maximum of the discrete Gaussian Free Field (DGFF) in dimension larger or equal to 3 is in the maximal domain of attraction of the Gumbel distribution. A finer description of the maximum can also be obtained, that is, the associated extremal process converges to a Poisson point process. These results holds both for the infinite-volume field as well as the field with zero boundary conditions. The proofs follow from an interesting application of the Stein-Chen method from Arratia et al. (1989).Joint work with Alessandra Cipriani (WIAS) and Rajat Subhra Hazra (Indian Statistical Institute)
Kolloquium: Sabine Jansen (Bochum)
Non-colliding Ornstein-Uhlenbeck bridges and symmetry breaking in a quantum 1D Coulomb system
Abstract:
Jellium is a model where negatively charged electrons move in a uniform neutralizing background of positive charge. Eugene Wigner conjectured that at low density, the electrons should crystallize, i.e., form a periodic lattice. We prove that in dimension 1, in a quantum mechanics setup, this actually happens for all temperatures and densities, thereby extending low-density results by Brascamp and Lieb (1975) and classical results by Aizenman and Martin (1980). The proof uses the Feynman-Kac formula to map the quantum model to asystem of non-colliding Ornstein-Uhlenbeck bridges, and then applies the Krein-Rutman theorem (an infinite-dimensional version of Perron-Frobenius). The talk is based on joint work with Paul Jung (University of Alabama at Birmingham).
17. Juni
RTG-Seminar: Matti Leimbach (TUB)
Porous medium equation with proliferation Abstract: Motivated by mathematical oncology, we present two PDE's, the viscous poruous medium equation and the Fisher-Kolmogorov-Petrovskii-Piskunov model (FKPP). Formally, we derive the viscous porous medium equation as the limit of the empirical measure of a system of interacting particles with intermediate interaction-range and large amplitude. We illustrate that also the FKPP model is a limit of a particle system with certain branching mechanism. At the end, we briefly discuss a combination of these two PDE's, the porous medium equation with proliferation. This is based on ongoing research with Franco Flandoli.
Kolloquium: Jan Swart (Prag)
Rank-based Markov chains, self-organized criticality, and order book dynamics. Abstract: In this talk, we will take a look at some systems of interacting particles on the real line, where the only spatial structure that is relevant for the dynamics is the relative order of the particles. Examples of such systems are the modified Bak-Sneppen model, introduced (as a variation of the original 1993 model) by Meester and Sarkar (2012), Barabási's (2005) queueing system and a variation on the latter due to Gabrielli and Caldarelli (2009), a model for the evolution of the state of an order book on a stock market, introduced by Stigler (1964) and independently by Luckock (2003), and a two models for canyon formation introduced by me (2014). All these systems employ a version of the rule "kill the lowest particle" and seem to exhibit self-organized criticality at a critical point that marks the boundary between an interval where all particles are eventually removed and an interval where particle stay in the system forever.
1. Juli
RTG-Seminar: Alexandros Saplaouras (TU Berlin)
Towards a robustness result for BSDEs with jumpsMotivated by the robustness of BSDEs with respect to the Brownian motion, see \cite{BDM}, we want to prove that the same holds when the BSDE is taken with respect to a square integrable, quasi-left-continuous martingale $M$. The robustness of a BSDE stands for the following property: having a suitable martingale approximation $M^n$ of $M$, then the solutions of the BSDEs driven by $M^n$, converge to the solution of the BSDE driven by $M$. In order to obtain the result, we need to overcome two intermediate problems. The first is to guarantee the existence and uniqueness of solutions of BSDEs driven by $M^n$. In this case, the predictable quadratic covariation of $M^n$ may have jumps, hence the Lebesgue-Stieltjes integral is not necessarily a continuous process. In this work we improve a general result of existence and uniqueness for BSDEs, see \cite{EKH}, where the Lebesgue-Stieltjes integral is with respect to a continuous, predictable and increasing process. Our improvement consists in allowing the integrator of the Lebesgue-Stieltjes integral having (suitably small) jumps, i.e. being a c\`adl\`ag, predictable and increasing process. The second problem consists in proving that the corresponding stochastic and Lebesgue-Stieltjes integrals with respect to $M^n$ and the predictable quadratic covariations $\pqc{M^n}$ respectively converge to the stochastic and Lebesgue-Stieltjes integral with respect to $M$ and the predictable quadratic covariation $\pqc{M}$ respectively. Once this second obstacle is overcome, we could proceed to proving the desired result. As a byproduct of this result, the convergence of the Euler scheme for BSDEs is obtained, where $M^n$ is the time discretization of $M$.
Kolloquium: Yuri Kifer, (Hebrew University, Jerusalem)
Further advances in nonconventional limit theorems
15. Juli (Beginn 16:30/17:30, Ort: U Potsdam)
RTG-Seminar: Eva Lang (TUB)
A multiscale analysis of traveling waves in stochastic neural fields
Kolloquium: Dirk Blömker (Augsburg) Stochastic dynamics near a change of stability (Amplitude- and Modulation-Equations)
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Fix an integer $k$. Let $X=G/P$ be a complex rational homogeneous variety. I assume here $G$ is a simply connected semi simple complex Lie group and $P=P_k$ is a maximal parabolic subgroup defined by dropping the $k$-th simple root (Assume we have chosen and fixed a Borel subgroup to avoid ambiguity, and we use Bourbaki convention about the order of simple roots).
I want to compute the cohomology of the homogeneous bundle $T_X\otimes L^{-\lambda_k}$ over $X$. Here, $T_X$ is the tangent bundle while $L^{-\lambda_k}$ is the line bundle corresponding to the 1-dimensional $P$-representation with character induced by $\lambda_k$. For example, if $X$ is a Grassmannian variety embedded into a projective space using Plücker embedding, then $L^{-\lambda_k}$ is just $\mathcal O(-1)$.
My strategy is to use Borel Weil Bott theorem. But $T_X$ is not irreducible in general (i.e. The $P$-representation $\mathfrak{g/p}$ is not irreducible), so is not $T_X\otimes L^{-\lambda_k}$. Hence, I have to find a $P$-representation filtration of $\mathfrak{g/p}$, say $0\subset s_1\subset s_2\subset\ldots\subset s_r=\mathfrak{g/p}$ with quotients $T_i$ irreducible $P$-representations. We shall use the same notation for the homogeneous vector bundles corresponding to $s_i, T_i$. My plan is: STEP I. compute the cohomology of $T_i\otimes L^{-\lambda_k}$ using Borel Weil Bott theorem for all $i$ since they are irreducible. STEP II. Using the filtration and step I to get the cohomology of $T_X\otimes L^{-\lambda_k}$.
STEP I is easily done by prudent computation, and STEP II can be done in most cases. But I meet some difficulties in step II for some special cases: I need to write the connection morphism down explicitly in these cases. I will use the following example to demonstrate my dilemma here.
From now on, let $G$ be the simply connected Lie group of type $B_l$ and $P=P_2$ be a maximal parabolic subgroup defined by dropping the second simple root. The the filtration of the tangent bundle of $X=G/P$ is $0\subset s_1\subset s_2=T_X$, and hence we have a short exact sequence $$ 0\to s_1\to T_X\to s_2/s_1\to 0$$ Tensoring $L^{-\lambda_2}$, we get another short exact sequence which we simply write as $$0\to s_1(-1)\to T_X(-1)\to s_2/s_1(-1)\to 0$$from which we have a long exact sequence. According to my computation using Borel Weil Bott theorem $$H^q(s_1(-1))=\mathbb C, q=1; 0, q\ne 1$$ $$H^q(s_2/s_1(-1))=\mathbb C, q=0; 0, q\ne 0$$
Hence, my long exact sequence looks like $$ 0\to H^0(T_X(-1))\to \mathbb C \stackrel{\delta}{\to} \mathbb C \to H^1(T_X)\to 0$$ Therefore, to determine what I want: $H^q(T_X(-1))$, I need to write down $\delta$ explicitly. Maybe there are other methods that can determine the cohomology of $T_X(-1)$ directly without $\delta$, so any idea is welcome and appreciated!
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Is it ever valid to include a two-way interaction in a model without including the main effects? What if your hypothesis is only about the interaction, do you still need to include the main effects?
In my experience, not only is it necessary to have all lower order effects in the model when they are connected to higher order effects, but it is also important to properly model (e.g., allowing to be nonlinear) main effects that are seemingly unrelated to the factors in the interactions of interest. That's because interactions between $x_1$ and $x_2$ can be stand-ins for main effects of $x_3$ and $x_4$. Interactions sometimes
seem to be needed because they are collinear with omitted variables or omitted nonlinear (e.g., spline) terms.
You ask whether it's ever valid. Let me provide a common example, whose elucidation may suggest additional analytical approaches for you.
The simplest example of an interaction is a model with one dependent variable $Z$ and two independent variables $X$, $Y$ in the form
$$Z = \alpha + \beta' X + \gamma' Y + \delta' X Y + \varepsilon,$$
with $\varepsilon$ a random term variable having zero expectation, and using parameters $\alpha, \beta', \gamma',$ and $\delta'$. It's often worthwhile checking whether $\delta'$ approximates $\beta' \gamma'$, because an algebraically equivalent expression of the same model is
$$Z = \alpha \left(1 + \beta X + \gamma Y + \delta X Y \right) + \varepsilon$$
$$= \alpha \left(1 + \beta X \right) \left(1 + \gamma Y \right) + \alpha \left( \delta - \beta \gamma \right) X Y + \varepsilon$$
(where $\beta' = \alpha \beta$, etc).
