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If you take the square root of the perimeter of a triangle and double it, you get the area of the triangle. The sides have integer lengths which are mutually different from each other. What are the lengths of its sides?
Let the side lengths of the triangle be $a,b,c$, and let $P = a+b+c$.
By Heron's formula, $$A = \sqrt{\frac P2\left(\frac P2-a\right)\left(\frac P2-b\right)\left(\frac P2-c\right)}$$
From the given information,
$$A = 2\sqrt{P}$$
Equating the two and check that $P\ne 0$,
$$\begin{align*} \sqrt{\frac 12\left(\frac P2-a\right)\left(\frac P2-b\right)\left(\frac P2-c\right)} &= 2\\ \left(\frac P2-a\right)\left(\frac P2-b\right)\left(\frac P2-c\right) &= 8\\ (P-2a)(P-2b)(P-2c) &= 64\\ (b+c-a)(c+a-b)(a+b-c) &= 64 \end{align*}$$
$a, b,c$ are all positive integers and are all different, so $P-2a, P-2b, P-2c$ are different positive integral factors of $64$.
First, all decomposed factor triplet with a $1$ is rejected. (Why?)
Then the only decomposition with different integers is $2\times4\times8$. Without loss of generality, let
$$\begin{align*} P-2a &= 2\\ P-2b &= 4\\ P-2c &= 8\\ 3P - 2(a+b+c) &= 2+4+8\\ P &= 14\\ a &= 6\\ b &= 5\\ c &= 3 \end{align*}$$
So the triangle is a $6-5-3$ triangle.
If we label the sides of the triangle $a,b,c$ then we know that the perimeter of the triangle is $$P=a+b+c$$
We can use Heron's Formula to calculate the area of a triangle when we know all three side lengths:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$s=\frac{a+b+c}2$$
Can you continue from here?
Without the restriction that the integers are different, there are two other solutions: 2, 9, 9 and 4, 4, 4. |
First of all, the axiom of countable choice says that given a countable family of non-empty sets, you can choose from each set
simultaneously. If you want to choose from one, then from another, then from another, and so on you need a strictly stronger form of choice called Dependent Choice, abbreviated as $\sf DC$.
To your questions, the definition of the $\aleph$ numbers uses absolutely no choice, although the proof that any of them is regular does use the axiom of choice (well, except $\aleph_0$). So it is consistent, for example, that $\aleph_1$ is singular. But assuming $\sf DC$ will prevent that. Whether or not you can have every cardinal $\geq\aleph_2$ singular with $\sf ZF+DC$ is still open.
The definition of $\beth$ numbers, on the other hand, goes out the window. The axiom of choice is equivalent to saying that the power set of a well-ordered set is well-ordered. So if the axiom of choice fails, there will be some ordinal $\alpha$ whose power set cannot be well-ordered. This means that at some point $\beth$ cardinals will not be $\aleph$ numbers. You can still talk about $\beth$ numbers, of course, as iterated powers of $\omega$, but that gives you significantly less information in that sense.
As for the continuum hypothesis, as noted by others it is still unprovable. But now you even get different forms of the continuum hypothesis. $2^{\aleph_0}=\aleph_1$ is no longer equivalent to "Every uncountable set of reals is equipollent with the reals themselves". Indeed, even without $\sf DC$, it is consistent that $2^{\aleph_0}$ and $\aleph_1$ are incomparable, and every uncountable set of reals has cardinality $2^{\aleph_0}$.
Finally, for large cardinals, most properties which are reserved for very large cardinals can be made compatible with $\aleph_1$ and $\sf ZF+DC$. For a more complete survey, look at What sort of large cardinal can $\aleph_1$ be without the axiom of choice?. |
While trying to solve the exercise below, I came up with a wrong conclusion, but I can't see why it's wrong. Also I'm accepting suggestions to get the right solution. This is the problem 17 from chapter 10 of Rudin's Functional Analysis.
Let $A$ be a Banach algebra. Suppose that the spectrum of $x\in A$ is not connected. Prove that $A$ contains a nontrivial idempotent $z$.
My attempt: Let $F_1, F_2$ be two disjoint closed non-empty sets in $\sigma(x)$ such that $F_1\cup F_2 = \sigma(x)$. There is a function $f$ defined over a neighborhood $\Omega$ of $\sigma(x)$ such that $f=1$ in $F_1$ and $f=0$ in $F_2$. Denote $$\tilde{f}(x) = \frac{1}{2\pi i}\int_\Gamma f(\lambda)(\lambda e-x)^{-1}\ d\lambda,$$which comes from the functional (or symbolic) calculus. $\Gamma$ is a contour of $\sigma(x)$ in $\Omega$.
The idea is to show that $\tilde{f}(x)$ is idempotent. This idea of proof was used in some books and I'm trying to follow it. My problem is this: it's clear that $f(\sigma(x)) = F_1$ from the very definition of $f$. I also know that $\sigma(\tilde{f}(x)) = f(\sigma(x)) = F_1$. But from this post, for instance, we have that the spcetrum of idempotents elements is $\{0,1\}$. If $\tilde{f}(x)$ would be idempotent, then $F_1=\{0,1\}$, but this is not necessarily the case.
Thank you for your help. |
In this scenario, the "joint dynamics" are trivially computed since the option value is a known deterministic function of stock price. For example, the mean of the option value for time $\tau$ is$$\mu_O = \int_0^\infty BSM( S_\tau ) p(S_\tau) dS_\tau$$which is best computed using quadrature as available in standard numerical libraries like scipy. The ...
I think an extremely interesting strand of research on this topic is represented by extensions of vine copulas with time-varying parameters.For vine copulas in general have a look at this site from the Technische Universität München:Vine Copula ModelsOne of their research projects, which is the most relevant in this context, is:Time varying vine copula ...
You can use copulas. The probability that B rises given A rises is $P(- R_B < 0 | - R_A < 0) = \frac{P(-R_B < 0, - R_A < 0)}{P(-R_A < 0)} = \frac{C(F_{-B}(0),F_{-A}(0))}{F_{-A}(0)}$.You can specify the marginals as a GARCH process and use either non parametric or parametric copulas to get your final conditional probability.
I have written R code for some time-varying bivariate fat-tailed copula functions (ripped off Patton's Matlab code) and played around with various optimizers.You can then use Rsolnp, nloptr, alabama or DEoptim packages to find an optimisation solution. Here is some R code where I play around with different optimisation algorithms. Note that the data2.csv ...
Once we start building time-varying copulas like Lopes suggests in that paper, I think we are better off venturing into the world of state space models. When viewed in a bayesian context, the similarities between the approaches are striking to me. The advantage of the copula, as I understand it, is that it is a quick and dirty way to understand the ...
if you agree that the marginal probability $P(u\le Y\le v)=F_Y(v)-F_Y(u)$, then your formula follows immediately, because next you simply plug the marginals into the copula.your 3rd equation for the joint probabilities is incorrect for $P(Z\le z,u\le Y\le v)$, I'm not sure where you got it from
Confidence intervals are applied to estimates to give a sense for potential error. If a, b, and c are R.V. as you described, they're independent by nature, making confidence intervals irrelevant. Either you're misunderstanding what was asked of you or your problem is misspecified (or you just misspecified it here).
A few initial observations but the quick answer is your Option 2.(1) Assuming both prepayment and default can occur only at discrete time-points is not strictly correct since a borrower has the right to payoff the loan in full at any time between payment dates.(2) Default is a nebulous concept - are you referring to the act of missing a payment itself, or ...
Suppose you would like to compute\begin{align}Q_1(x_1,x_2;B) &= \Bbb{E}[X_1\max(B-X_2,0)]\\Q_2(x_1,x_2;B) &= \Bbb{E}[X_2\max(X_1-B,0)]\end{align}where you know the marginal probability density functions $p_{X_1}(u)$ and $p_{X_2}(v)$.Let's start by focusing on $Q_1$. By definition, the expectation equivalently writes:$$ Q_1(x_1,x_2;B) = \... |
I understand how to derive the black scholes solution if $dS_t$ = $\mu S_tdt$ + $\sigma S_tdW_t$ and r is constant. The solution is c(t, x) = $xN(d_{+}(T - t), x))$ - K$e^{-r(T - t)}N(d\_(T - t), x))$ where $d_{+}(\tau, x)$ = $\frac{1}{\sigma\sqrt{\tau}}$ * $[log\frac{x}{K} + (r + \frac{1}{2}\sigma^2)\tau]$, $d\_(\tau, x) = d_{+}(\tau, x) - \sigma \sqrt{\tau}$
However, I need to find the solution when, $dS_t = \mu_{t}S_tdt + \sigma_{t}S_tdW_t$ and $r_t$ are deterministic functions of t. I was asked to guess the solution, so it must be a very close analogue to the solution above. I thought about integrating over time, but I haven't been able to verify that this works, and I do need to verify the solution.
Any help in figuring out what the form and how to go about verifying that it is a solution would be appreciated.
Update: Someone asked to see some extra work, here is my guess of what the solution should be: c(t, x) = $xN(d_{+}(T - t), x))$ - $Ke^{-\int_0^{T - t}r_udu}$N(d_(T - t), x)) where $d_{+}(\tau, x) = \frac{1}{\int_0^\tau \sigma_udu}$ * $[log\frac{x}{K} + \int_0^\tau (r + \frac{1}{2}\sigma^2)]$, $d\_(\tau, x) = d_{+}(\tau, x) - \int_0^\sqrt{\tau} \sigma_udu$.
I don't know if this guess is even correct, and if it is I need to verify that it is a solution the Black-Scholes PDE. |
I mean the basic principle is,
you're right, this setup is described by the math of classical electrical circuits as one which should drive an infinite current, and if you only have one charge to move then an infinite current necessitates it moving with infinite speed. If you do not give it a way to lose energy then it will continue to gain energy indefinitely.
Kirchhoff's voltage law can still be applied to such a condition but it is somewhat less informative. The voltage law comes from the fact that there is a scalar potential field $\varphi$ defined over space, it is just one field, it has only one value and no other. It is something like wondering why I cannot gain infinite energy by sledding down a hill: well, whatever speed you are getting on the way down, I can prove that you’re in for an equal-and-opposite trek up. So if this field is well-defined then Kirchhoff’s voltage law holds good.
This field $\varphi$ in turn comes from the Maxwell equations, which in my favorite units look like,$$\begin{align}\nabla \cdot E &= c\rho & \nabla\times E &= -\dot B\\\nabla \cdot B &= 0 & \nabla \times B &= J + \dot E\end{align}$$where $\dot X = \partial X/\partial w = c^{-1} \partial X/\partial t.$
First, the bottom-left one is used to rewrite $B = \nabla \times A$ for some “vector potential” $A$ and then the top-right one says $$\nabla \times (E + \dot A) = 0$$implying that there exists a scalar potential function $\varphi$ such that $$E + \dot A = -\nabla \varphi.$$So this is a mathematical theorem,
this scalar function $\varphi$ always exists. You can really say that there is a voltage drop of 6 volts from one side of the terminal to the other, and that is fine. Kirchhoff’s voltage law holds everywhere.
But notice what is not said: the connection to
a particle’s kinetic energy depends on work which depends on this electric field $E$ and its behavior over closed loops. And $\nabla\varphi$ vanishes over closed loops, but there is no reason that $\dot A$ must, and so a particle can gain/lose energy in a closed loop, if the vector potential is increasing/decreasing in time.
So, normally we can just step back from a circuit and assume that whatever the vector potential $A$ is, the circuit comes to some equilibrium value where it is not changing and $\dot A = 0$. Then $E = -\nabla \phi$ and we know that the thing must have lost all of its energy by the time that it comes back around the loop. That is how we know that the electron has lost its energy.
But, in the case that you are describing, we know that the Liénard–Wiechert potentials $\phi, A$ for a moving charge contain an $A$ which runs parallel to the charge, in rough proportion to its speed. As the charge is going faster and faster, this $A$ is increasing and increasing around the loop, and this factor of $\dot A$ is non-negligible.
Kirchhoff’s rule is not technically wrong, what is wrong is the assumption that energy
only changes from a change in potential, that these go hand-in-hand. The Maxwell equations do not say this, they say instead that there is this other thing, this vector-potential $A$, which must change in that scenario. |
In Blake 1986 'Mechanics of Flow-Induced Sound and Vibration V1: General Concepts', pg. 148 the author provides a helpful equation for the pressure radiated by organ pipes when the cavity is small compared to a wavelength, i.e., $\lambda_0 \gg$ cavity dimensions. (
Note this is the usual approximation for organ pipes, which are considered 'acoustically small'.)
$$\frac{\lvert p_{\ radiated}\rvert}{\lvert p_{\ cavity}\rvert} = \frac{\omega A_0}{c_0 \ r}$$
where $\omega$ is the resonant frequency of the pipe (
cavity),
$A_0$ is the area of the pipe's opening,
$r$ is the radius of the pipe, and
$c_0$ is the speed of sound in the pipe.
Given the same pipe and flow rate, $A_0$ and $r$ drop out of the equation.
The speed of sound in helium is obviously different than air, and so $c_0$ decreases. For a pipe of length $L$, the resonant frequency $\omega = \frac{c_0}{2L}$, so the decrease in sound speed found in the denominator is canceled by a matching decrease in the numerator. We are left with a factor of $2L$ which also will not change.
This suggests that the pressure radiated by the organ pipe should not change with the density of the gas. |
Suppose $Y_1,\dots,Y_n\mid\mu,\sigma^2 \sim \text{ iid } N(\mu,\sigma^2)$ and suppose the priors $\mu \mid \sigma^2 \sim N(\mu_0, \sigma^2 / \kappa_0)$ and $1/\sigma^2 \sim \text{gamma}(\nu_0/2, \nu_0 \sigma_0^2 / 2)$ are placed on the unknown parameters. Then \begin{align*} p(\sigma^2 \mid y_1,\dots,y_n) &\propto p(\sigma^2)p(y_1,\dots,y_n\mid\sigma^2) \\ &= p(\sigma^2) \int p(y_1,\dots,y_n \mid \mu,\sigma^2)p(\mu\mid\sigma^2)d\mu \end{align*} It turns out (see Hoff) with these priors that the marginal posterior distribution on $\sigma^2$ is $$1/\sigma^2\mid y_1,\dots,y_n \sim \text{gamma}(\nu_n/2,\nu_n\sigma_n^2/2)$$ where $$\nu_n = \nu_0 + n, \text{ and } \sigma_n^2 = \frac{1}{\nu_n}\left( \nu_0\sigma_0^2 + (n-1)s^2 + \frac{\kappa_0 n}{\kappa_0 + n} ( \bar{y}-\mu_0 )^2 \right)$$ where $\bar{y}$ and $s^2$ are the sample mean and unbiased sample variance. Thus we have that $$ \text{Var}\left[\frac{1}{\sigma^2} \mid y_1,\dots,y_n\right] = (\nu_n/2) / (\nu_n\sigma_n^2/2)^2 = \frac{2}{\nu_n\sigma_n^4}. $$
Now suppose we fix all hyperparameters except $\kappa_0$ and analyze the effect of increasing the variance of $\mu\mid\sigma^2$ (by decreasing $\kappa_0$). Since $f(\kappa_0)=\kappa_0 n / (\kappa_0 + n)$ is monotonically increasing as a function of $\kappa_0$, we see that \begin{align*} \kappa_{0,a} < \kappa_{0,b} &\Rightarrow \kappa_{0,a} n / (\kappa_{0,a} + n) < \kappa_{0,b} n / (\kappa_{0,b} + n) \\ &\Rightarrow \sigma_{n,a}^2 < \sigma_{n,b}^2 \\ &\Rightarrow \sigma_{n,a}^4 < \sigma_{n,b}^4 \\ &\Rightarrow \frac{2}{\nu_n\sigma_{n,b}^4} < \frac{2}{\nu_n\sigma_{n,a}^4}, \end{align*} i.e., increasing the variance of $\mu\mid\sigma^2$ results in an increase in the marginal posterior variance $\text{Var}\left[1/\sigma^2\mid y_1,\dots,y_n\right]$.
Intuitively, this result makes a lot of sense because if $p(\mu \mid \sigma^2)$ is more spread out, the averaging (integral) over $p(\mu \mid \sigma^2)$ above makes $p(\sigma^2 \mid y_1,\dots,y_n)$ more spread out.
This makes me think that this result should be able to be proved in general, i.e., given a sampling model $f_{Y\mid\theta,\phi}$ depending on two unknown parameters $\theta$ and $\phi$, and priors $f_{\theta\mid\phi}$ and $f_{\phi}$, an increase in $\text{Var}\left[ \theta\mid\phi\right]$ should result in an increase in $\text{Var}\left[ \phi \mid y_1,\dots,y_n \right]$. Does anyone know of a general result like this? |
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
That seems like what I need to do, but I don't know how to actually implement it... how wide of a time window is needed for the Y_{t+\tau}? And how on earth do I load all that data at once without it taking forever?
And is there a better or other way to see if shear strain does cause temperature increase, potentially delayed in time
Link to the question: Learning roadmap for picking up enough mathematical know-how in order to model "shape", "form" and "material properties"?Alternatively, where could I go in order to have such a question answered?
@tpg2114 For reducing data point for calculating time correlation, you can run two exactly the simulation in parallel separated by the time lag dt. Then there is no need to store all snapshot and spatial points.
@DavidZ I wasn't trying to justify it's existence here, just merely pointing out that because there were some numerics questions posted here, some people might think it okay to post more. I still think marking it as a duplicate is a good idea, then probably an historical lock on the others (maybe with a warning that questions like these belong on Comp Sci?)
The x axis is the index in the array -- so I have 200 time series
Each one is equally spaced, 1e-9 seconds apart
The black line is \frac{d T}{d t} and doesn't have an axis -- I don't care what the values are
The solid blue line is the abs(shear strain) and is valued on the right axis
The dashed blue line is the result from scipy.signal.correlate
And is valued on the left axis
So what I don't understand: 1) Why is the correlation value negative when they look pretty positively correlated to me? 2) Why is the result from the correlation function 400 time steps long? 3) How do I find the lead/lag between the signals? Wikipedia says the argmin or argmax of the result will tell me that, but I don't know how
In signal processing, cross-correlation is a measure of similarity of two waveforms as a function of a time-lag applied to one of them. This is also known as a sliding dot product or sliding inner-product. It is commonly used for searching a long signal for a shorter, known feature. It has applications in pattern recognition, single particle analysis, electron tomographic averaging, cryptanalysis, and neurophysiology.For continuous functions f and g, the cross-correlation is defined as:: (f \star g)(t)\ \stackrel{\mathrm{def}}{=} \int_{-\infty}^{\infty} f^*(\tau)\ g(\tau+t)\,d\tau,whe...
Because I don't know how the result is indexed in time
Related:Why don't we just ban homework altogether?Banning homework: vote and documentationWe're having some more recent discussions on the homework tag. A month ago, there was a flurry of activity involving a tightening up of the policy. Unfortunately, I was really busy after th...
So, things we need to decide (but not necessarily today): (1) do we implement John Rennie's suggestion of having the mods not close homework questions for a month (2) do we reword the homework policy, and how (3) do we get rid of the tag
I think (1) would be a decent option if we had >5 3k+ voters online at any one time to do the small-time moderating. Between the HW being posted and (finally) being closed, there's usually some <1k poster who answers the question
It'd be better if we could do it quick enough that no answers get posted until the question is clarified to satisfy the current HW policy
For the SHO, our teacher told us to scale$$p\rightarrow \sqrt{m\omega\hbar} ~p$$$$x\rightarrow \sqrt{\frac{\hbar}{m\omega}}~x$$And then define the following$$K_1=\frac 14 (p^2-q^2)$$$$K_2=\frac 14 (pq+qp)$$$$J_3=\frac{H}{2\hbar\omega}=\frac 14(p^2+q^2)$$The first part is to show that$$Q \...
Okay. I guess we'll have to see what people say but my guess is the unclear part is what constitutes homework itself. We've had discussions where some people equate it to the level of the question and not the content, or where "where is my mistake in the math" is okay if it's advanced topics but not for mechanics
Part of my motivation for wanting to write a revised homework policy is to make explicit that any question asking "Where did I go wrong?" or "Is this the right equation to use?" (without further clarification) or "Any feedback would be appreciated" is not okay
@jinawee oh, that I don't think will happen.
In any case that would be an indication that homework is a meta tag, i.e. a tag that we shouldn't have.
So anyway, I think suggestions for things that need to be clarified -- what is homework and what is "conceptual." Ie. is it conceptual to be stuck when deriving the distribution of microstates cause somebody doesn't know what Stirling's Approximation is
Some have argued that is on topic even though there's nothing really physical about it just because it's 'graduate level'
Others would argue it's not on topic because it's not conceptual
How can one prove that$$ \operatorname{Tr} \log \cal{A} =\int_{\epsilon}^\infty \frac{\mathrm{d}s}{s} \operatorname{Tr}e^{-s \mathcal{A}},$$for a sufficiently well-behaved operator $\cal{A}?$How (mathematically) rigorous is the expression?I'm looking at the $d=2$ Euclidean case, as discuss...
I've noticed that there is a remarkable difference between me in a selfie and me in the mirror. Left-right reversal might be part of it, but I wonder what is the r-e-a-l reason. Too bad the question got closed.
And what about selfies in the mirror? (I didn't try yet.)
@KyleKanos @jinawee @DavidZ @tpg2114 So my take is that we should probably do the "mods only 5th vote"-- I've already been doing that for a while, except for that occasional time when I just wipe the queue clean.
Additionally, what we can do instead is go through the closed questions and delete the homework ones as quickly as possible, as mods.
Or maybe that can be a second step.
If we can reduce visibility of HW, then the tag becomes less of a bone of contention
@jinawee I think if someone asks, "How do I do Jackson 11.26," it certainly should be marked as homework. But if someone asks, say, "How is source theory different from qft?" it certainly shouldn't be marked as Homework
@Dilaton because that's talking about the tag. And like I said, everyone has a different meaning for the tag, so we'll have to phase it out. There's no need for it if we are able to swiftly handle the main page closeable homework clutter.
@Dilaton also, have a look at the topvoted answers on both.
Afternoon folks. I tend to ask questions about perturbation methods and asymptotic expansions that arise in my work over on Math.SE, but most of those folks aren't too interested in these kinds of approximate questions. Would posts like this be on topic at Physics.SE? (my initial feeling is no because its really a math question, but I figured I'd ask anyway)
@DavidZ Ya I figured as much. Thanks for the typo catch. Do you know of any other place for questions like this? I spend a lot of time at math.SE and they're really mostly interested in either high-level pure math or recreational math (limits, series, integrals, etc). There doesn't seem to be a good place for the approximate and applied techniques I tend to rely on.
hm... I guess you could check at Computational Science. I wouldn't necessarily expect it to be on topic there either, since that's mostly numerical methods and stuff about scientific software, but it's worth looking into at least.
Or... to be honest, if you were to rephrase your question in a way that makes clear how it's about physics, it might actually be okay on this site. There's a fine line between math and theoretical physics sometimes.
MO is for research-level mathematics, not "how do I compute X"
user54412
@KevinDriscoll You could maybe reword to push that question in the direction of another site, but imo as worded it falls squarely in the domain of math.SE - it's just a shame they don't give that kind of question as much attention as, say, explaining why 7 is the only prime followed by a cube
@ChrisWhite As I understand it, KITP wants big names in the field who will promote crazy ideas with the intent of getting someone else to develop their idea into a reasonable solution (c.f., Hawking's recent paper) |
I was reading some stuff earlier today, and I wasn't sure how they changed the exponentials to trigs in this expression:
$$C_1x^{-1/4}\exp\left(\frac{2}{3}ix^{3/2}\right)+C_2x^{-1/4}\exp\left(-\frac{2}{3}ix^{3/2}\right)$$
$=$
$$A_1x^{-1/4}\sin\left(\frac{2}{3}x^{3/2}\right)+A_2x^{-1/4}\cos\left(\frac{2}{3}x^{3/2}\right)$$
Does it have something to do with euler's formula?
And can I change $$C_1x^{-1/2}\exp\left(\frac{i}{2x^2}\right)+C_2x^{-1/2}\exp\left(-\frac{i}{2x^2}\right)$$ to $$A_1x^{-1/2}\sin\left(\frac{1}{2x^2}\right)+A_2x^{-1/2}\cos\left(\frac{1}{2x^2}\right)?$$ |
A note on a superlinear and periodic elliptic system in the whole space
1.
Department of Mathematics, Yunnan Normal University, Kunming 650092 Yunnanage, China
2.
Office of Adult Education, Simao Teacher's College, Simao 665000 Yunnan, China
3.
Department of Mathematics, Yunnan Normal University, Kunming 650092 Yunnan
$ -\Delta u+V(x)u=g(x,v)$ in $R^N,$
$ -\Delta v+V(x)v=f(x,u)$ in $R^N,$
$ u(x)\to 0$ and $v(x)\to 0$ as $|x|\to\infty,$
where the potential $V$ is periodic and has a positive bound from below, $f(x,t)$ and $g(x,t)$ are periodic in $x$ and superlinear but subcritical in $t$ at infinity. By using generalized Nehari manifold method, existence of a positive ground state solution as well as multiple solutions for odd $f$ and $g$ are obtained.
Mathematics Subject Classification:Primary: 35J50; Secondary: 35J5. Citation:Shuying He, Rumei Zhang, Fukun Zhao. A note on a superlinear and periodic elliptic system in the whole space. Communications on Pure & Applied Analysis, 2011, 10 (4) : 1149-1163. doi: 10.3934/cpaa.2011.10.1149
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C. O. Alves, P. C. Carrião and O. H. Miyagaki,
[2] [3] [4]
T. Bartsch and D. G. De Figueiredo,
[5] [6] [7] [8]
D. G. De Figueiredo and Y. Ding,
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L. Jeanjean,
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P. L. Lions,
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A. Pistoia and M. Ramos,
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It's hard to say just from the sheet music; not having an actual keyboard here. The first line seems difficult, I would guess that second and third are playable. But you would have to ask somebody more experienced.
Having a few experienced users here, do you think that limsup could be an useful tag? I think there are a few questions concerned with the properties of limsup and liminf. Usually they're tagged limit.
@Srivatsan it is unclear what is being asked... Is inner or outer measure of $E$ meant by $m\ast(E)$ (then the question whether it works for non-measurable $E$ has an obvious negative answer since $E$ is measurable if and only if $m^\ast(E) = m_\ast(E)$ assuming completeness, or the question doesn't make sense). If ordinary measure is meant by $m\ast(E)$ then the question doesn't make sense. Either way: the question is incomplete and not answerable in its current form.
A few questions where this tag would (in my opinion) make sense: http://math.stackexchange.com/questions/6168/definitions-for-limsup-and-liminf http://math.stackexchange.com/questions/8489/liminf-of-difference-of-two-sequences http://math.stackexchange.com/questions/60873/limit-supremum-limit-of-a-product http://math.stackexchange.com/questions/60229/limit-supremum-finite-limit-meaning http://math.stackexchange.com/questions/73508/an-exercise-on-liminf-and-limsup http://math.stackexchange.com/questions/85498/limit-of-sequence-of-sets-some-paradoxical-facts
I'm looking for the book "Symmetry Methods for Differential Equations: A Beginner's Guide" by Haydon. Is there some ebooks-site to which I hope my university has a subscription that has this book? ebooks.cambridge.org doesn't seem to have it.
Not sure about uniform continuity questions, but I think they should go under a different tag. I would expect most of "continuity" question be in general-topology and "uniform continuity" in real-analysis.
Here's a challenge for your Google skills... can you locate an online copy of: Walter Rudin, Lebesgue’s first theorem (in L. Nachbin (Ed.), Mathematical Analysis and Applications, Part B, in Advances in Mathematics Supplementary Studies, Vol. 7B, Academic Press, New York, 1981, pp. 741–747)?
No, it was an honest challenge which I myself failed to meet (hence my "what I'm really curious to see..." post). I agree. If it is scanned somewhere it definitely isn't OCR'ed or so new that Google hasn't stumbled over it, yet.
@MartinSleziak I don't think so :) I'm not very good at coming up with new tags. I just think there is little sense to prefer one of liminf/limsup over the other and every term encompassing both would most likely lead to us having to do the tagging ourselves since beginners won't be familiar with it.
Anyway, my opinion is this: I did what I considered the best way: I've created [tag:limsup] and mentioned liminf in tag-wiki. Feel free to create new tag and retag the two questions if you have better name. I do not plan on adding other questions to that tag until tommorrow.
@QED You do not have to accept anything. I am not saying it is a good question; but that doesn't mean it's not acceptable either. The site's policy/vision is to be open towards "math of all levels". It seems hypocritical to me to declare this if we downvote a question simply because it is elementary.
@Matt Basically, the a priori probability (the true probability) is different from the a posteriori probability after part (or whole) of the sample point is revealed. I think that is a legitimate answer.
@QED Well, the tag can be removed (if someone decides to do so). Main purpose of the edit was that you can retract you downvote. It's not a good reason for editing, but I think we've seen worse edits...
@QED Ah. Once, when it was snowing at Princeton, I was heading toward the main door to the math department, about 30 feet away, and I saw the secretary coming out of the door. Next thing I knew, I saw the secretary looking down at me asking if I was all right.
OK, so chat is now available... but; it has been suggested that for Mathematics we should have TeX support.The current TeX processing has some non-trivial client impact. Before I even attempt trying to hack this in, is this something that the community would want / use?(this would only apply ...
So in between doing phone surveys for CNN yesterday I had an interesting thought. For $p$ an odd prime, define the truncation map $$t_{p^r}:\mathbb{Z}_p\to\mathbb{Z}/p^r\mathbb{Z}:\sum_{l=0}^\infty a_lp^l\mapsto\sum_{l=0}^{r-1}a_lp^l.$$ Then primitive roots lift to $$W_p=\{w\in\mathbb{Z}_p:\langle t_{p^r}(w)\rangle=(\mathbb{Z}/p^r\mathbb{Z})^\times\}.$$ Does $\langle W_p\rangle\subset\mathbb{Z}_p$ have a name or any formal study?
> I agree with @Matt E, as almost always. But I think it is true that a standard (pun not originally intended) freshman calculus does not provide any mathematically useful information or insight about infinitesimals, so thinking about freshman calculus in terms of infinitesimals is likely to be unrewarding. – Pete L. Clark 4 mins ago
In mathematics, in the area of order theory, an antichain is a subset of a partially ordered set such that any two elements in the subset are incomparable. (Some authors use the term "antichain" to mean strong antichain, a subset such that there is no element of the poset smaller than 2 distinct elements of the antichain.)Let S be a partially ordered set. We say two elements a and b of a partially ordered set are comparable if a ≤ b or b ≤ a. If two elements are not comparable, we say they are incomparable; that is, x and y are incomparable if neither x ≤ y nor y ≤ x.A chain in S is a...
@MartinSleziak Yes, I almost expected the subnets-debate. I was always happy with the order-preserving+cofinal definition and never felt the need for the other one. I haven't thought about Alexei's question really.
When I look at the comments in Norbert's question it seems that the comments together give a sufficient answer to his first question already - and they came very quickly. Nobody said anything about his second question. Wouldn't it be better to divide it into two separate questions? What do you think t.b.?
@tb About Alexei's questions, I spent some time on it. My guess was that it doesn't hold but I wasn't able to find a counterexample. I hope to get back to that question. (But there is already too many questions which I would like get back to...)
@MartinSleziak I deleted part of my comment since I figured out that I never actually proved that in detail but I'm sure it should work. I needed a bit of summability in topological vector spaces but it's really no problem at all. It's just a special case of nets written differently (as series are a special case of sequences). |
Rather than answer the question numerically I have outlined the four different cases,
reversible / irreversible and isothermal / adiabatic.
In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work done must therefore be at the expense of the internal energy. If the ‘system’ is a gas then its temperature will not remain constant during any expansion or compression.
In expansion the work done is $dw=-pdV$ and the change in internal energy $ dU=C_vdT$.
The heat change is zero then $dq=0$ which means from the First Law $$dU = dw$$and so$$ C_vdT = -pdV$$
Dividing both sides by
T and for one mole of an perfect gas $p=RT/V$ thus $$C_v\frac{dT}{T}=-R\frac{dV}{V}$$If the gas starts at $T_1,V_1$ and ends up at $T_2,V_2$ the last equation can be integrated (and rearranged) to give$$\ln\left (\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)^{R/C_v}$$or $$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
using the relationship $C_p = C_v+R$$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{(C_p-C_v)/C_v}$$Using the gas law this van be rewritten in a more useful form as$$ p_1V_1^{\gamma } = p_2V_2^{\gamma }$$where $\gamma = C_p/C_v$. This is also written as $pV^\gamma = const$.
The change in internal energy in an adiabatic process is$$\Delta U=C_v(T_2-T_1)$$
Adiabatic reversible
In a
reversible adiabatic change we use the formulas above to work out what happens. If n moles of a gas fill a container of volume $V_1$ at $p_1$ atm. and is expanded reversibly and adiabatically until it is in equilibrium at a final pressure $p_2$ we can calculate the final volume and temperature. The values of $C_p$ and $C_v$ are assumed to be known and are constant with temperature.
The equation to use is
$$ p_1V_1^{\gamma} = p_2V_2^{\gamma}$$and as only $V_2$ is unknown this can be calculated. The work done is $$w=nC_v(T_2-T_1)$$
Adiabatic irreversible
In an
irreversible adiabatic change if n moles of an perfect gas expands irreversibly from a pressure of $p_1$ against a constant external pressure $p_2$ the temperature drops from $T_1$ to $T_2$. We can calculate how much work is done and the final volume. The internal energy change is $\Delta U= nC_v(T_2-T_1)$and the work done is $w=p(V_2-V_1)$ and as $\Delta U = w$ (as the system is adiabatic) the final volume can be obtained. Similarly if the volumes are known the final temperature can be obtained.
Isothermal irreversible
In an
isothermal irreversible change the work done on suddenly allowing a perfect gas to expanding from $V_1$ to $V_2$ is determined by the final external pressure $p_2$ and is $$ q=-w =\int_{V_1}^{V_2} pdV = p_2(V_2-V_1)$$thus expansion into a vacuum does no work.
Isothermal reversible
A
reversible isothermal change performs the maximum possible amount of work, and assuming a perfect gas,$$ q=-w_{rev} =\int_{V_1}^{V_2} pdV=nRT \int_{V_1}^{V_2} \frac{1}{V}dV$$
$$w_{rev} =- nRT\ln\left(\frac{V_2}{V_1}\right)= - nRT\ln\left(\frac{p_1}{p_2}\right)$$ |
Adelman's theorem relativizes, i.e. for every oracle $O$ we have $\mathsf{BPP^O\subseteq P/Poly^O}$. To see why you have to go over the proof of Adelman's theorem. The proof goes through showing that given a probabilistic machine with exponentially low error probability, there exists a sequence of coin tosses which produces the correct answer for all (fixed length) inputs. This remains true in the relativized world, use that coin sequence as the advice, and query the oracle whenever necessary while simulating the probabilistic machine with the coins given in the advice.