Whence, if there's a reason to suppose $\left( \delta - \beta \gamma \right) \sim 0$, we can absorb it in the error term $\varepsilon$. Not only does this give a "pure interaction", it does so
without a constant term. This in turn strongly suggests taking logarithms. Some heteroscedasticity in the residuals--that is, a tendency for residuals associated with larger values of $Z$ to be larger in absolute value than average--would also point in this direction. We would then want to explore an alternative formulation
$$\log(Z) = \log(\alpha) + \log(1 + \beta X) + \log(1 + \gamma Y) + \tau$$
with iid random error $\tau$. Furthermore, if we expect $\beta X$ and $\gamma Y$ to be large compared to $1$, we would instead just propose the model
$$\log(Z) = \left(\log(\alpha) + \log(\beta) + \log(\gamma)\right) + \log(X) + \log(Y) + \tau$$
$$= \eta + \log(X) + \log(Y) + \tau.$$
This new model has just a single parameter $\eta$ instead of four parameters ($\alpha$, $\beta'$, etc.) subject to a quadratic relation ($\delta' = \beta' \gamma'$), a considerable simplification.
I am not saying that this is a necessary or even the only step to take, but I am suggesting that this kind of algebraic rearrangement of the model is usually worth considering whenever interactions alone appear to be significant.
Some excellent ways to explore models with interaction, especially with just two and three independent variables, appear in chapters 10 - 13 of Tukey's EDA.
While it is often stated in textbooks that one should never include an interaction in a model without the corresponding main effects, there are certainly examples where this would make perfect sense. I'll give you the simplest example I can imagine.
Suppose subjects randomly assigned to two groups are measured twice, once at baseline (i.e., right after the randomization) and once after group T received some kind of treatment, while group C did not. Then a repeated-measures model for these data would include a main effect for measurement occasion (a dummy variable that is 0 for baseline and 1 for the follow-up) and an interaction term between the group dummy (0 for C, 1 for T) and the time dummy.
The model intercept then estimates the average score of the subjects at baseline (regardless of the group they are in). The coefficient for the measurement occasion dummy indicates the change in the control group between baseline and the follow-up. And the coefficient for the interaction term indicates how much bigger/smaller the change was in the treatment group compared to the control group.
Here, it is not necessary to include the main effect for group, because at baseline, the groups are equivalent by definition due to the randomization.
One could of course argue that the main effect for group should still be included, so that, in case the randomization failed, this will be revealed by the analysis. However, that is equivalent to testing the baseline means of the two groups against each other. And there are plenty of people who frown upon testing for baseline differences in randomized studies (of course, there are also plenty who find it useful, but this is another issue).
The reason to keep the main effects in the model is for identifiability. Hence, if the purpose is statistical inference about each of the effects, you should keep the main effects in the model. However, if your modeling purpose is solely to predict new values, then it is perfectly legitimate to include only the interaction if that improves predictive accuracy.
this is implicit in many of answers others have given but the simple point is that models w/ a product term but w/ & w/o the moderator & predictor are just different models. Figure out
what each means given the process you are modeling and whether a model w/o the moderator & predictor makes more sense given your theory or hypothesis. The observation that the product term is significant but only when moderator & predictor are not included doesn't tell you anything (except maybe that you are fishing around for "significance") w/o a cogent explanation of why it makes sense to leave them out.
Arguably, it depends on what you're using your model for. But I've never seen a reason not to run and describe models with main effects, even in cases where the hypothesis is only about the interaction.
I will borrow a paragraph from the book
An introduction to survival analysis using Stata by M.Cleves, R.Gutierrez, W.Gould, Y.Marchenko edited by Stata press to answer to your question.
It is common to read that interaction effects should be included in the model only when the corresponding main effects are also included, but there is nothing wrong with including interaction effects by themselves. [...] The goal of a researcher is to parametrize what is reasonably likely to be true for the data considering the problem at hand and not merely following a prescription.
I would suggest it is simply a special case of model uncertainty. From a Bayesian perspective, you simply treat this in exactly the same way you would treat any other kind of uncertainty, by either:
Calculating its probability, if it is the object of interest Integrating or averaging it out, if it is not of interest, but may still affect your conclusions
This is exactly what people do when testing for "significant effects" by using t-quantiles instead of normal quantiles. Because you have uncertainty about the "true noise level" you take this into account by using a more spread out distribution in testing. So from your perspective the "main effect" is actually a "nuisance parameter" in relation to the question that you are asking. So you simply average out the two cases (or more generally, over the models you are considering). So I would have the (vague) hypothesis: $$\newcommand{\int}{\mathrm{int}}H_{\int}:\text{The interaction between A and B is significant}$$ I would say that although not precisely defined, this is the question you want to answer here. And note that it is not the verbal statements such as above which "define" the hypothesis, but the mathematical equations as well. We have some data $D$, and prior information $I$, then we simply calculate: $$P(H_{\int}|DI)=P(H_{\int}|I)\frac{P(D|H_{\int}I)}{P(D|I)}$$ (small note: no matter how many times I write out this equation, it always helps me understand the problem better. weird). The main quantity to calculate is the likelihood $P(D|H_{int}I)$, this makes no reference to the model, so the model must have been removed using the law of total probability: $$P(D|H_{\int}I)=\sum_{m=1}^{N_{M}}P(DM_{m}|H_{\int}I)=\sum_{m=1}^{N_{M}}P(M_{m}|H_{\int}I)P(D|M_{m}H_{\int}I)$$ Where $M_{m}$ indexes the mth model, and $N_{M}$ is the number of models being considered. The first term is the "model weight" which says how much the data and prior information support the mth model. The second term indicates how much the mth model supports the hypothesis. Plugging this equation back into the original Bayes theorem gives: $$P(H_{\int}|DI)=\frac{P(H_{\int}|I)}{P(D|I)}\sum_{m=1}^{N_{M}}P(M_{m}|H_{\int}I)P(D|M_{m}H_{int}I)$$ $$=\frac{1}{P(D|I)}\sum_{m=1}^{N_{M}}P(DM_{m}|I)\frac{P(M_{m}H_{\int}D|I)}{P(DM_{m}|I)}=\sum_{m=1}^{N_{M}}P(M_{m}|DI)P(H_{\int}|DM_{m}I)$$
And you can see from this that $P(H_{\int}|DM_{m}I)$ is the "conditional conclusion" of the hypothesis under the mth model (this is usually all that is considered, for a chosen "best" model). Note that this standard analysis is justified whenever $P(M_{m}|DI)\approx 1$ - an "obviously best" model - or whenever $P(H_{\int}|DM_{j}I)\approx P(H_{\int}|DM_{k}I)$ - all models give the same/similar conclusions. However if neither are met, then Bayes' Theorem says the best procedure is to average out the results, placing higher weights on the models which are most supported by the data and prior information.
Both
x and y will be correlated with xy (unless you have taken a specific measure to prevent this by using centering). Thus if you obtain a substantial interaction effect with your approach, it will likely amount to one or more main effects masquerading as an interaction. This is not going to produce clear, interpretable results. What is desirable is instead to see how much the interaction can explain over and above what the main effects do, by including x, y, and (preferably in a subsequent step) xy.
As to terminology: yes, β 0 is called the "constant." On the other hand, "partial" has specific meanings in regression and so I wouldn't use that term to describe your strategy here.
Some interesting examples that will arise once in a blue moon are described at this thread.
It is very rarely a good idea to include an interaction term without the main effects involved in it. David Rindskopf of CCNY has written some papers about those rare instances.
There are various processes in nature that involve only an interaction effect and laws that decribe them. For instance Ohm's law. In psychology you have for instance the performance model of Vroom (1964): Performance = Ability x Motivation.Now, you might expect finding an significant interaction effect when this law is true. Regretfully, this is not the case. You might easily end up with finding two main effects and an insignificant interaction effect (for a demonstration and further explanation see Landsheer, van den Wittenboer and Maassen (2006), Social Science Research 35, 274-294). The linear model is not very well suited for detecting interaction effects; Ohm might never have found his law when he had used linear models.
As a result, interpreting interaction effects in linear models is difficult. If you have a theory that predicts an interaction effect, you should include it even when insignificant. You may want to ignore main effects if your theory excludes those, but you will find that difficult, as significant main effects are often found in the case of a true data generating mechanism that has only a multiplicative effect.
My answer is: Yes, it can be valid to include a two-way interaction in a model without including the main effects. Linear models are excellent tools to approximate the outcomes of a large variety of data generating mechanisms, but their formula's can not be easily interpreted as a valid description of the data generating mechanism.
This one is tricky and happened to me in my last project. I would explain it this way: lets say you had variables A and B which came out significant independently and by a business sense you thought that an interaction of A and B seems good. You included the interaction which came out to be significant but B lost its significance. You would explain your model initially by showing two results. The results would show that initially B was significant but when seen in light of A it lost its sheen. So B is a good variable but only when seen in light of various levels of A (if A is a categorical variable). Its like saying Obama is a good leader when seen in the light of its SEAL army. So Obama*seal will be a significant variable. But Obama when seen alone might not be as important. (No offense to Obama, just an example.)