Additionally, $\mathsf{BPP^{\Sigma_i}}\subseteq \Sigma_{i+3}$. Again, you have to examine the details in the proof of $\mathsf{BPP\subseteq\Sigma_2}$. Suppose $L\in\mathsf{BPP^{\Sigma_i}}$, and $M$ is an oracle probabilistic Turing machine for $L$ with exponentially low error probability. You can use the same idea from $\mathsf{BPP\subseteq\Sigma_2}$, and claim that if $x\in L$ then the set of coin sequences who lead $M$ to accept is big, and in that case it can cover $\{0,1\}^m$ with a small number of shifts, where $m=poly(|x|)$ is the amount of coins used. Similarly, if $x\notin L$ you claim that the set of accepting coin sequences is small and cannot cover $\{0,1\}^m$ with a small number of shifts. All this remains true in the relativized world, as you only reasoned about the "BPP part" of $M$, i.e how it behaves with different coin tosses depending on whether or not the input is in $L$.
What's different is that when you find yourself in need to say "some vector $r\in\{0,1\}^m$ is obtained by shifting an accepting sequence", the part verifying that a sequence is indeed accepting will add additional $i+1$ quantifiers to the formula, after the initial two who talk about space shifting. To be more clear, your final formula has the form:
"$x\in L$ iff there exist shifts such that every vector $r\in\{0,1\}^m$ is obtained by performing one of those shifts on an accepting coin sequence for $x$", this is equivalent to:
"$x\in L$ iff there exists shifts such that every vector $r\in\{0,1\}^m$ can be shifted to generate an accepting coin sequence for $x$". Saying some vector $v$ is an accepting coin sequence requires expressing "$M$ accepts $v$" where $M$ is a poly time machine with access to a $\Sigma_i$ oracle. You can express this using a $\Sigma_{i+1}$ formula (this is analogous to $\mathsf{P^{\Sigma_i}\subseteq \Sigma_{i+1}}$).
The above shows that $\mathsf{BPP^{NP}\subseteq \Sigma_4\cap P/Poly^{NP}}$.As for your second question, I don't know if there are complete problems for $\mathsf{BPP^{\oplus P}}$, so this might make the class $\mathsf{BPP^{BPP^{\oplus P}}}$ ill defined. One interpretation is asking whether the containment holds for every oracle in that class. I doubt this is known, but no obvious important/terrible implication comes to mind. Note that disproving this containment is out of the question, as it would immediately imply $P\neq \oplus P$. |
Preprints (rote Reihe) des Fachbereich Mathematik Refine Year of publication 1996 (22) (remove)
293
Tangent measure distributions were introduced by Bandt and Graf as a means to describe the local geometry of self-similar sets generated by iteration of contractive similitudes. In this paper we study the tangent measure distributions of hyperbolic Cantor sets generated by contractive mappings, which are not similitudes. We show that the tangent measure distributions of these sets equipped with either Hausdorff or Gibbs measure are unique almost everywhere and give an explicit formula describing them as probability distributions on the set of limit models of Bedford and Fisher.
274
This paper investigates the convergence of the Lanczos method for computing the smallest eigenpair of a selfadjoint elliptic differential operator via inverse iteration (without shifts). Superlinear convergence rates are established, and their sharpness is investigated for a simple model problem. These results are illustrated numerically for a more difficult problem.
280
This paper develops truncated Newton methods as an appropriate tool for nonlinear inverse problems which are ill-posed in the sense of Hadamard. In each Newton step an approximate solution for the linearized problem is computed with the conjugate gradient method as an inner iteration. The conjugate gradient iteration is terminated when the residual has been reduced to a prescribed percentage. Under certain assumptions on the nonlinear operator it is shown that the algorithm converges and is stable if the discrepancy principle is used to terminate the outer iteration. These assumptions are fulfilled , e.g., for the inverse problem of identifying the diffusion coefficient in a parabolic differential equation from distributed data.
284
A polynomial function \(f : L \to L\) of a lattice \(\mathcal{L}\) = \((L; \land, \lor)\) is generated by the identity function id \(id(x)=x\) and the constant functions \(c_a (x) = a\) (for every \(x \in L\)), \(a \in L\) by applying the operations \(\land, \lor\) finitely often. Every polynomial function in one or also in several variables is a monotone function of \(\mathcal{L}\). If every monotone function of \(\mathcal{L}\)is a polynomial function then \(\mathcal{L}\) is called orderpolynomially complete. In this paper we give a new characterization of finite order-polynomially lattices. We consider doubly irreducible monotone functions and point out their relation to tolerances, especially to central relations. We introduce chain-compatible lattices and show that they have a non-trivial congruence if they contain a finite interval and an infinite chain. The consequences are two new results. A modular lattice \(\mathcal{L}\) with a finite interval is order-polynomially complete if and only if \(\mathcal{L}\) is finite projective geometry. If \(\mathcal{L}\) is simple modular lattice of infinite length then every nontrivial interval is of infinite length and has the same cardinality as any other nontrivial interval of \(\mathcal{L}\). In the last sections we show the descriptive power of polynomial functions of lattices and present several applications in geometry.
271
The paper deals with parallel-machine and open-shop scheduling problems with preemptions and arbitrary nondecreasing objective function. An approach to describe the solution region for these problems and to reduce them to minimization problems on polytopes is proposed. Properties of the solution regions for certain problems are investigated. lt is proved that open-shop problems with unit processing times are equivalent to certain parallel-machine problems, where preemption is allowed at arbitrary time. A polynomial algorithm is presented transforming a schedule of one type into a schedule of the other type.
282
Let \(a_1,\dots,a_m\) be independent random points in \(\mathbb{R}^n\) that are independent and identically distributed spherically symmetrical in \(\mathbb{R}^n\). Moreover, let \(X\) be the random polytope generated as the convex hull of \(a_1,\dots,a_m\) and let \(L_k\) be an arbitrary \(k\)-dimensional subspace of \(\mathbb{R}^n\) with \(2\le k\le n-1\). Let \(X_k\) be the orthogonal projection image of \(X\) in \(L_k\). We call those vertices of \(X\), whose projection images in \(L_k\) are vertices of \(X_k\)as well shadow vertices of \(X\) with respect to the subspace \(L_k\) . We derive a distribution independent sharp upper bound for the expected number of shadow vertices of \(X\) in \(L_k\).
285
On derived varieties (1996)
Derived varieties play an essential role in the theory of hyperidentities. In [11] we have shown that derivation diagrams are a useful tool in the analysis of derived algebras and varieties. In this paper this tool is developed further in order to use it for algebraic constructions of derived algebras. Especially the operator \(S\) of subalgebras, \(H\) of homomorphic irnages and \(P\) of direct products are studied. Derived groupoids from the groupoid \(N or (x,y)\) = \(x'\wedge y'\) and from abelian groups are considered. The latter class serves as an example for fluid algebras and varieties. A fluid variety \(V\) has no derived variety as a subvariety and is introduced as a counterpart for solid varieties. Finally we use a property of the commutator of derived algebras in order to show that solvability and nilpotency are preserved under derivation.
277
A convergence rate is established for nonstationary iterated Tikhonov regularization, applied to ill-posed problems involving closed, densely defined linear operators, under general conditions on the iteration parameters. lt is also shown that an order-optimal accuracy is attained when a certain a posteriori stopping rule is used to determine the iteration number. |
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Feynman-Kac for SPDEs with multiplicative noise
One application of the celebrated Feynman-Kac formula in probability theory gives a stochastic representation of solutions to Cauchy problems. Following the basic idea of the proof, the formula is first derived for terminal value problems of type \[ -\frac{\partial}{\partial t}u\,=\, {\cal A} u - k\times u + g,\quad u(x,T)=f(x), \] where $f$ is a function on $\mathbb{R}^d$, $k$ and $g$ are functions on $\mathbb{R}^d\times[0,T]$, and ${\cal A}$ is a second-order partial differential operator. Under the right conditions, the representation would read \[ u(x,t)\,=\, {\bf E}[f(X_T)\exp\{-\int_t^T k(X_s,s)\,ds\}+\int_t^T g(X_s,s)\exp\{-\int_t^s k(X_r,r)\,dr\}\,ds\,|\,X_t=x], \] for $(x,t)\in\mathbb{R}^d\times[0,T]$, where $X_t$ stands for the Markov process generated by ${\cal A}$, and ${\bf E}$ denotes the corresponding expectation operator.
This note is NOT about the conditions on $f,k,g, {\cal A}$ guaranteeing such a representation. This note is about the corresponding initial value problem with $k$ being a random space-time field. To concentrate on this issue, in what follows, $g$ is set to be zero. After guessing a formula for the initial value problem from the above formula by simple time-reversion, it is discussed why this formula needs further treatment when $k$ is a rather IRREGULAR random space-time field.
If $u(x,t)$ satisfies the initial value problem \[\frac{\partial}{\partial t}u\,=\, {\cal A} u - k(x,t)\times u,\quad u(x,0)=f(x),\tag{1}\label{ivp}\] and if ${\cal A}$ only acts on the $x$-variable then, for an arbitrary but fixed $t>0$, $\tilde{u}(x,t')=u(x,t-t')$ satisfies the terminal value problem \[-\frac{\partial}{\partial t'}\tilde{u}\,=\, {\cal A}\tilde{u} - k(x,t-t')\times \tilde{u},\quad \tilde{u}(x,t)=f(x),\] and hence \[ \tilde{u}(x,t')\,=\, {\bf E}[f(X_t)\exp\{-\int_{t'}^t k(X_s,t-s)\,ds\}\,|\,X_{t'}=x], \] for any $(x,t')\in\mathbb{R}^d\times[0,t]$, which gives the stochastic representation \[ u(x,t)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t k(X_s,t-s)\,ds\}\,|\,X_{0}=x]\tag{2}\label{FK}\] for the solution to the initial value problem, when $t'$ is taken to be zero.
Now assume that $k(x,t)$ is a random space-time field, for example, $k(x,t)=h(x)\times\dot{B}_t$, where $h$ denotes a deterministic function on $\mathbb{R}^d$, and $B=(B_t)_{t\ge 0}$ is a one-dimensional standard Brownian motion. Clearly, this makes (\ref{ivp}) a rather FORMAL stochastic partial differential equation (SPDE), because the product $k(x,t)\times u$ needs explanation, because $\dot{B}_t=\partial B_t/\partial t$ is not a function anymore.
However, when constructing both $X_t$ and $B_t$ as two independent processes on the same probability space, formula (\ref{FK}) could be read \[ u(x,t)(\omega)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t h(X_s)\dot{B}_{t-s}\,ds\}\,|\,X_{0}=x,B=\omega],\tag{3}\label{FK1}\] and this might give a solution to the SPDE, in some sense, if the integral $\int_{0}^t h(X_s)\dot{B}_{t-s}\,ds$ can be well-defined.
A first idea could be to treat the integral as a backward stochastic integral against Brownian motion. But this would cause an adaptedness-problem when verifying equation (\ref{ivp}) as the integral form of this equation would rather be obtained by integrating forward in time. So, one wants to write \[ \int_{0}^t h(X_{t-s})\dot{B}_s\,ds\quad\mbox{for}\quad \int_{0}^t h(X_s)\dot{B}_{t-s}\,ds \] which is a formal operation, only, since $\dot{B}$ is not a function. But this operation would have its proper meaning when $\dot{B}$ is approximated by a continuous function $\dot{B}_\varepsilon$ obtained by convoluting $\dot{B}$ against a mollifier. Then, \[ u_\varepsilon(x,t)(\omega)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t h(X_{t-s})\dot{B}_\varepsilon(s)\,ds\}\,|\,X_{0}=x,B=\omega]\tag{4}\label{FKapprox} \] solves \[ \frac{\partial}{\partial t}u_\varepsilon\,=\, {\cal A} u_\varepsilon - h(x)\dot{B}_\varepsilon(t)\times u_\varepsilon,\quad u_\varepsilon(x,0)=f(x), \] in mild sense, that is \[ u_\varepsilon(x,t)\,=\,\int p_t(x,y)f(y)\,dy+\int_0^t\!\!\!\int p_{t-s}(x,y)\,u_\varepsilon(y,s)\,h(y)dy\,\dot{B}_\varepsilon(s)ds, \] where $p_t(x,y)$ stands for the family of transition probability densities of the Markov process $X_t$. Note that $X_t$ is a time-homogeneous Markov process because its generator ${\cal A}$ does not act on time.
Taking $\varepsilon$ to zero in the last equation gives \[ u(x,t)\,=\,\int p_t(x,y)f(y)\,dy+\int_0^t\!\!\!\int p_{t-s}(x,y)\,u(y,s)\,h(y)dy\,\circ dB_s, \] where \[ u(x,t)(\omega)\,=\, {\bf E}[f(X_t)\exp\{-\int_{0}^t h(X_{t-s})\circ dB_s\}\,|\,X_{0}=x,B=\omega]\tag{5}\label{solStrato}\] is the $\varepsilon$-limit of (\ref{FKapprox}). Here $\circ dB_s$ denotes Stratonovich integration as this is what $\dot{B}_\varepsilon(s)ds$ would "produce" in the limit. All in all, (\ref{solStrato}) yields a mild solution to (\ref{ivp}) if the product $k(x,t)\times u$ with $k(x,t)=h(x)\times\dot{B}_t$ is understood in the sense of Stratonovich. If the SPDE (\ref{ivp}) is understood in the sense of Itô, then the integral under the exponential in (\ref{solStrato}) should be replaced by the corresponding Itô-integral minus correction-term $\frac{1}{2}\int_{0}^t h(X_{s})^2\,ds$.
Now, if the process $X_t$ is reversible, there is a trick to get rid of the backward time of the integrand under the exponential in (\ref{solStrato}). This trick is used by Bertini/Cancrini in
The Stochastic Heat Equation: Feynman-Kac Formula and Intermittence, Journal of Statistical Physics, Vol. 78, Nos 5/6, 1995, when ${\cal A}$ is the Laplacian and $k(x,t)$ is space-time white noise. They don't say that this trick is new but they don't give a reference, either. Below, the right-hand side of (\ref{solStrato}) is going to be transformed following Bertini/Cancrini's ideas (I do not know any earlier reference---comments on earlier references would be most welcome).
Since the processes $X_t$ and $B_t$ are independent, in what follows, the conditioning on $B$ is suppressed and the expectation is only taken with respect to the process $X_t$ writing ${\bf E}^X$ for the corresponding expectation operator. Using this notation, the right-hand side of (\ref{solStrato}) equals \begin{align*} &{\bf E}^X[f(X_t)\exp\{-\int_{0}^t h(X_{t-s})\circ dB_s\}\,|\,X_{0}=x]\\ =&\int f(y)\,{\bf E}^X[\exp\{-\int_{0}^t h(X_{t-s})\circ dB_s\}\,|\,X_{0}=x,X_t=y]\,p_t(x,y)\,dy. \end{align*} Obviously, for fixed $t$, the expectation is now taken with respect to an $X$-bridge satisfying $X_0=0$ and $X_t=y$. But the integration is with respect to the time-reversion of this bridge, and the time-reversed bridge has the same law as the corresponding forward-time bridge since the process $X_t$ is reversible in the sense of $p_t(x,y)=p_t(y,x)$. So, if ${\bf E}^b_{y,x;t}$ denotes the expectation with respect to an $X$-bridge satisfying $X_0=y$ and $X_t=x$, then the right-hand side of (\ref{solStrato}) becomes \[ \int dy\,f(y)\,p_t(y,x)\,{\bf E}^b_{y,x;t}[\exp\{-\int_{0}^t h(b_{s})\circ dB_s\}], \] which is the type of Feynman-Kac formula used by Bertini/Cancrini in the paper referenced above. They prove that their formula gives a solution to the corresponding SPDE. This proof is quite short, and I think that the validity of (3.21) on page 1394 of their paper can hardly be understood without mentioning the underlying principle of time-reversion I stressed above. |
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Even × Even = Even Even ÷ Even = Even ? \begin{aligned} \text{ Even } \times \text{ Even } &=& \text{ Even } \\\text{ Even } \div \text{ Even } &=& \text{ Even }?\end{aligned} Even × Even Even ÷ Even == Even Even ?
It is true that the product of two even numbers is always an even number, but is it also that the ratio of two even numbers is always an even number too?
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Plastids
Category : 11th Class
Plastids are semiautonomous organelles having DNA, RNA, Ribosomes and double membrane envelope. These are largest cell organelles in plant cell.
History
(1) Haeckel (1865) discovered plastid, but the term was first time used by Schimper (1883)
.
(2) A well organised system of grana and stroma in plastid of normal barley plant was reported by de Von Wettstein.
(3) Park and Biggins (1964
) gave the concept of quantasomes.
(4) The term chlorophyll was given by Pelletier and Caventou, and structural details were given by Willstatter and Stall.
(5) The term thylakoid was given by Menke (1962).
(6) Fine structure was given by Mayer.
(7) Ris and Plaut (1962) reported DNA in chloroplast and was called plastidome.
Types of plastids : According to Schimper , Plastids are of 3 types: Leucoplasts, Chromoplasts and Chloroplasts. Leucoplasts : They are colourless plastids which generally occur near the nucleus in nongreen cells and possess internal lamellae. Grana and photosynthetic pigments are absent. They mainly store food materials and occur in the cells not exposed to sunlight e.g., seeds, underground stems, roots, tubers, rhizomes etc. These are of three types.
(1)
Amyloplast : Synthesize and store starch grains. e.g., potato tubers, wheat and rice grains.
(2)
Elaioplast (Lipidoplast, Oleoplast) : They store lipids and oils e.g., castor endosperm, tube rose, etc.
(3)
Aleuroplast (Proteinoplast) : Store proteins e.g., aleurone cells of maize grains. Chromoplasts : Coloured plastids other than green are kown as chromoplasts . These are present in petals and fruits. These also carry on photosynthesis. These may arise from the chloroplasts due to replacement of chlorophyll by other pigments.
Green tomatoes and chillies turn red on ripening because of replacement of chlorophyll molecule in chloroplasts by the red pigment lycopene in tomato and capsanthin in chillies. Thus, chloroplasts are changed into chromatoplast.
All colours (except green) are produced by flavins, flavenoids and cyanin. Cyanin pigment is of two types one is anthocyanin (blue) and another is erythrocyanin (red). Anthocyanin are water soluble pigments and found in cell sap of vacoule.
Chloroplast : Discovered by Sachs and named by Schimper. They are greenish plastids which possess photosynthetic pigments. Number : It is variable. Number of chloroplast is 1 in Spirogyra indica, 2 in Zygnema, 16 in S.rectospora, up to 100 in mesophyll cells. The minimum number of one chloroplast per cell is found in Ulothrix and species of Chlamydomonas. Shape : They have various shapes
Cup shaped
Chlamydomonas sp.
Stellate shaped
Zygnema.
Collar or girdle shaped
Ulothrix
Spiral or ribbon shaped
Spirogyra
Reticulate
Oedogonium
Discoid
Voucheria
Size : It ranges from \[3-10\,\mu \,m\] (average \[5\,\mu \,m)\] in diameter. The discoid chloroplast of higher plants are \[4-10\,\mu \,m\] in length and \[2-4\,\mu \,m\] in breadth. Chloroplast of Spirogyra may reach a length of 1 mm. Sciophytes (Shade plant) have larger chloroplast. Chemical composition : Proteins 50 – 60%; Lipids 25 – 30%; Chlorophyll – 5- 10 %; Carotenoids (carotenes and xanthophylls) 1 –2%; DNA – 0.5%, RNA 2 – 3%; Vitamins K and E; Quinines, Mg, Fe, Co, Mn, P, etc. in traces. Ultrastructure : It is double membrane structure. Both membranes are smooth. The inner membrane is less permeable than outer but rich in proteins especially carrier proteins. Each membrane is \[90100\text{ }{AA}\] thick. The inter-membrane space is called the periplastidial space. Inner to membranes, matrix is present, which is divided into two parts.
(1)
Grana : Inner plastidial membrane of the chloroplast is invaginated to form a series of parallel membranous sheets, called lamellae, which form a number of oval – shaped closed sacs, called thylakoids. Thylakoids are structural and functional elements of chloroplasts.
Along the inner side of thylakoid membrane, there are number of small rounded para-crystalline bodies, called quantasomes (a quantasome is the photosynthetic unit).
Each quantasome contains about 230 chlorophyll molecules (160 chl. '
a' and 70 chl. ' b') and 50 carotenoid molecules.
In eukaryotic plant cells, a number of thylakoids are superimposed like a pile of coins to form a granum. The number of thylakoids in a granum ranges from 10-100 (average number is 20-50). Adjacent grana are interconnected by branched tubules, called stromal lamellae or Fret-channel or Fret membrane's.
(2)
Stroma : It is transparent, proteinaceous and watery substance. Dark reaction of photosynthesis occurs in this portion. Stroma is almost filled with “Rubisco” (about 15% of total enzyme, protein) enzyme \[C{{O}_{2}}\]is accepted by this enzyme. CO 2 assimilation results in carbohydrate formation. It has 20 – 60 copies of naked circular double stranded DNA.
Pigments of chloroplast Chlorophyll a : \[{{C}_{55}}\,{{H}_{72}}\,{{O}_{5}}{{N}_{4}}Mg\] (with methyl group) Chlorophyll b : \[({{C}_{55}}{{H}_{74}}{{O}_{6}}{{N}_{4}}Mg)\](with aldehyde group) Chlorophyll c : \[{{C}_{35}}{{H}_{32}}{{O}_{5}}\,{{N}_{4}}Mg\] Chlorophyll d : \[\alpha ,\beta ,\gamma \] Bacteriochlorophyll \[({{C}_{55}}{{H}_{74}}{{O}_{6}}{{N}_{4}}Mg)\] or chlorobium chlorophyll present in photosynthetic bacteria. These pigment are red in acidic and blue in alkaline medium. Carotenoids : These are hydrocarbons, soluble in organic solvents. These are of two types :
(1)
Carotenes : \[{{C}_{40}}{{H}_{56}}{{O}_{2}},\] derivatives of vitamin A. Carrot coloured \[\alpha ,\beta ,\gamma \] carotene, lycopene, etc. \[\beta -\]carotene most common.
(2)
Xanthophyll : \[{{C}_{40}}{{H}_{56}}{{O}_{2}},\]yellowish in colour, fucoxanthin, violaxanthin. Molar ratio of carotene and xanthophyll in young leaves is 2 : 1. Origin of chloroplast : Plastids, like the mitochondria, are self duplicating organelles. These develop from colourless precursors, called proplastids. They are believed to be evolved from endosymbiont origination. Functions
(1) It is the site of photosynthesis, (light and dark reaction).
(2) Photolysis of water, reduction of \[NADP\] to \[NADP{{H}_{2}}\] take place in granum.
(3) Photophosphorylation through cytochrome \[{{b}_{6}}\,f,\] plastocyanine and plastoquinone etc.
(4) They store starch or factory of synthesis of sugars.
(5) Chloroplast store fat in the form of plastoglobuli.
(6) They maintain the percentage of \[C{{O}_{2}}\]and \[{{O}_{2}}\] in atmosphere.
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Relative Velocity:
We encounter occasions where one or more objects move in a frame which is non-stationary with respect to another observer. For example, a boat crosses a river that is flowing at some rate or an airplane encountering wind during its motion. In all such instances, in order to describe the complete motion of the object, we need to consider the effect that the medium is causing on the object. While doing so, we calculate the relative velocity of the object considering the velocity of the particle as well as the velocity of the medium. Here, we will learn how to calculate the
relative velocity.
Let us consider two objects, A and B moving with velocities V
a and V b with respect to a common stationary frame of reference, say the ground, a bridge or a fixed platform.
The velocity of the object A relative to the object B can be given as,
\(V_{ab}=V_{a}-V_{b}\)
Similarly, the velocity of the object B relative to that of object a is given by,
\(V_{ba}=V_{b}-V_{a}\)
From the above two expressions, we can see that
\(V_{ab}=-V_{ba}\)
Although the magnitude of the both the relative velocities is equal to each other. Mathematically,
\(\left |V_{ab} \right |=\left |V_{ba} \right |\)
Examples of relative Velocity:
We can understand the concept of relative velocity more clearly with the help of the following example.
: A plane is traveling at velocity 100 km/hr, in the southward direction. It encounters wind traveling in the west direction at a rate of 25 km/hr. Calculate the resultant velocity of the plane. Example
Given, the velocity of the wind = V
w = 25km/hr
The velocity of the plane = V
a= 100 km/hr
The relative velocity of the plane with respect to the ground can be given as
The angle between the velocity of the wind and that of the plane is 90°. Using the Pythagorean theorem, the resultant velocity can be calculated as,
R
2= (100 km/hr) 2 + (25 km/hr) 2
R
2= 10 000 km 2/hr 2 + 625 km 2/hr 2
R
2= 10 625 km 2/hr 2
Hence,
R = 103.077 km/hr Using trigonometry, the angle made by the resultant velocity with respect to the horizontal plane can be given as,
\(tan\Theta =(\frac{window \: velocity}{aiprplane\: velocity})\\ \\ tan\Theta =(\frac{25}{100})\\ \\ \Theta =tan^{-1}\frac{1}{4}\\ \\ \Theta =14.0^{\circ}\)
Relative Velocity Problems
1)What is relative velocity?
Answer:
another object A.
2)A motorcycle travelling on the highway at a velocity of 120 km/h passes a car travelling at a velocity of 90 km/h. From the point of view of a passenger on the car, what is the velocity of the motorcycle?
Solution:
Let us represent the velocity of the motorcycle as
V A and the velocity of the car as V. B
Now, the velocity of the motorcycle relative to the point of view of a passenger is given as
V AB = V– A V B
Substituting the values in the above equation, we get
V AB = 120 km/h – 90 km/h = 30 km/h
Hence, the velocity of the motorcycle relative to the passenger of the car is 30 km/h.
3)A swimmer swimming across a river flowing at a velocity of 4 m/s swims at the velocity of 2 m/s. Calculate the actual velocity of the swimmer and the angle.
Solution:
The actual velocity of the swimmer can be found out as follows:\(V_{actual}=\sqrt{2^2+4^2}=4.47\,m/s\)
The angle is calculated as follows:\(\tan \Theta =\frac{2}{4}\) \(\Theta =\tan^{-1}\frac{2}{4}=26.57^{\circ}\)
4)
Solution: The ball lands at the point from which it was thrown, i.e. back to the thrower’s hand. It lands in front of the point from which it was thrown. It lands behind the point from which it was thrown. The ball will land to the left of the point from which it was thrown.
5. An aeroplane flies with a velocity of 450 m/s to the north, while an aeroplane B
travels at a velocity of 500 m/s to the south beside aeroplane A. Calculate the
relative velocity of the aeroplane A with respect to aeroplane B.
Solution:
is calculated as follows:
V AB = V– A V B V AB = 350 m/s – (–500 m/s) = 850 m/s
The velocity of aeroplane B is considered negative, as it flies in the opposite
direction to the of aeroplane A.
Stay tuned with BYJU’S to learn more about the concept of relative velocity, relative motion and other related topics.
Frequently Asked Questions on Relative Velocity What is relative velocity?
Relative velocity is defined as the velocity of an object B in the rest frame of another object A.
A motorcycle travelling on the highway at a velocity of 120 km/h passes a car travelling at a velocity of 90 km/h. From the point of view of a passenger on the car, what is the velocity of the motorcycle?
Let us represent the velocity of the motorcycle as VA and the velocity of the car as VB.
Now, the velocity of the motorcycle relative to the point of view of a passenger is given as, VAB = VA – VB
Substituting the values in the above equation, we get VAB = 120 km/h – 90 km/h = 30 km/h
Hence, the velocity of the motorcycle relative to the passenger of the car is 30 km/h. |
In a paper by Joos and Zeh, Z Phys B 59 (1985) 223, they say:This 'coming into being of classical properties' appears related to what Heisenberg may have meant by his famous remark [7]: 'Die "Bahn" entsteht erst dadurch, dass wir sie beobachten.'Google Translate says this means something ...
@EmilioPisanty Tough call. It's technical language, so you wouldn't expect every German speaker to be able to provide a correct interpretation—it calls for someone who know how German is used in talking about quantum mechanics.
Litmus are a London-based space rock band formed in 2000 by Martin (bass guitar/vocals), Simon (guitar/vocals) and Ben (drums), joined the following year by Andy Thompson (keyboards, 2001–2007) and Anton (synths). Matt Thompson joined on synth (2002–2004), while Marek replaced Ben in 2003. Oli Mayne (keyboards) joined in 2008, then left in 2010, along with Anton. As of November 2012 the line-up is Martin Litmus (bass/vocals), Simon Fiddler (guitar/vocals), Marek Bublik (drums) and James Hodkinson (keyboards/effects). They are influenced by mid-1970s Hawkwind and Black Sabbath, amongst others.They...
@JohnRennie Well, they repeatedly stressed their model is "trust work time" where there are no fixed hours you have to be there, but unless the rest of my team are night owls like I am I will have to adapt ;)
I think u can get a rough estimate, COVFEFE is 7 characters, probability of a 7-character length string being exactly that is $(1/26)^7\approx 1.2\times 10^{-10}$ so I guess you would have to type approx a billion characters to start getting a good chance that COVFEFE appears.
@ooolb Consider the hyperbolic space $H^n$ with the standard metric. Compute $$\inf\left\{\left(\int u^{2n/(n-2)}\right)^{-(n-2)/n}\left(4\frac{n-1}{n-2}\int|\nabla u|^2+\int Ru^2\right): u\in C^\infty_c\setminus\{0\}, u\ge0\right\}$$
@BalarkaSen sorry if you were in our discord you would know
@ooolb It's unlikely to be $-\infty$ since $H^n$ has bounded geometry so Sobolev embedding works as expected. Construct a metric that blows up near infinity (incomplete is probably necessary) so that the inf is in fact $-\infty$.
@Sid Eating glamorous and expensive food on a regular basis and not as a necessity would mean you're embracing consumer fetish and capitalism, yes. That doesn't inherently prevent you from being a communism, but it does have an ironic implication.
@Sid Eh. I think there's plenty of room between "I think capitalism is a detrimental regime and think we could be better" and "I hate capitalism and will never go near anything associated with it", yet the former is still conceivably communist.
Then we can end up with people arguing is favor "Communism" who distance themselves from, say the USSR and red China, and people who arguing in favor of "Capitalism" who distance themselves from, say the US and the Europe Union.
since I come from a rock n' roll background, the first thing is that I prefer a tonal continuity. I don't like beats as much as I like a riff or something atmospheric (that's mostly why I don't like a lot of rap)
I think I liked Madvillany because it had nonstandard rhyming styles and Madlib's composition
Why is the graviton spin 2, beyond hand-waiving, sense is, you do the gravitational waves thing of reducing $R_{00} = 0$ to $g^{\mu \nu} g_{\rho \sigma,\mu \nu} = 0$ for a weak gravitational field in harmonic coordinates, with solution $g_{\mu \nu} = \varepsilon_{\mu \nu} e^{ikx} + \varepsilon_{\mu \nu}^* e^{-ikx}$, then magic? |
This is an excellent series of questions, so I will have a go at part of it! I will start with an example from Verdeyen’s book (J. T. Verdeyen, Laser Electronics, Prentice-Hall, Inc., Englewood Cliffs, NJ, ©1981, Chapter 14). Assume $\lambda$ = 500 nm, quantum efficiency = 0.15, photomultiplier (PMT) gain = 1.68x$10^7$, transimpedance = 1 k$\Omega$ and RC= 100 $\mu$s. Assume $\Delta$t = 1 $\mu$s and the output of the RC LPF is sampled at the end of every 100 $\mu$s interval.Start at the bottom: assume an average of 2 photons/$\mu$s arrive at the PMT’s photocathode. Then the average incident optical power is 0.795 pW, the average photocurrent is 806 nA and the TIA’s average output voltage is 0.806 mV. The first figure below shows the Mueller calculus simulation models, with the above assumptions. (It is deliberately simple, for present purposes, but can be made much more complicated and realistic if desired.)
The light source produces Stokes vectors at the rate of 1/$\mu$s, and the intensity is proportional to the number of photons in that $\mu$s interval. The detector and transimpedance preamplifier (TIA) converts the incident light intensities to photocurrents and then to voltages. The FFT PSD block shows (second figure) that the noise is white.
Then the RC LPF filters the TIA’s outputs and its outputs are sampled every 100 $\mu$s and also processed to yield a PSD (third figure). The TIA's PSD is the average of 1000 8k FFT PSDs, the RC LPF's PSD is the average of just 100 8k FFT PSDs (Fig. 1B shows the data collection only, but not the separate FFT PSD processing), and they are all unilateral and mean-subtracted.
The fourth figure shows the raw TIA output (green), the RC LPF output (cyan) and the sampled RC LPF output (red). The quantization at low photon flux is clearly shown in the TIA output
Increasing the photon flux to 20 photons/$\mu$s gives the results in the fifth figure. Now the quantization is not so bad and the signal-to-noise ratio is higher, as expected. Increasing the photon flux further makes it reasonable to replace the Poisson distribution with its ‘envelope’ Gaussian distribution (not shown here). Hope this helps a bit: there is FAR more that can be said!
Note: I have corrected the first three figures to be consistent with the 'sample calculation' values given several paragraphs below. I will go through and update everything as needed or necessary!
The software I use is ExtendSim (commercial simulation software from Imagine That, Inc.), augmented with my free LightStone libraries of blocks (with full commented source code for all blocks). It is easy to construct time-series after the S&H and find the variance: I have been playing with just that since I saw your comment. Now for the specific question you ask, if photon flux is constant, at 2 photons/$\mu$s (= 20 photons/10$\mu$s), then the distinction is in the measurement interval, $\tau$. For the former, $\tau$ = 1 $\mu$s, so the mean number of photons in the measurement interval, m, is 2. For the latter, $\tau$ = 10 $\mu$s, m = 20 = 2 photons/$\mu$s x 10 $\mu$s. If the simulation step size, $\Delta$t, is constant at 1 $\mu$s, then the average signal magnitude in the larger $\tau$ will be 20/2 times larger than in the smaller $\tau$. The unilateral PSD will also be directly proportional to m, so it will be 10 times larger as well. Hence the variance at the output of the RC LPF will be 10 times larger (to the extent that the digital RC LPF ‘honors’ the analog RC LPF’s 1/4RC noise bandwidth), so the standard deviation will be $10^{1/2}$ times larger. So the SNR, as signal magnitude/noise standard deviation, is $10^{1/2}$ times larger, as expected for shot noise. By the way, in my field, SNRs are not usually done as signal power/noise power.
I ran the sims for the cases of m = 2 (with $\Delta$t = 1 $\mu$s) and m = 20 (with $\Delta$t = 1 $\mu$s and also with $\Delta$t = 10 $\mu$s), sampling (via the S&H) every 100 $\mu$s to get time-series. Each sim ran from 0 to 0.1 s and the S&H was sampled every 100 $\mu$s. I collected 1000 values for each series and discarded the data for the first 500 $\mu$s, because an RC LPF settles to 1% in 4.605 time constants. So each series is 995 S&H values, collected at 100 $\mu$s intervals.
Results summary: For m = 2, $\Delta$t = 1 $\mu$s, and $\tau$ = 1 $\mu$s, the average = 0.000807 V and standard deviation = 0.0000399 V. For m = 20, $\Delta$t = 10 $\mu$s, and $\tau$ = 10 $\mu$s, the average = 0.00808 V and the standard deviation = 0.000406 V. For m = 20, $\Delta$t = 1 $\mu$s, and $\tau$ = 10 $\mu$s, the average = 0.00807 V and the standard deviation = 0.000130 V. Note that 0.000130 V/0.0000399 V = 3.258, which is approximately $10^{1/2}$. N.B.: the ‘volt' units are fictitious, of course, since this is just a simulation!