F = m*a, force equals mass times acceleration.
It is not represented as F = m + a + ma, or some other linear combination of those parameters. Indeed, only the interaction between mass and acceleration would make sense physically.
If the variables in question are categorical, then including interactions without the main effects is just a reparameterizations of the model, and the choice of parameterization depends on what you are trying to accomplish with your model. Interacting continuous variables with other continuous variables ore with categorical variables is a whole different story. See: see this faq from UCLA's Institute for Digital Research and Education
Yes this can be valid, although it is rare. But in this case you still need to model the main effects, which you will afterward regress out.
Indeed, in some models, only the interaction is interesting, such as drug testing/clinical models. This is for example the basis of the Generalized PsychoPhysiological Interactions (gPPI) model:
y = ax + bxh + ch where
x/y are voxels/regions of interest and
h the block/events designs.
In this model, both
a and
c will be regressed out, only
b will be kept for inference (the beta coefficients). Indeed, both
a and
c represent spurious activity in our case, and only
b represents what cannot be explained by spurious activity, the interaction with the task.
The short answer:If you include interaction in the fixed effects, then the main effects are
automatically included whether or not you specifically include them in your code. The only difference is your parametrization, i.e., what the parameters in your model mean (e.g., are they group means or are they differences from reference levels).
Assumptions: I assume we are working in the general linear model and are asking when we can use the fixed effects specification $AB$ instead of $A + B + AB$, where $A$ and $B$ are (categorical) factors.
Mathematical clarification: We assume that the response vector $Y \sim \mathcal N(\xi , \sigma^2 I_n )$.If $X_A$, $X_B$ and $X_{AB}$ are the design matrices for the three factors, then a model with "main effects and interaction" corresponds to the restriction $\xi \in$ span$\{X_A, X_B, X_{AB}\}$.A model with "only interaction" corresponds to the restriction $\xi \in$ span$\{X_{AB}\}$.
However, span$\{X_{AB}\} =$ span$\{X_A, X_B, X_{AB}\}$. So, it's two different parametrizations of the same model (or the same family of distributions if you are more comfortable with that terminology).
I just saw that David Beede provided a very similar answer (apologies), but I thought I would leave this up for those who respond well to a linear algebra perspective.
Is it ever valid to include a two-way interaction without main effect?
Yes it can be valid and even necessary. If for example in 2. you would include a factor for main effect (average difference of blue vs red condition) this would make the model worse.
What if your hypothesis is only about the interaction, do you still need to include the main effects?
Your hypothesis might be true independent of there being a main effect. But the model might need it to best describe the underlying process. So yes, you should try with and without.
Note: You need to center the code for the "continuous" independent variable (measurement in the example). Otherwise the interaction coefficients in the modelwill not be symmetrically distributed (no coefficient for the first measurement in the example).
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Existence and non-existence results for variational higher order elliptic systems
Università degli Studi dell'Insubria, Dipartimento di Scienza e Alta Tecnologia, Via Valleggio 11, Como, 22100, Italy
$α ∈ \mathbb{N}$
$α ≥ 1$
$(-Δ)^{α} = -Δ((-Δ)^{α-1})$
$\begin{cases}\begin{aligned}(-Δ)^{α} u = H_v(u, v) \\(-Δ)^{α} v = H_u(u, v) \\\end{aligned} \text{ in } Ω \subset \mathbb{R}^N \\\frac{\partial^{r} u}{\partial ν^{r}} = 0, \, r = 0, \dots, α-1, \text{ on } \partial Ω \\\frac{\partial^{r} v}{\partial ν^{r}} = 0, \, r = 0, \dots, α-1, \text{ on } \partial Ω\end{cases}$
$Ω$
$N >2α$
$ν$
$\partial Ω$
$H ∈ C^1 (\mathbb{R}^2; \mathbb{R})$ Keywords:Polyharmonic operators, Lane-Emden systems, higher order Dirichlet boundary conditions, Hamiltonian systems, topological methods. Mathematics Subject Classification:35J48, 35J58, 35J50. Citation:Delia Schiera. Existence and non-existence results for variational higher order elliptic systems. Discrete & Continuous Dynamical Systems - A, 2018, 38 (10) : 5145-5161. doi: 10.3934/dcds.2018227
References:
[1]
R. Adams and J. Fournier,
[2] [3] [4] [5]
G. Caristi, L. D'Ambrosio and E. Mitidieri,
Representation formulae for solutions to some classes of higher order systems and related Liouville theorems,
[6] [7]
D. Cassani and C. Tarsi,
Existence of solitary waves for supercritical Schrödinger systems in dimension two,
[8]
D. Cassani and J. Zhang,
A priori estimates and positivity for semiclassical ground states for systems of critical Schrödinger equations in dimension two,
[9] [10] [11] [12] [13]
F. Gazzola, H. C. Grunau, and G. Sweers,
[14] [15] [16]
J. L. Lions and E. Magenes,
[17] [18] [19] [20] [21]
E. Mitidieri and S. I. Pohozaev,
A priori estimates and the absence of solutions of nonlinear partial differential equations and inequalities,
[22] [23]
P. Poláčik, P. Quittner and P. Souplet,
Singularity and decay estimates in superlinear problems via Liouville-type theorems, Ⅰ: Elliptic equations and systems,
[24] [25] [26]
B. Ruf, Superlinear elliptic equations and systems, in
[27] [28] [29] [30] [31] [32]
M. Struwe,
[33]
show all references
References:
[1]
R. Adams and J. Fournier,
[2] [3] [4] [5]
G. Caristi, L. D'Ambrosio and E. Mitidieri,
Representation formulae for solutions to some classes of higher order systems and related Liouville theorems,
[6] [7]
D. Cassani and C. Tarsi,
Existence of solitary waves for supercritical Schrödinger systems in dimension two,
[8]
D. Cassani and J. Zhang,
A priori estimates and positivity for semiclassical ground states for systems of critical Schrödinger equations in dimension two,
[9] [10] [11] [12] [13]
F. Gazzola, H. C. Grunau, and G. Sweers,
[14] [15] [16]
J. L. Lions and E. Magenes,
[17] [18] [19] [20] [21]
E. Mitidieri and S. I. Pohozaev,
A priori estimates and the absence of solutions of nonlinear partial differential equations and inequalities,
[22] [23]
P. Poláčik, P. Quittner and P. Souplet,
Singularity and decay estimates in superlinear problems via Liouville-type theorems, Ⅰ: Elliptic equations and systems,
[24] [25] [26]
B. Ruf, Superlinear elliptic equations and systems, in
[27] [28] [29] [30] [31] [32]
M. Struwe,
[33]
[1] [2]
Frank Arthur, Xiaodong Yan.
A Liouville-type theorem for higher order elliptic systems of Hé non-Lane-Emden type.
[3] [4] [5] [6]
Marian Gidea, Rafael De La Llave.
Topological methods in the instability problem of Hamiltonian systems.
[7]
Hatem Hajlaoui, Abdellaziz Harrabi, Foued Mtiri.
Liouville theorems for stable solutions of the weighted Lane-Emden system.
[8]
Jingbo Dou, Fangfang Ren, John Villavert.
Classification of positive solutions to a Lane-Emden type integral system with negative exponents.
[9]
Kazuyuki Yagasaki.
Higher-order Melnikov method and chaos
for two-degree-of-freedom Hamiltonian systems
with saddle-centers.
[10]
Guillaume Duval, Andrzej J. Maciejewski.
Integrability of Hamiltonian systems with homogeneous potentials of degrees $\pm 2$. An application of higher order variational equations.
[11]
Carmen Calvo-Jurado, Juan Casado-Díaz, Manuel Luna-Laynez.
Parabolic problems with varying operators and Dirichlet and Neumann boundary conditions on varying sets.
[12]
Robert Baier, Thuy T. T. Le.
Construction of the minimum time function for linear systems via higher-order set-valued methods.
[13]
Anna Kostianko, Sergey Zelik.
Inertial manifolds for 1D reaction-diffusion-advection systems. Part Ⅰ: Dirichlet and Neumann boundary conditions.
[14]
Sándor Kelemen, Pavol Quittner.
Boundedness and a priori estimates of solutions
to elliptic systems
with Dirichlet-Neumann boundary conditions.
[15]
Marek Fila, Hirokazu Ninomiya, Juan-Luis Vázquez.
Dirichlet boundary conditions can prevent blow-up in reaction-diffusion equations and systems.
[16]
Wei Dai, Zhao Liu, Guozhen Lu.
Hardy-Sobolev type integral systems with Dirichlet boundary conditions in a half space.
[17]
Maike Schulte, Anton Arnold.
Discrete transparent boundary conditions for the Schrodinger equation -- a compact higher order scheme.
[18]
Qiong Meng, X. H. Tang.
Solutions of a second-order Hamiltonian system with periodic boundary conditions.
[19] [20]
Tatsien Li, Bopeng Rao, Zhiqiang Wang.
Exact boundary controllability and observability for
first order quasilinear hyperbolic systems with a kind of nonlocal
boundary conditions.