Sample calculation: The unilateral PSD, at the output of the TIA, is m photons x 2 x $(0.0004037 V/photon)^2$ x $\Delta$t. For m = 2, this yields 6.52 x $10^{-13}$ $V^2$/Hz. Then the variance at the output of the digital RC LPF (if the noise bandwidth was actually 1/4RC), would be 6.52 x $10^{-13}$ $V^2$/Hz x 1 x 1/400 $\mu$s = 1.63 x $10^{-9}$ $V^2$, so the standard deviation would be 4.04 x $10^{-5}$. This is close to the above 3.99 x $10^{-5}$ V. Hope this helps.
Added 5/14/2019: For m = 2, $\Delta$t = 1 $\mu$s, and $\tau$ = 1 $\mu$s, I collected 100 files of 8k each, neglecting the first 1 ms (i.e., 10 RC times constants, to generously allow for settling of the RC LPF). I did this for 10 kHz sampling of the S&H and again for 100 kHz sampling of the S&H. For the 10 kHz sampling, the average variance of the 100 8k time series was 1.628 x $10^{-9}$ V and the PSD is shown below:
From the first PSD, the rms value was 4.035 x $10^{-5}$ V. This is the same, to 4 digits, as the square root of the variance.
For the 100 kHz sampling, the average variance of the 100 8k time series was 1.632 x $10^{-9}$ V and the PSD is shown below:
From the second PSD, the rms value was 4.039 x $10^{-5}$ V. The square root of the variance was 4.040 x $10^{-5}$ V. |
Do we take gravity = 9.8 m/s² for all heights when solving problems?
No, the value $9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ is an approximation that is only valid at or near the Earth's surface. You can go a few miles up or down and it'll still be good enough, but once you get any significant distance away from the surface of Earth, you would need to use a different value for gravitational acceleration. You can calculate the value from Newton's law of gravitation, $F = Gm_1m_2/r^2$, and you'll get
$$g = \frac{GM}{r^2} = \frac{3.99\times 10^{14}\ \mathrm{m^3/s^2}}{r^2}$$
where $M$ is the mass of the Earth and $r$ is the distance from the Earth's
center to the point for which you are doing the calculation.
To expand a little on David's point assume we move from the nominal "surface" where $g$ is $9.8\text{ m}/\text{s}^2$ to another point at radius $r + \Delta r$. How much does the acceleration of gravity change? $$ g = \frac{GM}{(r+\Delta r)^2} = \frac{GM}{r^2(1 + \Delta r/r)^2} $$ and as long as $\Delta r$ is small compared to $r$ we can reasonably approximate this as $$ g \approx \frac{GM}{r^2}\left(1 - 2\frac{\Delta r}{r}\right) . $$ Well, the radius of the Earth is about $6000 \text{ km}$ so the approximation is good at less than 1% error for around $30\text{ km}$ up or down from the nominal surface, which is all the land and sea floor, and a bit up and down from there.
It is also worth noticing that due to variations in the local mass density of the Earth the measured value of $g$ even at the surface can vary by several tenth of a percent.
$g$ becomes $ g \approx 9.7 \frac{m}{s^2}$ at a height of about 35km, so it would be ok to use the value $9.81$ for "down to earth" problems.
The relevant wikipedia article has lots of useful information, like for example the following approximation formula for different heights: $$ g_h=g_0\left(\frac{r_e}{r_e+h}\right)^2 $$ Where $g_h$, is the gravity measure at height $h$ above sea level; $r_e$, is the Earth's mean radius and $g_0$, is the standard gravity.
It might also be worth mentioning that $g$ isn't even constant over the earth's surface at sea level. Depending on the mass distribution and the shape (not perfectly spherical!) of the earth, different parts of the world have different $g$.
The approximation of 9.81 m/s^2 is a generalisation. The exact value is most likely different at a specific location, due to the distance from the centre of the earth to the point being evaluated.
The reference to "surface of the earth" is also a relative since the earth is known not to be perfectly round due to centrifugal forces making the radius greater at the equator.
Also, since the earth is spinning the same centrifugal forces have a slight influence on object mass at the evaluation point.
In metrology laboratories, the exact value for g is displayed for that exact location.
Acceleration due to gravity, g is not a universal constant like G. Its calculated by formula mentioned in previous answers. So, for a constant mass system, g depends only on r (distance between center of earth & object in problem). As r = R + h (R is radius of earth & h is height of object from surface) & R is constant, g depends mainly on height.
The relation: Increase the height, g will become less (as per formula) The value 9.8 m/s² is valid for the object at the surface of earth (at sea level). When height is small (with respect to radius of earth), the value is slightly less than 9.8 m/s². So, this variation can be neglected for a high school etc problems. When accuracy is important (due to scientific reasons etc), the value of g can't be 9.8 m/s².
Once again, This consideration is valid only for constant mass system. Plus, for relativistic systems, the formula isn't valid with constant space & time scale.
As we go above or below the surface of earth the value of g decreases since g is inversely proportional to height
protected by Qmechanic♦ Aug 19 '15 at 10:08
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During a discussion with my professor, he asked me about relativistic resistance transformation. Firstly I started with the formula $R=\frac{\rho L}{s}$, where $\rho$ is electric resistivity. So if some resistor moves with relativistic speed its length contracts. And this means that its resistance changes like its length. Its resistivity doesn't change because of Ohm's law $E=\rho I$. Current changes like an electric field and that means that $\rho=\text{const}$. But then the professor said that to find this transformation I can't use the first formula, and I should find it from some laws that have resistance in it. So I used $$P=I^2R \tag{2}$$ and $$R=\frac{U}{I} \tag{3}.$$ From formula (3) we find that resistance doesn't change because difference of potentials and current change the same way. But from formula (2) we find that resistance changes like $\frac{R}{\gamma^3} $ (where $\gamma$ is Lorentz factor). So basically I have three variants, but the professor said that I need to find one result. Any ideas?
Repeating what @Dale said in a slightly different way, if you want relativistic laws, it is better to work with relativisitic quantities.
The response of matter to applied electromagnetic field is generally expressed through charge density ($\rho$) and current density ($\mathbf{J}$). The 4D equivalent is the four-current:
$J^\mu=\left(c\rho, \mathbf{J}\right)^\mu$
Next, the electric and magnetic fields are not covariant quantities in SR, so use the electromagnetic tensor: $F^{\mu\nu}$ (https://en.wikipedia.org/wiki/Electromagnetic_tensor)
Now, conductivity, which in general is a tensor, arises as a result of postulating a linear relationship between the applied electric field and the current density.
So we could say:
$J^\mu=\sigma^{\mu}_{\nu\eta}F^{\nu\eta}$
Now this general definition hoovers up pretty much all the simple types of electromagnetic response. But that's the point, what looks like conductivity response to one observer will look very different to another one.
You could postulate that in your rest frame ($\bar{S}$) the response is solely that of the isotropic conductivity ($\sigma$), then the only non-zero components are ($c$ is the speed of light):
$\bar{\sigma}^{1}_{01}=-\bar{\sigma}_{10}=c\sigma/2$
$\bar{\sigma}_{02}=-\bar{\sigma}_{20}=c\sigma/2$
$\bar{\sigma}_{03}=-\bar{\sigma}^{3}_{30}=c\sigma/2$
Then consider the lab-frame relative to which the rest-frame is moving along x with speed $v$. You then transform your tensor $\sigma^\mu_{\eta\nu}=\frac{\partial x^\mu}{\partial \bar{x}^\alpha}\frac{\partial \bar{x}^\beta}{\partial x^\eta}\frac{\partial \bar{x}^\kappa}{\partial x^\nu}\bar{\sigma}^\alpha_{\beta\kappa}$. The only components to change are the ones involving x:
I get
$\sigma^{0}_{10}=-\sigma^{0}_{01}=\left(\frac{v}{c}\right)\gamma\cdot\frac{c\sigma}{2}$
$\sigma^{1}_{10}=-\sigma^{1}_{01}=\gamma\cdot\frac{c\sigma}{2}$
So you could say that in lab-frame the conductivity in x direction will change by $\gamma=1/\sqrt{1-\left(v/c\right)^2}$. This is the second equation, but that's not the end of the story. There is the first equation, which in non-relativistic terms would lead to:
$\rho=\frac{v\sigma}{c^2}\gamma\cdot E_x$
Where $E_x$ is the x-component of the electric field. So you would see a build-up of charge density due to applied electric field.
So, as you now see, the picture is more complex than just finding an equation of resistance $R$
So basically I have three variants, but professor that I need to find one result. Any ideas?
Circuit theory is inherently nonrelativistic and so there is no relativistic circuit theory. This should actually not be surprising since circuit theory does not use the concept of space at all and relativity is a theory of spacetime.
The issue is in the assumptions that circuit theory is based on. The three assumptions are: (1) there is no net charge on any circuit element, (2) there is no magnetic flux outside of any circuit element, (3) the circuit is small so that electromagnetic influences can be assumed to propagate instantaneously.
Assumption (3) is expressly prohibited by relativity, and since a current density in one frame is both a current and a charge density in another then (1) will generally not be satisfied in all frames. Therefore, it is fatally flawed to attempt to find a relativistic version of Ohms law based on the equations of circuit theory since those laws are inherently non relativistic.
However, you could take the form of Ohm’s law that is based directly on Maxwell’s equations: $\mathbf J = \sigma \mathbf E$. This law is based on Maxwell’s equations, which are completely relativistic. Since we know how $\mathbf J$ and $\mathbf E$ both transform then we can calculate how $\sigma$ transforms. |
There is a problem in checking whether the homography is OK.
The algorithm for checking correct homographies may interest someone, so I will write it down here:
1) Create a quadrilateral $ABDC$ with vertex coordinates (in homogenous coordinates):
$$\begin{eqnarray} A:& (-w/2,-h/2, 1.0) \\ B:& (w/2,-h/2, 1.0) \\ C:& (-w/2,h/2, 1.0) \\ D: &(w/2,h/2, 1.0) \end{eqnarray}$$
where $w,h$ are width and height of the image, respectively. If the whole image frame (a rectangle) is transformed to convex quadrilateral, than any convex quadrilateral within it will also be transformed "neatly".
2) Create transformed quadrilateral $A'B'D'C'$ in which every vertex is transformed using the computed homgraphy (e.g. $C'=HC$ ). From now on, all points will be converted to non-homogenous coordinates.
3) Compute vectors $\vec{u}$, $\vec{v}$ for parametric representation of diagonals:
$$\begin{eqnarray}d_{1} :& A+(D-A)s =A+ \vec{u}s \\ d_{2} :& B + (C-B)t=B+\vec{v}t \end{eqnarray}$$
The intersection of diagonal comes from $d_{1}=d_{2}$:
$$t=\frac{1}{d}\left[(B_{y}-A_{y})\vec{u}_{x} - (B_{x}-A_{x})\vec{u}_{y}\right]$$
$$s=\frac{1}{d}\left[(A_{x}-B_{x})\vec{v}_{y} - (A_{y}-B_{y})\vec{v}_{x}\right]$$
Then the convex quadrilateral satisfies $s,t \in (0,1)$.
In practice, one can introduce a
fudge factor to avoid not only non-convex and degenerate quadrilaterals, but also near-degenerate ones, like when three points are near colinear. So the test can be modified such that $s,t \in (\lambda, 1.0-\lambda)$, where lambda is a small number (e.g. $\lambda=0.01$).
Older problem, fixed in the above algorith:
I found the problem here - having a certain homography, the test can pass for a smaller quadrilateral, but not for the larger one. This is why some "ill" homographies passed through.
The green square represents a source image, the orange is a transformed one. As you can see, the left hand one is convex, but starts deforming as the source is larger:
Finally even larger source yield non conver quadrilateral:
I made a mistake with scaling. The points in homogenous coordinates $(x,y,w)$ were scaled in $x$ and $y$ direction, but in $w$. This is why the same transform produced different quadrilaterals.
I have corrected the algorithm accordingly. |
In what follows, we have a level $N \geq 3$, and the modular curve $X(N)$, and the invertible sheaf $\omega$ on $X(N)$ such that the global sections of $\omega^{\otimes k}$ correspond to modular forms of weight $k$ and level $N$.
In this letter, Serre looks at an exact sequence of sheaves:
$0 \rightarrow \omega^{k - p + 1} \rightarrow \omega^k \rightarrow \mathcal{S}_k \rightarrow 0$
where the second arrow is given by the multiplication with the Hasse invariant $A$, which is a modular form of weight $p-1$.
Question 1. Serre says that since $A$ vanishes with multiplicity 1 at the supersingular points, it follows that $\mathcal{S}_k$ is 0 outside of the supersingular points, and of dimension 1 at the supersingular points. What does he mean by that exactly? Does he means the stalks $\mathcal{S}_k, x$ at various points $x$ are either ${0}$ or $1$-dim $\mathbb{F}_p$-vector spaces? Question 2. We let $S_k$ denote the global sections of $\mathcal{S}_k$. Serre shows that $S_k$ depends only on $k \pmod{p^2 - 1}$. He does it in the following way (or how i understand it): $S_k$ is supported at the supersingular locus, so he looks at the stalk at a point corresponding to a supersingular elliptic curve $E$. He says that $E$ has a canonical (and functorial) structure on $\mathbb{F}_{p^2}$ such that the Frobenius acts as $-p$. This means that the tangent space to $E$ also has a canonical $\mathbb{F}_{p^2}$ structure, and its $p^2-1$ tensor power has a canonical basis. This basis lets us identify the global differentials on $E$ of order $k$ with those of order $k + p^2 - 1$, and that this identification is respected by isogenies.
What I don't understand, is what he means here by "canonical basis", or how this basis lets us identify these two spaces of differentials $\omega^k(E)$ and $\omega^{k+p^2 -1}(E)$. Is it only $\omega^{p^2-1}(E)$ which has a "canonical" basis? Or why can't we simiarly identify $\omega^k$ and $\omega^{k'}$ for random values of $k$ and $k'$?
I would appreciate it if someone can explain this part of the proof to me.
Edit: This proof also appears in Edixhoven's paper on the weights in Serre's conjecture, and in Alex Ghitza's PhD thesis, but the explanations there do not give any further clarifications, they just repeat what Serre says. So maybe I am missing something very easy? |
In this question it is shown that being able to compute reciprocals (together with sums and differences) is enough to do do multiplication in a field of characteristic $\ne 2$. That made me wonder:
Can we formulate a set of field axioms that are based on the reciprocal function rather than a multiplication operation? Something like
A
fieldis an abelian group $G$ (written additively) together with a designated element $1\ne 0$ and an involution $x\mapsto\frac{1}{x}$ of $G\setminus\{0\}$ such that $\frac 11=1$ ???
Obviously we could derive a sufficient set of axioms by translating the usual field axioms using the algorithm from the earlier question, but that would end up being very ugly (not least because there are several cases that one must handle specially in order not to divide by zero). Is there a nicer way to characterize involutions that arise as the reciprocal of some field?
Partial answers (for example, ones that work only for characteristic 0) would also be interesting. |
If we arbitrarily choose a point from each orthant in $\mathbb{R}^n$, that is we choose $2^n$ points in total, how do we prove that 0 is in the convex hull of these $2^n$ points? It seems obvious, but when I sit down and start thinking about it, I couldn't definitely find a set of $\lambda_i, i = 1,2,\cdots,2^n$ such that $0 \leq \lambda_i \leq 1, \sum_i \lambda_i = 1$ and $0 = \sum_i \lambda_i x_i$, where $x_i, i = 1,2,\cdots,2^n$ are any set of points satisfying the condition.
Induct on dimension.
In the $n=1$ base case, you have a positive number and a negative number; $0$ can be represented as a convex combination of them.
For $n>1$: Given your $2^n$ points, one from each orthant, divide them into 2 sets of size $2^{n-1}$, where the first set has points whose last coordinates are positive and the second set where the last coordinates are negative. By the inductive hypothesis there is a convex combo of the first set of points such that the first $n-1$ coordinates vanish, and similarly for the second set. The last coordinates of these two convex combos are positive and negative, so there is a convex combo of them that is zero.
Take in pairs the points that differ in sign for the coordinate $n$ (and only that one) and form the convex combinations such that that coordinate is cancelled.
Now you have $2^{n-1}$ points in each orthants of the diminished space.
The results holds because the convex combination of linear combinations is a convex combination. |
Extraordinary claims require extraordinary proofs which really is the reason why this sorts of discussion is important. Similarly, sometimes, you are so blinded to some sorts of a truth and are faced with something so different that you can misread entirely what is being said. If you read this morning's entry, you might get a feel that I am little ambivalent about the true interesting nature of a paper entitled Statistical physics-based reconstruction in compressed sensing by Florent Krzakala, Marc Mézard, François Sausset, Yifan Sun, Lenka Zdeborová. Let's put this in perspective, our current understanding so far is that the universal phase transition observed by Donoho and Tanner seems to be seen with all the solvers featured here, that there are many ensembles for which it fits (not just Gaussian, I remember my jaw dropping when Jared Tanner showed it worked for the ensembles of Piotr Indyk, Radu Berinde et al) and that the only way to break it is to now consider structured sparsity as shown by Phil Schniter at the beginning of the week. In most people's mind, the L_1 solvers are really a good proxy to the L_0 solvers since even greedy solvers (the closest we can find to L_0 solvers) seem to provide similar results. Then there are results like the ones of Shrinivas Kudekar and Henry Pfister. ( Figure 5 of The Effect of Spatial Coupling on Compressive Sensing) that look like some sort of improvement (but not a large one). In all, a slight improvement over that phase transition could, maybe, be attributed to a slightly different solver, or ensemble (measurement matrices). So this morning I made the point that given what I
understoodabout the graphs displayed in the article, it may be at besta small improvement over the Donoho-Tanner phase transition known to hold for not only Gaussian but other types of matrices and for different kinds of solvers, including greedy algorithms and SL0 (that simulate some sorts of L_0 approach). At bestis really an overstatement but I was intrigued mostly because of the use of an AMP solver, so I fired off an inquisitive e-mail on the subject to the corresponding author: Dear Dr. Krzakala, ... I briefly read your recent paper on arxiv with regards to your statistical physics based reconstruction capability and I am wondering if your current results are within the known boundary of what we know of the phase transition found by Donoho and Tanner or if it is an improvement on it. I provided an explanation of what I meant in today's entry (http://nuit-blanche.blogspot.com/2011/09/this-week-in-compressive-sensing.html). If this is an improvement, I'd love to hear about it. If it is is not an improvement, one wonders if some of the deeper geometrical findings featured by the Donoho-Tanner phase transition have a bearing on phase transition on real physical systems.Best regards,Igor.
The authors responded quickly with:
Dear Igor, Thanks for writing about our work in your blog.Please notice, however, that our axes in the figure you show are not the same as those of Donoho and Tanner. For a signal with N components, we define \rho N as the number of non-zeros in the signal, and \alpha N as the number of measurements. In our notation Donoho and Tanner's parameters are rho_DT = rho/alpha and delta_DT = alpha. We are attaching our figure plotted in Donoho and Tanner's way. Our green line is then exactly the DT's red line (since we do not put any restriction of the signal elements), the rest is how much we can improve on it with our method. Asymptotically (N\to \infty) our method can reconstruct exactly till the red line alpha=rho, which is the absolute limit for exact reconstruction (with exhaustive search algorithms).So we indeed improve a lot over the standard L1 reconstruction!We will be of course very happy to discuss/explain/clarify details if you are interested.With best regardsFlorent, Marc, Francois, Yifan, and Lenka
The reason I messed up reading the variables is because I was probably not expecting something that stunning.
Thank you for Florent Krzakala, Marc Mézard, François Sausset, Yifan Sun, Lenka Zdeborová for their rapid feedback.
Liked this entry ? subscribe to the Nuit Blanche feed, there's more where that came from |
I would appreciate if somebody could run this over and see if it works out? any suggestions or pointers would be appreciated. I denote the standard eta function $\eta$ by $\zeta^{*}$. I have not used big O notation and just used general well behaved functions. I do not wish to express the full error term, but instead ,just the principal part.
Behaviour of $\zeta(s)$ near $1$
From Abel's Theorem we can see that when $s=1$, $ \zeta^{*}(1) = \log(2)$. Now looking at $(1-2^{1-s})$ we can write it in terms of an exponential like so, \begin{equation} 1-2^{1-s} = 1 - e^{(1-s)\log(2)} \end{equation}
The power series expansion of $e^{z}$ is,
\begin{equation} e^{z}= \sum_{n=0}^{\infty} \frac{z^{n}}{n!}\\ \Rightarrow 1-2^{1-s} = - e^{\log(2)(s-1)}= - \sum_{n=0}^{\infty} \frac{((1-s)\log(2))^{n}}{n!} \end{equation}
We can ignore the term when $n=0$ due to it being zero and sum from $n=1$ instead,
\begin{equation} 1-2^{1-s} = 0 - \sum_{n=1}^{\infty} \frac{(1-s)^{n}\log(2)^{n}}{n!} \end{equation}
Expanding this sum and multiplying in the negative sign we have,
\begin{equation} 1-2^{1-s}= (s-1) \Bigg( \log(2) - \frac{\log(2)^{2}}{2!}(s-1) + \cdot \cdot \cdot \Bigg ) \end{equation}
Factorizing the $\log(2)$ term out, \begin{equation*} (s-1)\log(2)\Bigg [ 1 - \bigg( \frac{\log(2)}{2!}(s-1) + \frac{\log(2)}{3!}(s-1)^{2} - \cdot \cdot \cdot \bigg ) \Bigg ] \end{equation*}
By the geometric series formula, for $|s| < 1$, \begin{equation} \frac{1}{\bigg[1 - \bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \bigg ) \bigg ] }= 1 + \Bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \Bigg ) + \Bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \Bigg )^{2} + \cdot \cdot \cdot \end{equation} The terms of this geometric series decrease rapidly, so we are only interested in keeping the first terms while letting a well-behaved and analytic function $g$ represent the remaining terms as a function in $s$.
\begin{equation} \frac{1}{\bigg[1 - \bigg( \frac{\log(2)}{2!}(s-1) + \cdot \cdot \cdot \bigg ) \bigg ] } = 1 + \frac{\log(2)(s-1)}{2} + (s-1)^{2}\cdot g(s). \end{equation}
We can now return to $\frac{1}{1-2^{1-s}}$, and express it in terms of what we have learned. \begin{equation} \frac{1}{1-2^{1-s}} = \frac{1}{\log(2)(s-1)} \Bigg( 1 + \frac{\log(2)(s-1)}{2} + (s-1)^{2}\cdot g(s) \Bigg ) = \frac{1}{\log(2)} \cdot \Bigg [ \frac{1}{s-1} + \frac{\log(2)}{2} + (s-1)g(s)\Bigg ] \end{equation}
We can now study $\zeta(s)$ when $s$ is near to $1$. \begin{equation} \zeta(s) = \frac{\zeta^{*}(s)}{1-2^{1-s}} = \frac{\zeta^{*}(s)}{\log(2)} \cdot \Bigg [ \frac{1}{s-1} + \frac{\log(2)}{2} + (s-1)g(s)\Bigg ] = \frac{\zeta^{*}(s)}{\log(2)} \cdot \frac{1}{s-1} + \frac{\zeta^{*}(s)}{2 \log(2)} \log(2) + \frac{\zeta^{*}(s)(s-1)g(s)}{\log(2)} \end{equation}
As we know already, $\zeta^{*}(1) = \log(2)$ is analytic, so $\zeta^{*}(s)$ can be expanded as a series around $1$, \begin{equation} \zeta^{*}(s) = \log(2) + (s-1)a_1 + (s-1)^{2}a_2 + \cdot \cdot \cdot = \log(2) + (s-1) h(s) \end{equation} for a well behaved and analytic $h$.
Near $s=1$ and by just looking at the principal terms, \begin{equation} \zeta(s) = \frac{\zeta^{*}(s)}{1-2^{1-s}} = \frac{ \log(2) +(s-1)h(s) }{\log(2)(s-1)} = \frac{1}{s-1} + \frac{h(s)}{\log(2)} \end{equation} |
What I want to achieve
The framework ROOT can create 2D histogram plots with colored boxes indicating count rate that looks something like:
My question is really only:
can I produce this kind of 2D histogram through PGFPlots? The rest of this post describes my current findings and attempts.
As of writing quite recently, ROOT got a TikZ output engine called
TTeXDump, that in turn would generate:
This is closing in on good graphical quality. The axis label texts are easily manually modified to TeX syntax (or it could be done in ROOT before the export), but there are other issues:
Placement of labels, ticks, etc., are all done by raw coordinates (sample TikZ code output from
TTeXDumpdescribing the above image). To e.g. center the
xlabel below the axis is thus not trivial, which makes it tricky to conform to a graphical layout that coincides with other plots made directly with PGFPlots.
Since all graphical entities are statically defined, scaling will not yield transparent results.
and probably other things.
Own attempts
I have made some attempts at generating the plot with PGFPlots from exported ROOT data. There are, however, several details that I do not get right, and perhaps there is a more obvious solution.
By dumping the histogram bin data in the form
xcenter ycenter weight
in the file
scatter.csv (Pastebin link of data), the following code gives the below result:
\documentclass{article}\usepackage{tikz}\usepackage{pgfplots}\pgfplotsset{compat=newest}\usepackage{siunitx}\begin{document}\begin{tikzpicture} \begin{axis}[ xlabel={$\theta$ /($\pi$ rad)}, ylabel={Energy /\si{\MeV}}, enlarge x limits=.02, enlarge y limits=.02, minor tick num=4, xticklabel style={/pgf/number format/fixed},% exponential axis notation looks bad in this case colorbar, scatter/use mapped color={% draw=mapped color, fill=mapped color }] \addplot[ scatter, scatter src=explicit, only marks, mark=square*, ] file{scatter.csv}; \end{axis}\end{tikzpicture}\end{document}
Fine-tuning of text, tick marks, colormap, etc., is easily done after this, but there are some issues with the data rendering:
The "bin" dimensions are emulated by the marker size, but it would be nice to input the bin numbers directly to set the marker sizes. This would also need to be done asymmetrically, since there will be a different amount of bins in the
xand
ydirection. I have looked at
mark=cube*with
cube/size xand
cube/size y, but have not successfully been able to change the mark dimensions. Currently, the marks are symmetric, which might look alright at a quick glance, but actually marks overlap in non-trivial ways and it is a deal breaker in itself.
The
enlarge x axisand
enlarge y axisvalues are inserted after inspection, to avoid the marks from protruding outside the axis. Rather, the axis distance should be automatically calculated from the marker size.
The data point markers are on top of the axis and tick marks, which is not optimal here. Force “axis on top” for plotmarks in pgfplots has some info on this, with somewhat convoluted solutions. Is there an even better way to be found here?
Is the scatter plot approach taken above perhaps the wrong one? Some other thoughts have been:
Perhaps I could use the scatter plot data directly and perform the binning with PGFPlots instead of exporting calculated bins. I can not find a way to do this though, and there is a potential risk of running into the memory limit.
The initial ROOT output PDF could be stripped of axis, tick marks and titles, keeping only the graph surface. Then I could use
\addplot figureto include this and paint the axis back on with PGFPlots. Numerical axis limits could be extracted from ROOT, so the scale should be able to be correctly reproduced. I have not looked into calibrating the colormap scale in PGFPlots from max/min values, but that should also be possible. There would perhaps be some alignment issues to solve.
It would help automation if I could use the
TTeXDumpoutput, strip the statically defined axis, ticks, etc., and just use the generated TikZ commands for painting the graph body. I can not see a trivial way to combine this with
\addplotthough.
The data output could have been defined in bin numbers instead of explicit coordinates, i.e.
1 10 11.0
for
xbin 1,
ybin 10 with value 11, instead of the current:
0.04580479262184749 0.0755985979686503 11.0
Since we also have the axis limits and bin amounts defined, this should really be all the info we need to build the histogram, but I do not find it to be trivial to perform.
Conclusion
That might seem like a lot of questions, but as mentioned initially, the kernel is really only a single one:
can I produce this kind of 2D histogram through PGFPlots? |
Spectral internal transmittance τ
i (λ) is the ratio of the light intensities at the outlet Φ out (λ) and inlet Φ in (λ) of the flow cell.
$\tau_{i} \left( \lambda \right) = \frac{\phi_{\text{out}} \left( \lambda \right) }{\phi_{\text{in}} \left( \lambda \right) }$
Fig. 51: Internal spectral transmittance
Unlike the transmittance, in this case the reflection losses occurring at the flow cell windows have to be eliminated. In practice, this is done by running a comparative measurement of the carrier medium without the absorbent product. |
One disadvantage of the fact that you have posted 5 identical answers (1, 2, 3, 4, 5) is that if other users have some comments about the website you created, they will post them in all these place. If you have some place online where you would like to receive feedback, you should probably also add link to that. — Martin Sleziak1 min ago
BTW your program looks very interesting, in particular the way to enter mathematics.
One thing that seem to be missing is documentation (at least I did not find it).
This means that it is not explained anywhere: 1) How a search query is entered. 2) What the search engine actually looks for.
For example upon entering $\frac xy$ will it find also $\frac{\alpha}{\beta}$? Or even $\alpha/\beta$? What about $\frac{x_1}{x_2}$?
*******
Is it possible to save a link to particular search query? For example in Google I am able to use link such as: google.com/search?q=approach0+xyz Feature like that would be useful for posting bug reports.
When I try to click on "raw query", I get curl -v https://approach0.xyz/search/search-relay.php?q='%24%5Cfrac%7Bx%7D%7By%7D%24' But pasting the link into the browser does not do what I expected it to.
*******
If I copy-paste search query into your search engine, it does not work. For example, if I copy $\frac xy$ and paste it, I do not get what would I expect. Which means I have to type every query. Possibility to paste would be useful for long formulas. Here is what I get after pasting this particular string:
I was not able to enter integrals with bounds, such as $\int_0^1$. This is what I get instead:
One thing which we should keep in mind is that duplicates might be useful. They improve the chance that another user will find the question, since with each duplicate another copy with somewhat different phrasing of the title is added. So if you spent reasonable time by searching and did not find...
In comments and other answers it was mentioned that there are some other search engines which could be better when searching for mathematical expressions. But I think that as nowadays several pages uses LaTex syntax (Wikipedia, this site, to mention just two important examples). Additionally, som...
@MartinSleziak Thank you so much for your comments and suggestions here. I have took a brief look at your feedback, I really love your feedback and will seriously look into those points and improve approach0. Give me just some minutes, I will answer/reply to your in feedback in our chat. — Wei Zhong1 min ago
I still think that it would be useful if you added to your post where do you want to receive feedback from math.SE users. (I suppose I was not the only person to try it.) Especially since you wrote: "I am hoping someone interested can join and form a community to push this project forward, "
BTW those animations with examples of searching look really cool.
@MartinSleziak Thanks to your advice, I have appended more information on my posted answers. Will reply to you shortly in chat. — Wei Zhong29 secs ago
We are open-source project hosted on GitHub: http://github.com/approach0Welcome to send any feedback on our GitHub issue page!
@MartinSleziak Currently it has only a documentation for developers (approach0.xyz/docs) hopefully this project will accelerate its releasing process when people get involved. But I will list this as a important TODO before publishing approach0.xyz . At that time I hope there will be a helpful guide page for new users.
@MartinSleziak Yes, $x+y$ will find $a+b$ too, IMHO this is the very basic requirement for a math-aware search engine. Actually, approach0 will look into expression structure and symbolic alpha-equivalence too. But for now, $x_1$ will not get $x$ because approach0 consider them not structurally identical, but you can use wildcard to match $x_1$ just by entering a question mark "?" or \qvar{x} in a math formula. As for your example, enter $\frac \qvar{x} \qvar{y} $ is enough to match it.
@MartinSleziak As for the query link, it needs more explanation, technologically the way you mentioned that Google is using, is a HTTP GET method, but for mathematics, GET request may be not appropriate since it has structure in a query, usually developer would alternatively use a HTTP POST request, with JSON encoded. This makes developing much more easier because JSON is a rich-structured and easy to seperate math keywords.
@MartinSleziak Right now there are two solutions for "query link" problem you addressed. First is to use browser back/forward button to navigate among query history.
@MartinSleziak Second is to use a computer command line 'curl' to get search results from particular query link (you can actually see that in browser, but it is in developer tools, such as the network inspection tab of Chrome). I agree it is helpful to add a GET query link for user to refer to a query, I will write this point in project TODO and improve this later. (just need some extra efforts though)
@MartinSleziak Yes, if you search \alpha, you will get all \alpha document ranked top, different symbols such as "a", "b" ranked after exact match.
@MartinSleziak Approach0 plans to add a "Symbol Pad" just like what www.symbolab.com and searchonmath.com are using. This will help user to input greek symbols even if they do not remember how to spell.
@MartinSleziak Yes, you can get, greek letters are tokenized to the same thing as normal alphabets.
@MartinSleziak As for integrals upper bounds, I think it is a problem on a JavaScript plugin approch0 is using, I also observe this issue, only thing you can do is to use arrow key to move cursor to the right most and hit a '^' so it goes to upper bound edit.
@MartinSleziak Yes, it has a threshold now, but this is easy to adjust from source code. Most importantly, I have ONLY 1000 pages indexed, which means only 30,000 posts on math stackexchange. This is a very small number, but will index more posts/pages when search engine efficiency and relevance is tuned.
@MartinSleziak As I mentioned, the indices is too small currently. You probably will get what you want when this project develops to the next stage, which is enlarge index and publish.
@MartinSleziak Thank you for all your suggestions, currently I just hope more developers get to know this project, indeed, this is my side project, development progress can be very slow due to my time constrain. But I believe its usefulness and will spend my spare time to develop until its publish.
So, we would not have polls like: "What is your favorite calculus textbook?" — GEdgar2 hours ago
@GEdgar I'd say this goes under "tools." But perhaps it could be made explicit. — quid1 hour ago
@quid I think that the type of question mentioned in GEdgar's comment is closer to book-recommendations which are valid questions on the main. (Although not formulated like that.) I also think that his comment was tongue-in-cheek. (Although it is a bit more difficult for me to detect sarcasm, as I am not a native speaker.) — Martin Sleziak57 mins ago
"What is your favorite calculus textbook?" is opinion based and/or too broad for main. If at all it is a "poll." On tex.se they have polls "favorite editor/distro/fonts etc" while actual questions on these are still on-topic on main. Beyond that it is not clear why a question which software one uses should be a valid poll while the question which book one uses is not. — quid7 mins ago
@quid I will reply here, since I do not want to digress in the comments too much from the topic of that question.
Certainly I agree that "What is your favorite calculus textbook?" would not be suitable for the main. Which is why I wrote in my comment: "Although not formulated like that".
Book recommendations are certainly accepted on the main site, if they are formulated in the proper way.
If there will be community poll and somebody suggests question from GEdgar's comment, I will be perfectly ok with it. But I thought that his comment is simply playful remark pointing out that there is plenty of "polls" of this type on the main (although ther should not be). I guess some examples can be found here or here.
Perhaps it is better to link search results directly on MSE here and here, since in the Google search results it is not immediately visible that many of those questions are closed.