2018 Impact Factor: 1.143
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Why does one consider the dual of the Steenrod algebra?
There are really two reasons, both of which are already alluded to in Eric Peterson's answer and in Pierre's answer.
Formally one can think about mod p cohomology as being a functor to the category of modules over the Steenrod Algebra. Similarly one can think about mod p homology as being a functor to the category of comodules over the dual to the Steenrod Algebra.
These seem formally the same, and the first seems preferable since most of us are happier thinking about modules than about comodules. But:
1) Certain kinds of formulas can be written down more easily for the dual of the Steenrod Algebra than for the Steenrod Algebra itself. In particular, the coproduct on the dual of the Steenrod Algebra has a formula that is simpler for some applications than understanding the non-commutative product on the Steenrod Algebra, via (for example) the Adem relations.
As an example, the Milnor basis of the Steenrod Algebra is defined by noting that the dual to the Steenrod Algebra is a polynomial algebra. The Milnor basis to the Steenrod Algebra is then the dual basis to the monomial basis of the dual algebra.
2) The Steenrod Algebra is the (Hopf) algebra of cohomology operations. By contrast, the dual to the Steenrod Algebra is the (Hopf) algebra of homology co-operations.
When constructing the Adams spectral sequence for other theories (other than mod p homology or cohomology) it is more convenient to use the homology version with the algebra of homology co-operations. This is because the cohomology of a product often involves completions, whereas the homology of a product does not.
Once one has incorporated this way of thinking about things it becomes convenient even for the ordinary Adams spectral sequence. It removes the confusion caused by contravariant functors, and it makes maps between Adams spectral sequences (for different homology theories) more transparent.
Hal Sadofsky's answer is great, and you should accept it.
If you want a more obscure point of view, one which is purely algebraic, then (at least at the prime 2), the dual of the Steenrod algebra represents the group of strict automorphisms of the additive formal group. That is, for any commutative $\mathbf{F}_2$-algebra $R$, the additive formal group for $R$ is just the group $(R,+)$, and an automorphism of this is determined by a power series $f(x)$ with coefficients in $R$ such that $f(x+y) = f(x) + f(y)$. The group operation is composition of power series. So such automorphisms are just power series of the form$$r_0 x + r_1 x^2 + r_2 x^4 + \dots + r_{n} x^{2^n} + \dots,$$where each $r_i$ is in $R$ and $r_1$ is in $R^\times$. Such an automorphism is called
strict if $r_1=1$. Now if $A$ is the dual of the mod 2 Steenrod algebra, then it's easy to check that$$\text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R)$$is in bijection with the set of strict automorphisms, as above, where given a map $A \to R$, each coefficient $r_{n}$ of the power series is determined by the image of the polynomial generator $\xi_n$ (or maybe its antipode $\zeta_n$). It's fun to then verify that composition of power series is induced by the coproduct in $A$: the group operation $$\text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R) \times \text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R) \to \text{Hom}_{\mathbf{F}_2-\text{alg}}(A, R)$$is induced by an $\mathbf{F}_2$-algebra map $$A \to A \otimes A,$$and it's precisely the coproduct of $A$, as described by Milnor.
At odd primes you can do this with just the polynomial part of the dual of the Steenrod algebra.
There is also a connection between $A$ and complete binary trees...
Decided to move a link in one of my comments to an answer - just for convenience.
Some detailed information about close connection between the Milnor dual of the Steenrod algebra and the automorphism group scheme of the additive group scheme is given in the paper "On Realizations of the Steenrod Algebras" by Lebedev and Leites (Journal of Prime Research in Mathematics 2 (2007) 1–13)
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Difference between revisions of "Stiffened equation of state"
m (Added some internal links)
(Quite a number of chages; fixed LANL reference)
Line 1: Line 1: −
The '''Stiffened equation of state''' is a simplified form of the Grüneisen equation of state <ref>[http://catalog.lanl.gov/F/
+
The '''Stiffened equation of state''' is a simplified form of the Grüneisen equation of state <ref>[http://catalog.lanl.gov/F/-?func=&doc_library=LNL01&doc_number=000518513&=&=&=Francis H. Harlow and Anthony A. Amsden "Fluid Dynamics", Los Alamos Report Number LA-4700 page 3 (1971)]</ref>.
When considering water under very high pressures (typical applications are underwater explosions, extracorporeal shock wave lithotripsy, and sonoluminescence) the stiffened [[Equations of state|equation of state]] is often used:
When considering water under very high pressures (typical applications are underwater explosions, extracorporeal shock wave lithotripsy, and sonoluminescence) the stiffened [[Equations of state|equation of state]] is often used:
−
:<math> p=\rho(\gamma-1)e-
+
:<math> p=\rho(\gamma-1)e- p^</math>
where <math>e</math> is the internal energy per unit mass, given by (Eq. 15 in <ref>[http://dx.doi.org/10.1016/S0045-7930(02)00021-X H. Paillère, C. Corre, and J. R. Garcı́a Cascales "On the extension of the AUSM+ scheme to compressible two-fluid models", Computers & Fluids '''32''' pp. 891-916 (2003)]</ref>):
where <math>e</math> is the internal energy per unit mass, given by (Eq. 15 in <ref>[http://dx.doi.org/10.1016/S0045-7930(02)00021-X H. Paillère, C. Corre, and J. R. Garcı́a Cascales "On the extension of the AUSM+ scheme to compressible two-fluid models", Computers & Fluids '''32''' pp. 891-916 (2003)]</ref>):
Line 8: Line 8:
:<math> e = \frac{C_p}{\gamma}T + \frac{p^0}{p} </math>
:<math> e = \frac{C_p}{\gamma}T + \frac{p^0}{p} </math>
−
where <math>C_p</math> is the [[heat capacity]] at constant [[pressure]]. <math>\gamma</math> is an empirically determined constant typically taken to be about 6.1, and <math>p^
+
where <math>C_p</math> is the [[heat capacity]] at constant [[pressure]]. <math>\gamma</math> is an empirically determined constant typically taken to be about 6.1, and <math>p^</math> is another constant, representing the molecular attraction between [[water]] molecules. The magnitude of the later correction is about 2 gigapascals (20,000 atmospheres).
− +
the [[speed of sound]] in water is given by
+
<math>c^2=\gamma p+p^\</math>
+
.
Thus water behaves as though it is an [[ideal gas]] that is ''already'' under about 20,000 atmospheres (2 GPa) pressure, and explains why water is commonly assumed to be incompressible: when the external pressure changes from 1 atmosphere to 2 atmospheres (100 kPa to 200 kPa), the water behaves as an ideal gas would when changing from 20,001 to 20,002 atmospheres (2000.1 MPa to 2000.2 MPa).
Thus water behaves as though it is an [[ideal gas]] that is ''already'' under about 20,000 atmospheres (2 GPa) pressure, and explains why water is commonly assumed to be incompressible: when the external pressure changes from 1 atmosphere to 2 atmospheres (100 kPa to 200 kPa), the water behaves as an ideal gas would when changing from 20,001 to 20,002 atmospheres (2000.1 MPa to 2000.2 MPa).
This equation mispredicts the heat capacity of water but few simple alternatives are available for severely nonisentropic processes such as strong shocks.
This equation mispredicts the heat capacity of water but few simple alternatives are available for severely nonisentropic processes such as strong shocks.
+ + + + + +
==References==
==References==
<references/>
<references/>
[[Category:equations of state]]
[[Category:equations of state]]
Revision as of 13:42, 10 October 2013
The
Stiffened equation of state is a simplified form of the Grüneisen equation of state [1].When considering water under very high pressures (typical applications are underwater explosions, extracorporeal shock wave lithotripsy, and sonoluminescence) the stiffened equation of state is often used:
where is the internal energy per unit mass, given by (Eq. 15 in
[2]):
where is the heat capacity at constant pressure. is an empirically determined constant typically taken to be about 6.1, and is another constant, representing the molecular attraction between water molecules. The magnitude of the later correction is about 2 gigapascals (20,000 atmospheres).
,
from which the value of $p^*$ may be computed given all the other variables.
Thus water behaves as though it is an ideal gas that is
already under about 20,000 atmospheres (2 GPa) pressure, and explains why water is commonly assumed to be incompressible: when the external pressure changes from 1 atmosphere to 2 atmospheres (100 kPa to 200 kPa), the water behaves as an ideal gas would when changing from 20,001 to 20,002 atmospheres (2000.1 MPa to 2000.2 MPa).
This equation mispredicts the heat capacity of water but few simple alternatives are available for severely nonisentropic processes such as strong shocks.
It is useful to notice that, given this equation of state, the adiabatic law is modified from its ideal form:
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I am teaching a Calculus class and we are finishing up power/Taylor series this week. The last section of the chapter is on applications, but the only ones listed there are approximating non-rational numbers like $\sqrt{1.02}$ and computing limits like $\lim_{x\to 0}\frac{\sin x}{x}$. I would like to find better examples that may or may not have a quick physical application (I cannot assume they know any physics beyond what I can explain). So, my question is, do any of you know some applications of Taylor series that I could spend maybe about half an hour to forty five minutes doing? They needn't be physical applications, just interesting. I have already done Euler's formula. Also, we do not deal with any remainder theorems in this class. Thanks.