Of course, I might be wrong - it is possible that GEdgar's comment was meant seriously.
I have seen for the first time on TeX.SE. The poll there was concentrated on TeXnical side of things. If you look at the questions there, they are asking about TeX distributions, packages, tools used for graphs and diagrams, etc.
Academia.SE has some questions which could be classified as "demographic" (including gender).
@quid From what I heard, it stands for Kašpar, Melichar and Baltazár, as the answer there says. In Slovakia you would see G+M+B, where G stand for Gašpar.
But that is only anecdotal.
And if I am to believe Slovak Wikipedia it should be Christus mansionem benedicat.
From the Wikipedia article: "Nad dvere kňaz píše C+M+B (Christus mansionem benedicat - Kristus nech žehná tento dom). Toto sa však často chybne vysvetľuje ako 20-G+M+B-16 podľa začiatočných písmen údajných mien troch kráľov."
My attempt to write English translation: The priest writes on the door C+M+B (Christus mansionem benedicat - Let the Christ bless this house). A mistaken explanation is often given that it is G+M+B, following the names of three wise men.
As you can see there, Christus mansionem benedicat is translated to Slovak as "Kristus nech žehná tento dom". In Czech it would be "Kristus ať žehná tomuto domu" (I believe). So K+M+B cannot come from initial letters of the translation.
It seems that they have also other interpretations in Poland.
"A tradition in Poland and German-speaking Catholic areas is the writing of the three kings' initials (C+M+B or C M B, or K+M+B in those areas where Caspar is spelled Kaspar) above the main door of Catholic homes in chalk. This is a new year's blessing for the occupants and the initials also are believed to also stand for "Christus mansionem benedicat" ("May/Let Christ Bless This House").
Depending on the city or town, this will be happen sometime between Christmas and the Epiphany, with most municipalities celebrating closer to the Epiphany."
BTW in the village where I come from the priest writes those letters on houses every year during Christmas. I do not remember seeing them on a church, as in Najib's question.
In Germany, the Czech Republic and Austria the Epiphany singing is performed at or close to Epiphany (January 6) and has developed into a nationwide custom, where the children of both sexes call on every door and are given sweets and money for charity projects of Caritas, Kindermissionswerk or Dreikönigsaktion[2] - mostly in aid of poorer children in other countries.[3]
A tradition in most of Central Europe involves writing a blessing above the main door of the home. For instance if the year is 2014, it would be "20 * C + M + B + 14". The initials refer to the Latin phrase "Christus mansionem benedicat" (= May Christ bless this house); folkloristically they are often interpreted as the names of the Three Wise Men (Caspar, Melchior, Balthasar).
In Catholic parts of Germany and in Austria, this is done by the Sternsinger (literally "Star singers"). After having sung their songs, recited a poem, and collected donations for children in poorer parts of the world, they will chalk the blessing on the top of the door frame or place a sticker with the blessing.
On Slovakia specifically it says there:
The biggest carol singing campaign in Slovakia is Dobrá Novina (English: "Good News"). It is also one of the biggest charity campaigns by young people in the country. Dobrá Novina is organized by the youth organization eRko. |
Research talks;Number Theory
For a non-principal Dirichlet character $\chi$ modulo $q$, the classical Pólya-Vinogradov inequality asserts that
$M (\chi) := \underset{x}{max}$$| \sum_{n \leq x}$$\chi(n)| = O (\sqrt{q} log$ $q)$. This was improved to $\sqrt{q} log$ $log$ $q$ by Montgomery and Vaughan, assuming the Generalized Riemann hypothesis GRH. For quadratic characters, this is known to be optimal, owing to an unconditional omega result due to Paley. In this talk, we shall present recent results on higher order character sums. In the first part, we discuss even order characters, in which case we obtain optimal omega results for $M(\chi)$, extending and refining Paley's construction. The second part, joint with Alexander Mangerel, will be devoted to the more interesting case of odd order characters, where we build on previous works of Granville and Soundararajan and of Goldmakher to provide further improvements of the Pólya-Vinogradov and Montgomery-Vaughan bounds in this case. In particular, assuming GRH, we are able to determine the order of magnitude of the maximum of $M(\chi)$, when $\chi$ has odd order $g \geq 3$ and conductor $q$, up to a power of $log_4 q$ (where $log_4$ is the fourth iterated logarithm).
For a non-principal Dirichlet character $\chi$ modulo $q$, the classical Pólya-Vinogradov inequality asserts that $M (\chi) := \underset{x}{max}$$| \sum_{n \leq x}$$\chi(n)| = O (\sqrt{q} log$ $q)$. This was improved to $\sqrt{q} log$ $log$ $q$ by Montgomery and Vaughan, assuming the Generalized Riemann hypothesis GRH. For quadratic characters, this is known to be optimal, owing to an unconditional omega result due to Paley. In this talk, we ...
11L40 ; 11N37 ; 11N13 ; 11M06
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2.Prove that no number in this sequence 11,111,1111,11111,........... is a perfect square 2. \text{Prove that no number in this sequence } 11,111,1111,11111,........... \\ \text{ is a perfect square} 2.Prove that no number in this sequence 11,111,1111,11111,........... is a perfect square
Give different methods of solving this. This was a problem to the OPC 2. The OPC 2 has already ended.
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Don't get me wrong, but I feel like most of the problems are from "an excursion in mathematics"
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Well I have heard a lot about that book. But these problems I have thought up and one is a famous one from IMO.
OK Allright :)
write
1⋯1⏟n times=a2 \underbrace {1 \cdots 1} _ {n \textrm{ times}}= a^2n times1⋯1=a2
Of course aaa must be odd (and greater than 111 since our sequence begins with 111111, in other words n≥2n \geq 2n≥2 ).Consider the even b=a−1b = a -1b=a−1 . We have
1⋯1⏟n times=b2−2b+1 \underbrace {1 \cdots 1} _ {n \textrm{ times}}= b^2 -2b +1n times1⋯1=b2−2b+1
1⋯1⏟n−1 times×10=b2−2b \underbrace {1 \cdots 1} _ {n-1 \textrm{ times}}\times 10= b^2 - 2bn−1 times1⋯1×10=b2−2b
bbb is even so 444 divides the LHS, but clearly not th RHS. Contradiction.
The numbers in the sequence can be written as 111....108+3111....108+3111....108+3 now since the last two digits, i.e. 080808, is divisible by 444, the numbers are of the form of 4k+34k+34k+3. Now note that a perfect square, say a2a^2a2 when written in modulo 444 ⟹ a2=0,1,2(mod4)\implies a^2=0,1,2\pmod4⟹a2=0,1,2(mod4) but not =3(mod4)=3\pmod4=3(mod4) which contradicts to the form of the numbers in the sequence.
any no will have last two digits 11 and therefore cannot be a perfect square .i hope u get it .if u dont just ask
You mean any number with last 2 digits 11, 31, 51, 71, 91 cannot be a perfect square?
Yup, It can't be.
absolutely.
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Here is one definition of the proportional odds model: $$\frac{S(t;{\bf Z})}{1-S(t;{\bf Z})}=e^{-{\bf Z}^T\pmb{\beta}}\frac{S_0(t;{\bf Z})}{1-S_0(t;{\bf Z})},$$ where ${\bf Z}$ = covariates, $\pmb{\beta}$ = regression coefficients, $T$ = survival time, $S(\cdot)$ = survival function, $S_0(\cdot)$ = baseline survival function. This equation is clearly equivalent to $\mbox{logit}[S(t;{\bf Z})]=\mbox{logit}[S_0(t)]-{\bf Z}^T\pmb{\beta}$.
Let $H(t)=-\mbox{logit}[S_0(t)]$. Now the book I'm reading claims, "If $H(t)$ is strictly increasing, [the first equation] can be equivalently written as $H(T)=-{\bf Z}^T\pmb{\beta}+W$, where the random variable $W$ follows a standard logistic distribution".
Can someone explain to me why this statement is true? First of all, I know that by definition $H(t)$ is monotone increasing, but I don't see why we need $H(t)$ to be
strictly increasing. Then, it seems to me like the part where he says "$W$ follows a standard logistic distribution" is just plain wrong; why in the world should $-\mbox{logit}[S(t;{\bf Z})]$ follow a logistic distribution?! |
In David Chandler's 'intro to statistical mechanics' he states that for an ideal gas at high-temperature
$$ \langle n_j\rangle=\langle N\rangle\frac{e^{-\beta \epsilon_j}}{\sum e^{-\beta \epsilon_j}} $$
Which I can believe from intuition, but he losses me on the derivation.
Starting with the general form for the occupancy of a boson or Fermion gas:
$$ \langle n_j \rangle=[e^{\beta (\epsilon_j-\mu)} \pm 1]^{-1} $$
Then at high-temperature, $\beta \rightarrow 0$,
[ page 101 (b) ]
$$ e^{\beta(\epsilon_j-\mu)}>>1 $$
So the assumption can be made that
$$ \langle n_j \rangle=e^{-\beta (\epsilon_j-\mu)} $$
Which would make sense if $e^{\beta(\epsilon_j-\mu)}>>1$, but it sure seems like that $ e^{\beta(\epsilon_j-\mu)}\approx 1$ in that case. Is there another assumption that's made here?
He does say:
Note that if this eqn is true for all $\epsilon_j$, then $-\beta \mu >>1 $
so, $\mu \rightarrow -\infty$ at high-temperature? That doesn't strike me as something that's obvious.
The rest of the derivation:
$$ \langle N \rangle = \sum \langle n_j \rangle =\sum e^{-\beta(\epsilon_j-\mu)}=e^{\beta \mu} \sum e^{-\beta \epsilon_j} \\ e^{\beta \mu} = \frac{\langle N \rangle}{\sum e^{-\beta \epsilon_j}} $$
using $\langle n_j \rangle = e^{-\beta(\epsilon_j - \mu)}$
$$ \langle n_j\rangle=\langle N\rangle\frac{e^{-\beta \epsilon_j}}{\sum e^{-\beta \epsilon_j}} $$ |
I am thinking I can't be the only one encountering this. I am trying to do maximum likelihood estimation on a probit model, i.e. trying to find the most optimal fit for three parameters in my case through the following equation:
$$l\left(\beta; y, X\right) = \sum_{i=1}^N \left[y_i \ln\left(F\left(x_i \beta\right)\right) + \left(1 - y_i\right) \ln \left(1-F\left(x_i \beta\right)\right)\right]$$
$y_i$ can take the value 0 or 1 depending on whether there has been an event or not, and then some form of probability inside the $\ln$ function. So in my case I am just computing a large grid of parameters (which is used as input into the $F$ function), and then doing this computation over $i$ number of cases.
However, my problem is that some parameter sets (in the grid) will return either 1 or 0 as probability. That is unavoidable. And in those $i$-th cases (depending on whether the corresponding $y_i$ is 1 or 0) the result will be $\log(0)$, which is not something you would want I imagine.
How do one deal with this problem, or should it be dealt with at all ? My first assumption was to first calculate the $F$ function for all cases $i$ for all parameters in the grid. Then figure out which specific elements/indexes have the value 1 or 0, and subtract all combined indexes from all $i$ cases in the grid, so I wouldn't have 1 or 0 anywhere, and all cases has had those particular indexes removed all over the board. But that seems like some kind of weird solution to me, so that's what I am here for. |
We use the notation $\chi_{q}(n,\cdot)$ to identify Dirichlet characters $\Z\to \C$, where $q$ is the modulus, and $n$ is the index, a positive integer coprime to $q$ that identifies a Dirichlet character of modulus $q$ as described below. The LMFDB label $\texttt{q.n}$, with $1\le n < \max(q,2)$ uniquely identifies $\chi_{q}(n,\cdot)$.
Introduced by Brian Conrey, this labeling system is based on an explicit isomorphism between the multiplicative group $(\Z/q\Z)^\times$ and the group of Dirichlet characters of modulus $q$ that makes it easy to recover the order, the conductor, and the parity of a Dirichlet character from its label, or to induce characters.
As an example, $\chi_q(1, \cdot)$ is always trivial, $\chi_q(m,\cdot)$ is real if $m^2=1\bmod q$, and for all $m,n$ coprime to $q$ we have $\chi_q(m,n)=\chi_q(n,m)$.
For prime powers $q=p^e$ we define $\chi_q(n,\cdot)$ as follows:
For each odd prime $p$ we choose the least positive integer $g_p$ which is a primitive root for all $p^e$, and then for $n \equiv g_p^a $ mod $p^{e}$ and $m \equiv g_p^{b} $ mod $p^{e}$ coprime to $p$ we define $$ \chi_{p^e}(n, m) = \exp\left(2\pi i \frac{a b}{\phi(p^{e})} \right). $$
$\chi_2(1, \cdot)$ is trivial, $\chi_4(3, \cdot)$ is the unique nontrivial character of modulus $4$, and for larger powers of $2$ we choose $-1$ and $5$ as generators of the multiplicative group. For $e > 2$, if $$ n \equiv \epsilon_a 5^a \pmod{2^e} $$ and $$ m \equiv \epsilon_b 5^b \pmod{2^e} $$ with $\epsilon_a, \epsilon_b \in \{\pm 1\}$, then \[ \chi_{2^e}(n, m) = \exp\left(2 \pi i \left(\frac{(1 - \epsilon_a)(1 - \epsilon_b)}{8} + \frac{ab}{2^{e-2}}\right)\right). \]
For general $q$, the function $\chi_q(n, m)$ is defined multiplicatively: $\chi_{q_1 q_2}(n, m) := \chi_{q_1}(n, m)\chi_{q_2}(n, m)$ for all coprime positive integers $q_1$ and $q_2$. The Chinese remainder theorem implies that this definition is well founded and that every Dirichlet character can be defined in this way. In particular, every Dirichlet character $\chi$ of modulus $q$ can be written uniquely as a product of Dirichlet characters of prime power modulus. |
If we have some reaction $\ce{A(aq) + B(aq) -> AB(aq)}$ and right now we have less $\ce{A}$ and $\ce{B}$ then we would have at equilibrium, is there ever a time where the amount of $\ce{A}$ and $\ce{B}$ would overshoot the equilibrium of $\ce{A}$ and $\ce{B}$ if the reaction proceeded without any new sources of energy or reactants. In other words, will the concentrations of $\ce{A}$ and $\ce{B}$ always increase towards their equilibrium concentrations but never actually reach those points or will they sometimes overshoot that concentration a large amount (not just an amount that can be ascribed to chance) but then fall back towards the equilibrium concentration.
In normal experiments vast numbers of molecules are involved, $10^{18}$ would not be untypical, and as a result the kinetic behaviour observed as equilibrium is approached is rather like that of a highly over-damped mechanical system. If, however, molecules could be observed one by one at each of many small time intervals then considering a very small sample it is possible that overshoot could be observed. As an example consider a cis-trans isomerisation $\ce{A <=>}B $ that is initiated by a very short light pulse of perhaps femtosecond duration. Consider this only as possible mechanism with which to start a reaction so that we might study it.
The molecule when excited into its excited state has a certain finite lifetime, its reciprocal is the rate constant. If we observe just a single molecule we cannot know exactly when it will react but only that if we repeatedly measure the times it remains in the excited state and generate a histogram of these times that an exponential decay is produced from which a lifetime can be measured. Now if we observe just a few molecules, say 10, it is possible, albeit with a small probability, that all of them will react within a small time, say, 10% of the lifetime, thus we have a situation that in such a small time interval after the reaction starts that all the molecules are now products. The 'equilibrium' at this time is now product 10, reactants 0. The product also has a lifetime with which it transforms back into reactants in just the same manner. And so depending on pure chance the s 'snap-shot' ratio of reactant to products can vary widely.
Of course the chance of all molecule reacting in the same short time-span is small, and if we observed the same molecules many times over the 'normal' equilibrium constant would be observed. The same is true if we observed very many molecules that reacted in the same short time interval as did our initial 10. Thus this variability is not observed in practice because by observing over longer times and/or larger numbers of molecules the average properties are obtained.
Not really. Overshooting doesn't happen, rather the concentrations
gradually approaches the equilibrium concentration as an . asymptote
You'll soon see what I mean.
As an example, for simplicity, lets consider:
$$\ce{P <=>[k_1][k_2] Q}$$
I will assume here that the forward and backward reactions are first order, so we get the rate to be:
$$\text{Rate}_{\text{forward}}=k_1\ce{[P]}$$
and
$$\text{Rate}_{\text{backward}}=k_2\ce{[Q]}$$
As a result the net rate would be:
$$\frac{d\ce{[P]}}{dt}=-k_1\ce{[P]}+k_2\ce{[Q]}$$
Let's take the initial concentration of $\ce{P}$ to be $\ce{[P]_0}$, and the initial concentration of $\ce{Q}$ to be zero.
Suppose at any time $t$, the concentration of the reactant is $\ce{[P]}$. Thus, $\ce{[Q]} = \ce{[P]_0} - \ce{[P]}$. Now, we get the rate law to be:
$$\frac{d\ce{[P]}}{dt} = k_2\ce{[P]_0} - (k_1 + k_2)\ce{[P]}$$
Integrating this from $\ce{[P]_0}$ to $\ce{[P]}$:
$$\int_{\ce{[P]_0}}^{\ce{[P]}}\frac{d\ce{[P]}}{k_2{[P]_0}-(k_1 + k_2)\ce{[P]}}=\int_0^tdt$$
Solving the integral, we get:
$$\frac{-1}{(k_1 + k_2)}\log_e{\left(\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}\right)}=t$$
$$\frac{(k_1 + k_2)\ce{[P]} - k_2\ce{[P]_0}}{k_1\ce{[P]_0}}=e^{-(k_1 + k_2)t}$$
Finally, on simplifying it, we get our integrated rate equation as:
$$\ce{[P]}=\ce{[P]_0}\frac{k_2 + k_1e^{-(k_1+k_2)t}}{k_1+k_2}$$
Let's plot this on a graph to understand what this is actually saying:
You can clearly observe that the concentrations never overshoot the asymptotes (black dotted line). The asymptotes denote the equilibrium concentrations, so it's evident that the reaction tends toward the equilibrium concentration. |
If I write FourierSinSeries[x, x, 10]
Mathematica will output the first 10 terms of the Fourier Sine Series f(x) = x defined for 0 < x < Pi.