Not quite on the power of Taylor's series, but you could use the Taylor series for $f(x) = e^x$ to show that $e$ is irrational.
Here is an interesting application of power series; unfortunately one would need to bother with the remainder to make it really interesting.
$$\arctan(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n+1} \,.$$
Plug in $x= \frac{1}{\sqrt 3}$ ($1$ would also work but one would need to explain why this formula also holds at the end point of the interval). We get:
$$\frac{\pi}{6} = \frac{1}{\sqrt{3}}\sum_{n=0}^\infty \frac{(-1)^n}{3^n2n+1} \,.$$
The right side is an alternating series which converges very fast, thus you can use it to calculate $\pi$ with 5-6 digits. And it is alternating, which means you could use the Alternating Series error estimate.
You can also do the same for the Taylor series of $e^x$.
I presume the students would know how to sum (finite/infinite) geometric series. You can tell them that the infinite geometric series can be seen as the Taylor expansion of the function $\frac{1}{1-x}$: $$ \frac{1}{1-x} = 1+ x + x^2 + x^3 + \cdots. $$ Integrating term by term (not always valid!), $$ -\ln (1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots, $$ which is nothing but the Taylor expansion of the function $\ln(1+x)$. (Of course, you can also derive the Taylor expansion directly, if you prefer.) Finally, plugging in $x=-1$ (once again, not really valid!), we get the sum of the alternating harmonic series: $$ \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots. $$
There are many unjustified steps in the above derivation, but it gives a cool demonstration of power series and Taylor expansions.
You can show that a certain function $f$ does not have any zero between $-1$ and $1$, for instance, by showing that $1/f$ is analytic with a radius of analyticity at least $1$ (and therefore finite).
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as stated: why other assets' value can be determined by taking into consideration their expected cash flow (CF)? I read an argument which refers to arbitrage, but I wonder is there an additional simple argument from a theoretical point of view. thanks,
closed as unclear what you're asking by chollida, Bob Jansen♦ Apr 5 '17 at 17:14
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
Discouting a cash-flow to get its present value only works for non random cash-flows. In the option case, the cash-flow (the option's pay-off) is unknown as it depends on the value of the underlying at maturity, value that you don't know. Therefore you can't simply discount the option pay-off/cash-flow to get the option price/present value of the cash-flow.
You have to discount first indeed and then take an expectation (with respect to the risk neutral measure) of the discounted payoff to get the option's price. This is general, and for non-random future cash-flow (fix payments) you recover the discounting method. That is what essentialy states the fundamental theorem of asset pricing which is connected to the notion of arbitrage (to the non existence of it, precisely).
To be more theoretical, under the hypothesis that arbitrage do not exist, there exists a numéraire $N$ and a probability measure $\mathbf{Q}^N$ associated to it such that each tradable asset (of expiry $T$) price $X$ is a (local) martingale under the numéraire $N$, that is $X/N$ is a (local) martingale : $$\forall t\in [0,T], X_t = N_t \mathbf{E}^{\mathbf{Q}^N}\left[ \left.\frac{X_T}{N_T}\right| \mathscr{F}_t\right].$$ (Please note I don't give full precision here, just rough ideas allowing to get a formula for the price.) In particular, today's price at $t=0$ is the expectation $$X_0 = N_0 \mathbf{E}^{\mathbf{Q}^N}\left[ \frac{X_T}{N_T}\right].$$
If your asset pays a
known cash-flow $c$ at $T$ then $$X_0 = N_0 c \mathbf{E}^{\mathbf{Q}^N}\left[ \frac{1}{N_T}\right].$$
Often the measure $N$ is the bank-account numéraire measure (also called risk neutral measure), the numéraire being $N_t = e^{\int_0^t r_s ds}$ where $r_s$ is the time $s$ instantaneous interest rate, and then $$X_0 = c\mathbf{E}^{\mathbf{Q}^N}\left[ e^{-\int_0^T r_s ds}\right].$$ In this case $\mathbf{E}^{\mathbf{Q}^N}\left[ e^{-\int_0^T r_s ds}\right]$ is the price today of the zero-coupon of maturity $T$ (the product that pays you $1$ at $T$), this price is noted $P_{0,T}$ and called the discount factor of maturity $T$, so that simply $$X_0 = c P_{0,T}.$$
You see : to get the price today of a
known (that is, non random) cash-flow $c$ payed at $T$, you simply multiply $c$ by the discount factor $P_{0,T}$.
Of course, it can be more complex, think of call options for instance, for which for a strike $K$ you'd have $$\textrm{Call price}_{t=0} = k\mathbf{E}^{\mathbf{Q}^N}\left[ e^{-\int_0^T r_s ds}\left( S_T - K \right)_{+}\right]$$ would $S$ be the underlying.
For a neat introduction to all of this and to quantitative finance, you have the following references :
chapter 2 of Pierre Henry-Labordère's "Analysis, Geometry, and Modeling in Finance: Advanced Methods in Option Pricing"
chapter 1 of volume 1 ("Foundations and Vanilla Models") of Andersen's and Piterbarg's "Interest Rate Modeling"
They can be a bit "rough", but there are really worth the pain.
I guess the existence of a market is the gist of the question. It is easier to understand if you compare the
insurance industry with the investment banking industry:
Broadly speaking, when an insurer underwrites an insurance policy, it is unable to hedge its risk by subscribing the same insurance policy it just underwrote $-$ there exists reinsurance treaties that allow insurers to offload part of their risk, but they tend to be aggregate (
i.e.on a portfolio of policies basis) and non-linear (see non-proportional reinsurance article in Wikipedia if you are interested). Hence, a risk management approach based on expected cash flows and the law of large numbers is a sensible approach.
On the contrary, investment banks can hedge their derivative exposure by entering a trade which corresponds to a contrary, offsetting position to the one they just underwrote.
The main difference between these 2 situations is the existence of a (reasonably) complete, (reasonably) liquid and (reasonably) well-functioning market, which leads to the following claims:
When there is a functioning market in the risk you trade, your price must be determined by the. absence of arbitrage When there is no functioning market for this risk, your price must rely on your. expected cash flowsand the law of large numbers
So, getting back to the purely financial realm and forgetting about insurance, I guess you can apply this type of reasoning to:
Other assets for which there is no sufficiently liquid market for them, such as Commercial Real Estate; In general, corporate finance projects, companies investment plans, etc. which have traditionally been evaluated through the discounted cash flows method.
To finish off, I will include some excerpts from the introduction (Chapter 1) of Baxter's and Rennie's excellent book
Financial Calculus':
With markets where the stock can be bought and sold freely and arbitrarily positive and negative amounts of stock can be maintained without cost, trying to trade forward using the strong law would lead to disaster […].
[…]
But the existence of an arbitrage price, however surprising, overrides the strong law. To put it simply, if there is an arbitrage price, any other price is too dangerous to quote.
[…]
The strong law and expectation give the wrong price for forwards. But in a certain sense, the forward is a special case. The construction strategy $-$ buying the stock and holding it $-$ certainly wouldn’t work for more complex claims. The standard call option which offers the buyer the right but not the obligation to receive the stock for some strike price agreed in advance certainly couldn’t be constructed this way. If the stock price ends up above the strike, then the buyer would exercise the option and ask to receive the stock – having it salted away in a drawer would then be useful to the seller. But if the stock price ends up below the strike, the buyer will abandon the option and any stock owned by the seller would have incurred a pointless loss.
Thus maybe a strong-law price would be appropriate for a call option, and until 1973, many people would have agreed. Almost everything appeared safe to price via expectation and the strong law, and only forwards and close relations seemed to have an arbitrage price. Since 1973, however, and the infamous Black-Scholes paper, just how wrong this is has slowly come out. Nowhere in this book will we use the strong law again. […] All derivatives can be built from the underlying $-$ arbitrage lurks everywhere.
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Difference between revisions of "Pole (of a function)"
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An isolated singular point $a$ of single-valued character of an analytic function $f(z)$ of the complex variable $z$ for which $\abs{f(z)}$ increases without bound when $z$ approaches $a$: $\lim_{z\rightarrow a} f(z) = \infty$. In a sufficiently small punctured neighbourhood $V=\set{z\in\C : 0 < \abs{z-a} < R}$ of the point $a \neq \infty$, or $V'=\set{z\in\C : r < \abs{z} < \infty}$ in the case of the point at infinity $a=\infty$, the function $f(z)$ can be written as a Laurent series of special form: \begin{equation} \label{eq1} f(z) = \sum_{k=-m}^\infty c_k (z-a)^k,\quad \text{'"`UNIQ-MathJax15-QINU`"', '"`UNIQ-MathJax16-QINU`"', '"`UNIQ-MathJax17-QINU`"'}, \end{equation} or, respectively, \begin{equation} \label{eq2} f(z) = \sum_{k=-m}^\infty \frac{c_k}{z^k},\quad \text{'"`UNIQ-MathJax18-QINU`"', '"`UNIQ-MathJax19-QINU`"', '"`UNIQ-MathJax20-QINU`"'}, \end{equation} with finitely many negative exponents if $a\neq\infty$, or, respectively, finitely many positive exponents if $a=\infty$. The natural number $m$ in these expressions is called the order, or multiplicity, of the pole $a$; when $m=1$ the pole is called simple. The expressions \ref{eq1} and \ref{eq2} show that the function $p(z)=(z-a)^mf (z)$ if $a\neq\infty$, or $p(z)=z^{-m}f(z)$ if $a=\infty$, can be analytically continued to a full neighbourhood of the pole $a$, and, moreover, $p(a) \neq 0$. Alternatively, a pole $a$ of order $m$ can also be characterized by the fact that the function $1/f(z)$ has a zero of multiplicity $m$ at $a$.
A point $a=(a_1,\ldots,a_n)$ of the complex space $\C^n$, $n\geq2$, is called a pole of the analytic function $f(z)$ of several complex variables $z=(z_1,\ldots,z_n)$ if the following conditions are satisfied: 1) $f(z)$ is holomorphic everywhere in some neighbourhood $U$ of $a$ except at a set $P \subset U$, $a \in P$; 2) $f(z)$ cannot be analytically continued to any point of $P$; and 3) there exists a function $q(z) \not\equiv 0$, holomorphic in $U$, such that the function $p(z) = q(z)f(z)$, which is holomorphic in $U \setminus P$, can be holomorphically continued to the full neighbourhood $U$, and, moreover, $p(a) \neq 0$. Here also $$ \lim_{z\rightarrow a}f(z) = \lim_{z\rightarrow a}\frac{p(z)}{q(z)} = \infty; $$ however, for $n \geq 2$, poles, as with singular points in general, cannot be isolated.
References
[Sh] B.V. Shabat, "Introduction of complex analysis", 2, Moscow (1976) (In Russian) Comments References
[Ah] L.V. Ahlfors, "Complex analysis", McGraw-Hill (1979) pp. Chapt. 8 [GrFr] H. Grauert, K. Fritzsche, "Several complex variables", Springer (1976) (Translated from German) [Ra] R.M. Range, "Holomorphic functions and integral representation in several complex variables", Springer (1986) pp. Chapt. 1, Sect. 3 How to Cite This Entry:
Pole (of a function).
Encyclopedia of Mathematics.URL: http://www.encyclopediaofmath.org/index.php?title=Pole_(of_a_function)&oldid=25728
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In the Hamilton-Jacobi equation, we take the partial time derivative of the action. But the action comes from integrating the Lagrangian over time, so time seems to just be a dummy variable here and hence I do not understand how we can partial differentiate $S$ with respect to time? A simple example would also be helpful.
The action functional and Hamilton's principal function are two different mathematical objects related to the same physical quantity.
The action along a trajectory $\gamma:[t_1,t_2]\rightarrow Q$ is given by $$ S[\gamma] = \int_{t_1}^{t_2}L(\gamma(t'),\dot\gamma(t'),t')dt' $$ whereas the principal function is the solution of the Hamilton-Jacobi equation $$ H(q,\nabla S(q,t),t) + \frac{\partial S}{\partial t}(q,t) = 0 $$
If you denote by $\gamma_{q,t}$ the solution of the Euler-Lagrange equations with $$ \gamma_{q,t}(t_0)=q_0\\ \gamma_{q,t}(t)=q $$ then $$ S(q,t):=S[\gamma_{q,t}]=\int_{t_0}^{t}L(\gamma_{q,t}(t'),\dot\gamma_{q,t}(t'),t')dt' $$ will solve the Hamilton-Jacobi equation.
On the flip side, for the principal function we have the following $$ \frac{d}{dt}S(\gamma(t),t)=L(\gamma(t),\dot\gamma(t),t) $$ and thus $$ S[\gamma]=S(\gamma(t_2),t_2)-S(\gamma(t_1),t_1) $$
Note that the last two equations only hold for trajectories with $$ \frac{\partial L}{\partial\dot q}(\gamma(t),\dot\gamma(t),t) = \nabla S(\gamma(t),t) $$
Geometrically, the choice of integration constants of the principal function selects a leave of a foliation of phase space, which corresponds to the choice of initial condition $\gamma_q(t_0)=q_0$ from above.
I) At least three different quantities in physics are customary called an action and denoted with the letter $S$.
The (off-shell) action $$\tag{1}S[q]~:=~ \int_{t_i}^{t_f}\! dt \ L(q(t),\dot{q}(t),t) $$ is a
functionalof the full position curve/path $q^i:[t_i,t_f] \to \mathbb{R}$ for alltimes $t$ in the interval $[t_i,t_f]$. See also this question. (Here the words on-shelland off-shellrefer to whether the equations of motion (eom) are satisfied or not.)
If the variational problem $(1)$ with well-posed boundary conditions, e.g. Dirichlet boundary conditions $$\tag{2} q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_i,$$ has a unique extremal/classical path $q_{\rm cl}^i:[t_i,t_f] \to \mathbb{R}$, it makes sense to define an on-shell action $$ \tag{3} S(q_f;t_f;q_i,t_i) ~:=~ S[q_{\rm cl}], $$ which is a
functionof the boundary values. See e.g. MTW Section 21.1.
The Hamilton's principal function $S(q,\alpha, t)$ in Hamilton-Jacobi equation is a
functionof the position coordinates $q^i$, integration constants $\alpha_i$, and time $t$, see e.g. H. Goldstein, Classical Mechanics,chapter 10. The total time derivative $$\tag{4} \frac{dS}{dt}~=~ \dot{q}^i \frac{\partial S}{\partial q^i}+ \frac{\partial S}{\partial t}$$ is equal to the Lagrangian $L$ on-shell, as explained here. As a consequence, the Hamilton's principal function $S(q,\alpha, t)$ can be interpreted as an action on-shell.
II)
Example: A non-relativistic free particle in 1 dimension.
The off-shell action is $$\tag{5} S[q]~=~ \frac{m}{2}\int_{t_i}^{t_f}\! dt \ \dot{q}(t)^2. $$
If we assume Dirichlet boundary conditions (2), the unique classical trajectory $q_{\rm cl}$ has constant velocity $$\tag{6}\dot{q}_{\rm cl}~=~\frac{q_f-q_i}{t_f-t_i}.$$ The Dirichlet on-shell action (3) is $$ \tag{7} S(q_f,t_f;q_i,t_i) ~=~ \frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}. $$
The Hamilton's principal function, i.e. a solution to Hamilton-Jacobi equation, is $$\tag{8} S(q,E,t)~=~\pm\sqrt{2m E} q - Et, $$ where $E$ is an integration constant (=the total energy). Due to the interpretation of Hamilton's principal function as a type 2 generator of canonical transformations, the partial derivative $$\tag{9}Q~:=~ \frac{\partial S}{ \partial E}~\stackrel{(8)}{=}~\pm\sqrt{\frac{m}{2E}}q -t $$ must be a constant of motion. In other words, the position $q(t)$ is, as expected, an affine function of time $t$. This implies that the velocity is constant $$\tag{10} \dot{q} ~\approx~\pm\sqrt{\frac{2E}{m}}, $$ where the "$\approx$" symbol means equality modulo eom. The total time derivative of the Hamilton's principal function (8) is equal to the Lagrangian (=the kinetic energy) on-shell $$\tag{11} \frac{dS}{dt}~\stackrel{(8)}{=}~ \pm\sqrt{2m E} \dot{q} -E ~\stackrel{(10)}{\approx}~E.$$
Let us now compare point 2 and 3. With the Dirichlet boundary conditions (2), the energy becomes $$ \tag{12} E~=~ \frac{m}{2} \cdot \left(\frac{q_f-q_i}{t_f-t_i}\right)^2. $$ A comparison of eqs. (6) and (10) shows that we should use the plus (minus) branch of the solution (8) if $q_f\geq q_i$ ($q_f\leq q_i$), respectively. It is straightforward to check that the difference in the Hamilton's principal function becomes the on-shell action (7), $$ \tag{13} S(q_f,E,t_f)-S(q_i,E,t_i)~\stackrel{(8)+(12)}{=}~\frac{m}{2} \cdot \frac{(q_f-q_i)^2}{t_f-t_i}~\stackrel{(7)}{=}~ S(q_f,t_f;q_i,t_i).\qquad $$
I think the other two answers are overkill. The simpler answer is that time $t$ is not a dummy variable. The integration of $L$ over time here is an indefinite integration, so if we have $L(q,\dot{q},t)$, and we want to integrate it over time $t$, the result is $$\int_{t_i}^tL(q,\dot{q},\tau)d\tau$$ here $\tau$ is the dummy variable but $t$ is not, and the result is a function of $t$.
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Every geometry textbook has formulas for the circumference (\(C = 2 \pi r\)) and area (\(A = \pi r^2\)) of a circle. But where do these come from? How can we prove them?
Well, the first is more a definition than a theorem: the number \(\pi\) is usually
defined as the ratio of a circle’s circumference to its diameter: \(\pi = C/(2r)\). Armed with this, we can compute the area of a circle. Archimedes’ idea (in 260 BCE) was to approximate this area by looking at regular \(n\)-sided polygons drawn inside and outside the circle, as in the diagram below. Increasing \(n\) gives better and better approximations to the area.
Look first at the inner polygon. Its perimeter is slightly less than the circle’s circumference, \(C = 2 \pi r\), and the height of each triangle is slightly less than \(r\). So when reassembled as shown, the triangles form a rectangle whose area is just under \(C/2\cdot r = \pi r^2\). Likewise, the outer polygon has area just larger than \(\pi r^2\). As \(n\) gets larger, these two bounds get closer and closer to \(\pi r^2\), which is therefore the circle’s area.
Archimedes used this same idea to approximate the number \(\pi\). Not only was he working by hand, but the notion of “square root” was not yet understood well enough to compute with. Nevertheless, he was amazingly able to use 96-sided polygons to approximate the circle! His computation included impressive dexterity with fractions: for example, instead of being able to use \(\sqrt{3}\) directly, he had to use the (very close!) approximation \(\sqrt{3} > 265/153\). In the end, he obtained the bounds \( 3\frac{10}{71} < \pi < 3\frac{1}{7} \), which are accurate to within 0.0013, or about .04%. (In fact, he proved the slightly stronger but uglier bounds \(3\frac{1137}{8069} < \pi < 3\frac{1335}{9347}\). See this translation and exposition for more information on Archimedes’ methods.)
These ideas can be pushed further. Focus on a circle with radius 1. The area of the regular \(n\)-sided polygon inscribed in this circle can be used as an approximation for the circle’s area, namely \(\pi\). This polygon has area \(A_n = n/2 \cdot \sin(360/n)\) (prove this!). What happens when we double the number of sides? The approximation changes by a factor of $$\frac{A_{2n}}{A_n} = \frac{2\sin(180/n)}{\sin(360/n)} = \frac{1}{\cos(180/n)}.$$ Starting from \(A_4 = 2\), we can use the above formula to compute \(A_8,A_{16},A_{32},\ldots\), and in the limit we find that $$\pi = \frac{2}{\cos(180/4)\cdot\cos(180/8)\cdot\cos(180/16)\cdots}.$$ Finally, recalling that \(\cos(180/4) = \cos(45) = \sqrt{\frac{1}{2}}\) and \(\cos(\theta/2) = \sqrt{\frac{1}{2}(1+\cos\theta)}\) (whenever \(\cos(\theta/2) \ge 0\)), we can rearrange this into the fun infinite product $$\frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}}}} \cdots$$ (which I found at Mathworld). (It’s ironic that this formula for a
circle uses so many square roots!)
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Assume the amount of evidence against a defendant in a criminal trial is an exponential random variable $X$. If the defendant is innocent, then $X$ has mean $1$, and if the defendant is guilty, then $X$ has mean $2$. The defendant will be ruled guilty if $X>c$, where $c$ is a suitably chosen constant.
If the judge wants to be $95\%$ certain that an innocent man will not be convicted, what should the value of $c$ be? For this $c$ value, what is the probability that a guilty defendant will be convicted? Assume before the trial begins, you believe the defendant to be guilty with probability $10\%$. If the defendant is convicted, what is your updated belief about the probability of their guilt?
For this question, I am not sure as to where to begin for finding the initial $c$. I know that $\lambda_{\text{innocent}}=1$ and $\lambda_{\text{guilty}}=0.5$. my thought was to compute $0.95=\int(\lambda e^{-\lambda x})dx$ from $0$ to $c$ and solving for $c$ in both the guilty and innocent cases. I obtained negative values for both, which I assume are wrong. Am I on the right track or is there another approach I am not seeing?
After $c$ is known, finding the probability the guilty defendant will be convicted should be the straight forward integral for an exponential distribution from $0$ to $c$?
If the defendant is convicted, would be probability of their guilty be increased to $100\%$? or $0.1\times0.95$?
Thank you in advance for any help in understanding the problem!
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Typesetting and Converting Mathematics¶
There are two main uses for MathJax:
Typesetting all the mathematics within a web page, and Converting a string containing mathematics into another form.
In version 2, MathJax could perform the first function very well, but it was much harder to do the second. MathJax version 3 makes both easy to do. Both these tasks are described below.
Typesetting Math in a Web Page¶
MathJax makes it easy to typeset all the math in a web page, and in fact it will do this automatically when it is first loaded unless you configure it not to. So this is one of the easiest actions to perform in MathJax; if your page is static, there is nothing to do but load MathJax.
If your page is dynamic, and you may be adding math after the page isloaded, then you will need to tell MathJax to typeset the mathematicsonce it has been inserted into the page. There are two methods fordoing that:
MathJax.typeset() and
MathJax.typesetPromise().
The first of these,
MathJax.typeset(), typesets the page, anddoes so immediately and synchronously, so when the call finishes, thepage will have been typeset. Note, however, that if the math includesactions that require additional files to be loaded (e.g., TeX inputthat uses require, or that includes autoloaded extensions), thenan error will be thrown. You can use the
try/catch command totrap this condition.
The second,
Mathjax.typesetPromise(), performs the typesettingasynchronously, and returns a promise that is resolved when thetypesetting is complete. This properly handles loading of externalfiles, so if you are expecting to process TeX input that can includerequire or autoloaded extensions, you should use this form oftypesetting. It can be used with
await as part of a larger
async function.
Both functions take an optional argument, which is an array of elements whose content should be processed. An element can be either an actual DOM element, or a CSS selector string for an element or collection of elements. Supplying an array of elements will restrict the typesetting to the contents of those elements only.
Handling Asynchronous Typesetting¶
It is generally a bad idea to try to perform multiple asynchronoustypesetting calls simultaneously, so if you are using
MathJax.typesetPromise() to make several typeset calls, youshould chain them using the promises they return. For example:
MathJax.typesetPromise().then(() => { // modify the DOM here MathJax.typesetPromise();}).catch((err) => console.log(err.message));
This approach can get complicated fast, however, so you may want to maintain a promise that can be used to chain the later typesetting calls. For example,
let promise = Promise.resolve(); // Used to hold chain of typesetting callsfunction typeset(code) { promise = promise.then(() => {code(); return MathJax.typesetPromise()}) .catch((err) => console.log('Typeset failed: ' + err.message)); return promise;}
Then you can use
typeset() to run code that changes the DOMand typesets the result. The
code() that you pass it does theDOM modifications and returns the array of elements to typeset, or
null to typeset the whole page. E.g.,
typeset(() => { const math = document.querySelector('#math'); math.innerHTML = '$$\\frac{a}{1-a^2}$$'; return math;});
would replace the contents of the element with
id="math" with thespecified fraction and have MathJax typeset it (asynchronously).Because the
then() call returns the result of
MathJax.typesetPromise(), which is itself a promise, the
then() will not resolve until that promise is resolved; i.e.,not until the typesetting is complete. Finally, since the
typeset() function returns the
promise, you can use
await in an
async function to wait for the typesetting tocomplete:
await typeset(...);
Note that this doesn’t take the initial typesetting that MathJaxperforms into account, so you might want to use
MathJax.startup.promise in place of
promise above.I.e., simply use
function typeset(code) { MathJax.startup.promise = MathJax.startup.promise .then(() => {code(); return MathJax.typesetPromise()}) .catch((err) => console.log('Typeset failed: ' + err.message)); return MathJax.startup.promise;}
This avoids the need for the global
promise variable, andmakes sure that your typesetting doesn’t occur until the initialtypesetting is complete.
Resetting Automatic Equation Numbering¶
The TeX input jax allows you to automatically number equations. When modifying a page, this can lead to problems as numbered equations may be removed and added; most commonly, duplicate labels lead to issues.
You can reset equation numbering using the command
MathJax.texReset([start])
where
start is the number at which to start equation numbering.
If you have inserted new content, that may require the entire page to be reprocessed in order to get the automatic numbering, labels, and references to be correct. In that case, you can do
MathJax.startup.document.state(0);MathJax.texReset();MathJax.typeset();
to force MathJax to reset the page to the state it was before MathJax processed it, reset the TeX automatic line numbering and labels, and then re-typeset the contents of the page from scratch.
Loading MathJax Only on Pages with Math¶
The MathJax combined configuration files are large, and so you maywish to include MathJax in your page only if it is necessary. If youare using a content-management system that puts headers and footersinto your pages automatically, you may not want to include MathJaxdirectly, unless most of your pages include math, as that would loadMathJax on
all your pages. Once MathJax has been loaded, it shouldbe in the browser’s cache and load quickly on subsequent pages, butthe first page a reader looks at will load more slowly. In order toavoid that, you can use a script like the following one that checks tosee if the content of the page seems to include math, and only loadsMathJax if it does. Note that this is not a very sophisticated test,and it may think there is math in some cases when there really isn’tbut it should reduce the number of pages on which MathJax will have tobe loaded.
Create a file called
check-for-tex.js containing the following:
(function () { var body = document.body.textContent; if (body.match(/(?:\$|\\\(|\\\[|\\begin\{.*?})/)) { if (!window.MathJax) { window.MathJax = { tex: { inlineMath: {'[+]': [['$', '$']]} } }; } var script = document.createElement('script'); script.src = 'https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js'; document.head.appendChild(script); }})();
and then use
<script src="check-for-tex.js" defer></script>
in order to load the script when the page content is ready. Notethat you will want to include the path to the location where youstored
check-mathjax.js, that you should change
tex-chtml.js to whatever component file you want to use, and thatthe
window.MathJax value should be set to whatever configurationyou want to use. In this case, it just adds dollar signs to thein-line math delimiters. Finally, adjust the
body.match() regularexpression to match whatever you are using for math delimiters.
This simply checks if there is something that looks like a TeX in-line or displayed math delimiter, and loads MathJax if there is. If you are using different delimiters, you will need to change the pattern to include those (and exclude any that you don’t use). If you are using AsciiMath instead of TeX, then change the pattern to look for the AsciiMath delimiters.
If you are using MathML, you may want to use
if (document.body.querySelector('math')) {...}
for the test instead (provided you aren’t using namespace prefixes, like <m:math>).
Converting a Math String to Other Formats¶
An important use case for MathJax is to convert a string containing mathematics (in one of the three forms that MathJax understands) and convert it into another form (either MathML, or one of the output formats that MathJax supports). This was difficult to do in MathJax version 2, but easy to do in version 3.
When MathJax startup up, it creates methods for converting from theinput format(s) to the output format(s) that you have loaded, and toMathML format. For example, if you have loaded the MathML input jaxand the SVG output jax (say by using the
mml-svg component), thenMathJax will create the following conversion methods for you:
MathJax.mathml2svg(math[,options])
MathJax.mathml2svgPromise(math[,options])
MathJax.mathml2mml(math[,options])
MathJax.mathml2mmlPromise(math[,options])
If you had loaded the TeX input jax as well, you would also get fourmore methods, with
tex in place of
mathml.
As the names imply, the
Promise functions perform the conversionasynchronously, and return promises, while the others operatesynchronously and return the converted form immediately. The firsttwo functions (and any others like them) produce DOM elements as theresults of the conversion, with the promise versions passing that totheir
then() functions as their argument (see the section onAsynchronous Conversion below), and the non-promise versions returningthem directly. You can insert these DOM elements into the documentdirectly, or you can use their
outerHTML property to obtaintheir serialized string form.
The functions that convert to MathML produce serialized MathML stringsautomatically, rather than DOM elements. (You can use the browser’s
DOMParser object to convert the string into a MathML DOM treeif you need one.)
Conversion Options¶
All four of these functions require an argument that is the mathstring to be converted (e.g., the serialized MathML string, or in thecase of
tex2chtml(), the TeX or LaTeX string). You can alsopass a second argument that is an object containing options thatcontrol the conversion process. The options that can be included are:
display, a boolean specifying whether the math is in display-mode or not (for TeX input). Default is
true.
em, a number giving the number of pixels in an
emfor the surrounding font. Default is
16.
ex, a number giving the number of pixels in an
exfor the surrounding font. Default is
8.
containerWidth, a number giving the width of the container, in pixels. Default is 80 times the
exvalue.
lineWidth', a number giving the line-breaking width in
emunits. Default is a very large number (100000), so effectively no line breaking.
scale, a number giving a scaling factor to apply to the resulting conversion. Default is 1.
For example,
let html = MathJax.tex2chtml('\\sqrt{x^2+1}', {em: 12, ex: 6, display: false});
would convert the TeX expression
\sqrt{x^2+1} to HTML as anin0line expression, with
em size being 12 pixels and
ex sizebeing 6 pixels. The result will be a DOM element containing the HTMLfor the expression. Similarly,
let html = MathJax.tex2chtml('\\sqrt{x^2+1}', {em: 12, ex: 6, display: false});let text = html.outerHTML;
sets
text to be the serialized HTML string for the expression.
Obtaining the Output Metrics¶
Since the
em,
ex, and
containerWidth alldepend on the location where the math will be placed in the document(they are values based on the surrounding text font and the containerelements width), MathJax provides a method for obtaining these valuesfrom a given DOM element. The method
MathJax.getMetricsFor(node, display)
takes a DOM element (
node) and a boolean (
display), indicatingif the math is in display mode or not, and returns an objectcontaining all six of the options listed above. You can pass thisobject directly to the conversion methods discussed above. So you cando something like
let node = document.querySelector('#math');let options = MathJax.getMetricsFor(node, true);let html = MathJax.tex2svg('\\sqrt{x^2+1}', options);node.appendChild(html);
in order to get get the correct metrics for the (eventual) location ofthe math that is being converted. Of course, it would be easier tosimply insert the TeX code into the page and use
MathJax.typeset() to typeset it, but this is just an exampleto show you how to obtain the metrics from a particular location inthe page.
Note that obtaining the metrics causes a page refresh, so it is expensive to do this. If you need to get the metrics from many different locations, there are more efficient ways, but these are advanced topics to be dealt with elsewhere.
Obtaining the Output Stylesheet¶
The output from the SVG and CommonHTML output jax both depend on CSS stylesheets in order to properly format their results. You can obtain the SVG stylesheet element by calling
MathJax.svgStylesheet();
and the HTML stylesheet from
MathJax.chtmlStylesheet();
The CommonHTML output jax CSS can be quite large, so the output jaxtries to minimize the stylesheet by including only the styles that areactually needed for the mathematics that has been processed by theoutput jax. That means you should request the stylesheet only
afteryou have typeset the mathematics itself.
Moreover, if you typeset several expressions, the stylesheet will include everything needed for all the expressions you have typeset. If you want to reset the stylesheet, then use
MathJax.startup.output.clearCache();
if the output jax is the CommonHTML output jax. So if you want toproduce the style sheet for a single expression, issue the
clearCache() command just before the
tex2chtml() call.
Asynchronous Conversion¶
If you are converting TeX or LaTeX that might use require to load extensions, or where extensions might be autoloaded, you will either need to use one of the “full” components that include all the extensions, or preload all the extensions you need if you plan to use the synchronous calls listed above. Otherwise, you can use the promise-based calls, which handle the loading of extensions transparently.
For example,
let node = document.querySelector('#math');let options = MathJax.getMetricsFor(node, true);MathJax.tex2chtmlPromise('\\require{bbox}\\bbox[red]{\\sqrt{x^2+1}}', options) .then((html) => { node.appendChild(html); let sheet = document.querySelector('#MJX-CHTML-styles'); if (sheet) sheet.parentNode.removeChild(sheet); document.head.appendChild(MathJax.chtmlStylesheet()); });
would get the metrics for the element with
id="math", convertthe TeX expression using those metrics (properly handling theasynchronous load needed for the
\require command); then when theexpression is typeset, it is added to the document and the CHTMLstylesheet is updated.
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I am new in the field of signal processing. What's the different between Hamming and Hanning window? When we use the former and latter?
Max is right that the difference between Hamming and Hann windows are small. (BTW, I am a proponent of the movement to totally do away with them term
"Hanning". There is no Dr. Hanning nor Mr./Ms. Hanning that the window is named after.)
The Hamming window is 92% Hann window and 8% rectangular window. Hamming found out that he was able to reduce the height of the maximum side lobe by doing that.
Another issue that I just thought of: The Hann window (and some others) is a
complementary window; that is that the latter half of the window of one frame adds to the first half of the following frame to 1. For analysis, the property of being complementary is usually not important, but for synthesis (or reconstruction), this property is salient.
So if you like symmetry about 0, the Hann window having even length $N$, is
$$ w\left( \tfrac{n}{N} \right) = \begin{cases} \tfrac12 + \tfrac12 \cos\left( 2 \pi \tfrac{n}{N} \right) \qquad & |n| \le \frac{N}{2} \\ 0 \qquad & |n| \ge \frac{N}{2} \\ \end{cases} $$
The property of being
complementary is:
$$ \sum\limits_{m=-\infty}^{+\infty} w\left( \tfrac{n-mH}{N} \right) = 1 $$
then
$$\begin{align} x[n] &= x[n] \times 1 \\ &= x[n] \times \sum\limits_{m=-\infty}^{+\infty} w\left( \tfrac{n-mH}{N} \right) \\ &= \sum\limits_{m=-\infty}^{+\infty} x[n] w\left( \tfrac{n-mH}{N} \right) \\ &= \sum\limits_{m=-\infty}^{+\infty} x_m[n-mH] \\ \end{align}$$
where
$$\begin{align} x_m[n-mH] & \triangleq x[n] w\left( \tfrac{n-mH}{N} \right) \\ x_m[n] & = x[n+mH] w\left( \tfrac{n}{N} \right) \\ \end{align}$$
So $x_m[n]$ is a finite-length frame of signal of length $N$ and represents the audio in the neighborhood of $x[n+mH]$ or $m$ hops from the beginning. $H = \frac{N}{2}$ is the frame hop displacement.
The Short-Time Fourier Transform (STFT) is: $$\begin{align} X_m[k] &= \sum\limits_{n=-\tfrac{N}{2}}^{\tfrac{N}{2}-1} x_m[n] e^{-j 2 \pi nk/N} \\ \\ &= \sum\limits_{n=-\tfrac{N}{2}}^{\tfrac{N}{2}-1} x[n+mH] w\left( \tfrac{n}{L} \right) e^{-j 2 \pi nk/N} \\ \\ &= \sum\limits_{n=0}^{N-1} \hat{x}_m[n] e^{-j 2 \pi nk/N} \\\end{align}$$
where
$$ \hat{x}_m[n] = \begin{cases} x_m[n] \qquad & 0 \le n < \tfrac{N}{2} \\ \\ x_m[n-N] \qquad & \tfrac{N}{2} \le n < N \\ \end{cases}$$
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