How do I output the the Fourier Sine Series for a different domain, say 0 < x < 2 Pi?
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Use
{a,b}->
{1,1/2} for
FourierParameters
FourierSinSeries[x, x, 10, FourierParameters -> {1, 1/2}]
This is because the Fourier sine series is defined as $b_k= \frac{2}{L} \int_0^L f(x) \sin\left( \frac{k \pi x}{L} \right) dx$ and Mathematica uses
so for $L=2\pi$, if we use $b=\frac{1}{2}$ and $a=1$ then Mathematica definition becomes $b_k= \frac{1}{\pi} \int_0^{2\pi} f(x) \sin\left( \frac{k x}{2} \right) dx$ which is what you get when $L=2\pi$ |
I think you need the Nernst equation. This is something we were just introduced to today in my class, so I’m not so sure about the notation you’re using in your question at the top, but you should be able to correct my answer to fit your needs if you understand the equation I’m about to implement.
$$E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{zF}\ln Q$$
Where $E^\circ_\text{cell}$ = what the $E_\text{cell}$ normally is, under STP with the solutes at a concentration of $1\ \mathrm{mol\ dm^{−3}}$.
$$E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}$$
Since, the reduction potential of copper is lower, it will be the reducing agent in this reaction.
$$E^\circ_\text{cell} = -0.7993\ \mathrm{V} + 0.518\ \mathrm{V}$$
$$E^\circ_\text{cell} = -0.281\ \mathrm{V}$$
Where $R$ is the familiar gas law constant:
$$R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$$
$T$ is the absolute temperature (it’s always assumed to be $298.15\ \mathrm{K}$):
$$T = 298.15\ \mathrm{K}$$
$z$ is the number of moles of electrons transferred in the half-reaction:
$$z = 1$$
And $F$ is Faraday’s constant, which is measured in coulombs per mole:
$$F = 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}$$
And $Q$ is the also familiar reaction quotient, but as I discovered, it takes some work to calculate, and you have to use the quadratic equation.
$$Q=\frac{[\ce{red}]}{[\ce{ox}]}$$$$Q=\frac{[\ce{Cu+}]}{[\ce{Ag+}]}$$
$$K_{\text{sp},\ce{CuI}}=10^{-12}=[\ce{Cu+}][\ce{I-}]$$$$1.0 × 10 ^{-12} = [X][0.090 + X]$$Do the quadratic formula$$X = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$$$[\ce{Cu^+}] = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$$$K_{\text{sp},\ce{AgCl}} = 1.8 \times 10^{-10}\ \mathrm{mol\ l^{-1}} = [\ce{Ag+}][\ce{Cl-}]$$$$1.8 \times 10^{-10} = [X][0.120 + X]$$Do the quadratic formula$$X = 1.50 \times 10 ^{-9}\ \mathrm{mol\ l^{-1}}$$$$[\ce{Ag+}]=1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}$$
$$Q=\frac{1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}}{1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}}$$
$$Q=0.00740$$
So, plugging this all into the Nernst equation, the answer should be:
$$E_\text{cell}=-0.281\ \mathrm{V}-\frac{8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 298.15\ \mathrm{K}}{1 \times 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}}\ln(0.00740)$$
$$E_\text{cell}=-0.155\ \mathrm{V}$$
I should add that since $R$ is a constant, $F$ is a constant, and $T$ is usually something like $298.15\ \mathrm{K}$ or just not measured in elementary chemistry, the equation can just be
$$E_\text{cell}=E^\circ_\text{cell}-\frac{0.0257}{z}\ln Q$$
Which will give the same answer:
$$E_\text{cell}=-0.281\ \mathrm{V}-0.0257\ \mathrm{V}\ln(0.00740)$$$$E_\text{cell}=-0.155\ \mathrm{V}$$ |
Title Topological "observables" in semiclassical field theories Publication Type Journal Article Year of Publication 1992 Authors Nolasco, M, Reina, C Journal Phys. Lett. B 297 (1992) 82-88 Abstract
We give a geometrical set up for the semiclassical approximation to euclidean field theories having families of minima (instantons) parametrized by suitable moduli spaces ${\mathcal{M}}$. The standard examples are of course Yang-Mills theory and non-linear $\sigma$-models. The relevant space here is a family of measure spaces $\tilde{\mathcal{N}} \rightarrow \mathcal{M}$, with standard fibre a distribution space, given by a suitable extension of the normal bundle to $\mathcal{M}$ in the space of smooth fields. Over $\tilde{\mathcal{N}}$ there is a probability measure $d\mu$ given by the twisted product of the (normalized) volume element on $\mathcal{M}$ and the family of gaussian measures with covariance given by the tree propagator $C_\phi$ in the background of an instanton $\phi \in \mathcal{M}$. The space of "observables", i.e. measurable functions on ($\tilde{\mathcal{N}},\, d\mu$), is studied and it is shown to contain a topological sector, corresponding to the intersection theory on $\mathcal{M}$. The expectation value of these topological "observables" does not depend on the covariance; it is therefore exact at all orders in perturbation theory and can moreover be computed in the topological regime by setting the covariance to zero.
URL http://hdl.handle.net/1963/3541 DOI 10.1016/0370-2693(92)91073-I Topological "observables" in semiclassical field theories
Research Group: |
Markov's principle is a statement of constructive arithmetic that allows classical reasoning on formulas of the shape $\exists x P$ when $P$ is a recursive predicate: $\neg \neg \exists x P \to \exists x P$
Its formulation is well known in the context of arithmetic, and it is well known that adding it to Heyting Arithmetic gives rise to a constructive system: when $A \lor B$ is provable, either $A$ is provable or $B$ is provable; when $\exists x \, A(x)$ is provable, there is a $t$ such that $A(t)$ is provable.
However I think it is not clear how one could formulate it in the context of pure intuitionistic first order logic (if it does make sense at all).
Various sources ([1], [2]) state it as "$\neg \neg \exists x P \to \exists x P$ for $P$ $\forall, \to$-free". Another tempting formulation would be $\forall x (P \lor \neg P) \to \neg \neg \exists x P \to \exists x P$ for any propositional $P$.
As far as I can see, these two formulations are not comparable. Both these axioms, though, share the property that if we add them to pure intuitionistic first order logic we still obtain a constructive system (this is proved in [1] for the first axiom; I couldn't find references for the second axiom, but a proof can be obtained with a very similar argument). Is there a more general analog of Markov's rule for first-order logic which preserves the disjunction property and other proof-theoretic properties of constructive systems? Or alternatively, is there some other source justifying the choice of the formalization used in [1,2] for Markov's principle?
[1] H. Herbelin, An intuitionistic logic that proves Markov's principle https://hal.inria.fr/inria-00481815/ [2] U. Berger, A Computational Interpretation of Open Induction http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=1319627 |
In this tutorial you will learn how to set up and execute Purpose:
calculations of the magneto-optical Kerr effect (
exciting
). As an example, we calculate the
MOKE
spectrum for ferromagnetic fcc-Ni.
MOKE 0. Define relevant environment variables and download scripts Read the following paragraphs before starting with the rest of this tutorial!
Before starting, be sure that relevant environment variables are already defined as specified in
. How to set environment variables for tutorials scripts
From now on the symbol
will indicate the shell prompt. $
Important note: All input parameters are in atomic units! 1. Previous knowledge
The evaluation of the
spectrum involves a spin polarized ground state calculation followed by a MOKE
calculation. Therefore, before starting, it is recommended to be familiar with the following related tutorials:
TDDFT 2. Theoretical background of the Magneto-Optical Kerr Effect
When linearly polarized light is reflected on the surface of a ferromagnetic material, the reflected beam will be, in general, elliptically polarized with its polarization plane rotated with respect to the incident beam. The magnitude of both the polarization rotation $\theta_K$ and the ellipticity $\gamma_K$ are given by the complex Kerr angle: $\phi_K(\omega)= \theta_K(\omega)+i \gamma_K(\omega)$ which can be obtained from the dielectric tensor $\epsilon_{\alpha\beta}$. In particular, for the so-called polar configuration present when both incident light and the magnetic moment of the material are perpendicular to the surface, the Kerr angle is given by:(1)
A proper description of the
requires the knowledge of the current-current response function. However, in the MOKE
framework the longitudinal contraction of the dielectric tensor is obtained from the density-density response function, which lacks a part of the transverse response contained in the current-current response. This term, which is non-zero only when time-reversal symmetry is broken, is called the anomalous Hall conductivity (
TDDFT
) and constitutes an essential ingredient for describing magneto-orbit effects. Therefore, in the
AHC
calculation this term must be added to obtain the correct
TDDFT
spectrum. In order to do that, we must set the attribute
MOKE
to
ahc "true"in the
element, as shown in the example input file of this example (see below).
tddft 3. The MOKE spectrum of Nickel
In order to generate the
spectrum for nickel, we will use the following input file ( MOKE
) to run
input.xml
:
exciting <input> <title>FM fcc Ni</title> <structure speciespath="$EXCITINGROOT/species"> <crystal scale="6.652"> <basevect> 0.5 0.5 0.0 </basevect> <basevect> 0.5 0.0 0.5 </basevect> <basevect> 0.0 0.5 0.5 </basevect> </crystal> <species speciesfile="Ni.xml" rmt="2.3"> <atom coord="0.0 0.0 0.0"/> </species> </structure> <groundstate do="skip" xctype="LDA_PW" rgkmax="7.0" epsengy="1.0d-4"> <spin reducebf="0.5" spinorb="true"/> </groundstate> <xs xstype="TDDFT" ngridk="4 4 4" vkloff="0.097 0.273 0.493" dfoffdiag="true" dogroundstate="fromscratch" maxscl="200" bfieldc="0.0 0.0 -2.0" broad="0.02" tevout="true"> <tddft fxctype="RPA" drude="0.18 0.001" ahc="true"/> <qpointset> <qpoint> 0.0 0.0 0.0 </qpoint> </qpointset> <energywindow intv="0.00 0.32" points="150"/> </xs> </input> Make sure to set
to the correct $EXCITINGROOT
excitingroot directory in the
attribute using the command
speciespath
$ SETUP-excitingroot.sh
Please be aware that for time-saving reasons, here we have chosen computational parameters which do not correspond to those obtained from convergence tests. However, with them we are able to reproduce the main spectral features for the system under consideration.
Before running
, let's examine the input file displayed above. First, we consider the exciting
element. Inside it we find the
groundstate
element as explained in
spin
. Most importantly, we must set the
Spin-polarized calculations
attribute to
spinorb
, because the spin-orbit coupling is essential to obtain the Kerr rotation. As you can notice, we
"true" skipthe ground-state calculation. The reason for this is that the self-consistency loop will be triggered (from scratch) by the
element, where a fine
xs
-grid is defined (see below). Nevertheless, we need to set the
k
element to define several parameters, such as the type of exchange-correlation functional or the
groundstate
element and its attributes. Notice that the convergence threshold for the total energy,
spin
, is set to
epsengy
, which is a larger value than the default one.
"1.0d-4"
Next, we consider the
element. Most of the attributes and convergence issues for that element are discussed in xs
. Therefore, we refer the reader to it in case of doubts. However, we have in this case a few additional parameters: We set the attribute
Excited states from TDDFT
=
dogroundstate
, meaning that we will trigger the ground-state calculation from the
"fromscratch"
element, therefore, there is no need in this case to have a previous groundstate run. We set also
xs
to
maxscl
and
"200"
to a finite value in the
bfieldc
-direction. Both attributes have exactly the same meaning as their homonyms in the
z
element. Inside the
groundstate
element, we must set
tddft
to
ahc
in order to include the anomalus Hall conductivity term to the
"true"
response functions, as explained in the theoretical section.
TDDFT
In order to run
create the exciting
from the example above and run the scripts
input.xml
$ EXECUTE-single.sh k444
After the completion of the run, all results will be written inside the directory
. In particular for the k444
, the relevant file is called
MOKE
. In order to visualize the
MOKE_ label.OUT
spectrum, move to the directory
MOKE k444
$ cd k444
and run the following script:
$ PLOT-MOKE-spectrum.py
The resulting plot is saved in the files
and PLOT.png
, which should look like in the following image:
PLOT.ps 4. Convergence issues
In the calculation of optical properties, two of the most important convergence parameters are the size of the
-point grid and the number of empty states included in the calculation of the response function. They are controlled, respectively, by the attributes k
and
ngridk
of the
nempty
element. It is important to note that to obtain convergent optical spectra one has to use rather large
xs
-point sets. In addition, the attribute
k
plays now not only the role of a convergence parameter for spin-polarized calculations. In this case, one has to be sure that the excited (empty) states are covering the energy range where the
nempty
spectrum is searched. Additionally, as we are dealing with a metal system, the smearing of the
MOKE
-points (controlled by the attributes
k
and
stype
of the
swidth
element) can play an important role to reach convergence.
groundstate Exercise In this tutorial we are using the default value for the attribute. Check if this selection is enough to obtain a reasonable spectrum in the energy range between 0 and 8 eV (Hint: Inspect the file nempty ). EIGVAL_QMT001.OUT Perform the complete convergence test to get the value of the magnetic moment of FM fcc-Ni and compare it with the experimental value of ~0.6 $\mu_{\tt B}$. Use the default value for the attribute and repeat the calculations. Is there any difference in the calculated spectra? epsengy Converge the spectrum with respect to MOKE size. ngridk As a matter of fact, the is mainly due to presence of the spin-orbit coupling. One can readily check it by recalculating the spectrum by setting MOKE = spinorb . "false" Literature J. Kunes, Physica Scripta T109, 116 (2004) D. Sangalli, A. Marini, and A. Debernardi, Phys. Rev. B 86, 125139 (2012) |
(I will not be surprised if this problem has been solved and/or has a trivial solution – I just do not know the right terminology to google for it.)
So the problem is as follows. I have an $m \times n$ matrix consisting of zeroes and ones. I treat the rows $\mathbf r_1, \mathbf r_2, \dotsc, \mathbf r_m$ of the matrix as vectors in $\mathbb R^n$ (or, to be more precise, in $\mathbb R_{\geq 0}^n$) and want to generate linear combinations that give vectors with non-negative entries:
$$ \alpha_1 \mathbf r_1 + \alpha_2 \mathbf r_2 + \dotsb \alpha_m \mathbf r_m = \mathbf h = (h_1, h_2, \dotsc, h_n) \in \mathbb R_{\geq 0}^n \,, $$ where the coefficients of the combination can be bothe positive, zeroes, and negative: $\alpha_1, \alpha_2, \dotsc, \alpha_m \in \mathbb R$.
In other words, I want to find many vectors from $\operatorname{span} \{\mathbf r_1, \mathbf r_2, \dotsc, \mathbf r_m\}$ that have non-negative entries. It goes without saying, that number of such vectors is infinite. But I am paricularly interested in generating vectors
with different supports — this requires that some of $\alpha_1, \alpha_2, \dotsc, \alpha_m$ are sometimes negative. The routine can be (or should be?) randomised, and for my real task $n$ is usually around 100–500.
P.S. The support of the vector is a set of non-zero positions: $$ \operatorname{supp} (\mathbf h) = \{ i : h_i \neq 0 \} \,. $$ |
I hope you can help me and this is the right forum to ask.
In the process of building and programming my own Quadcopter, I encountered the term Euler angles. I took some time to understand them and then wondered why they are used in multicopter systems.
In my understanding Euler angles are used to rotate a point or vector in a coordinate system/ to express that rotation. I now wonder why i should use Euler angles to compute the orientation of the quadcopter as I could easily(at least i think so) compute the angles by themself, like$$\theta = arctan(y/z)$$$$\phi = arctan(x/z)$$(just using accelerometer, where $x, y, z$ are axis accelerations and $\theta, \phi$ are pitch and roll, respectively. In the actual implementation I do
not only use the accelerometer, this is just simplified to make the point clear).
Where exactly are Euler angles used? Are they only used to convert desired trajectory in the earth frame to desired trajectory in the Body frame?
I would be very glad if anyone could point this out and explain the concept/ why and where they are used further.
To clarify: I do know that Euler angles encounter gimbal lock, that they are three rotations about $x, y, z$ axis and how they generally work(I think). @Christo gave a very good explanation.
My question now is, why are
they used? Isn't it counterproductive to apply the yaw rotation, then pitch, and then roll?
-Earth frame X,Y,Zrotation about Z(psi)->Frame 1 x', y', z'rotation about y'(theta)->Frame 2 x'', y'', z''rotation about x''(phi)->Body frame x, y, z
and vice versa.
Why? I would just have said:
pitch = angles between X and xroll = angle between Y and yyaw = angle between x-y-projection of the magnetic field-vector and the starting vector(yaw is kinda different).
(Notice the difference between uppercase and lowercase, look at the Earth-to-Body-Frame for notation). Tied with this i wonder why the correct formula for pitch($\theta)$ should be $$\theta = \tan^{-1}\left(-f_x/\sqrt{f_y^2+f_z^2}\right)$$ I would have thought $$\theta = \tan^{-1}\left(-f_x/f_z\right)$$ suffices. Maybe I have some flaw in my knowledge or a piece of the puzzle is still missing.
I hope this is understandable, if not feel free to ask. If this gets too crowded, I can always ask another question, just make me aware of it.
If anyone could explain how to use quaternions to express orientation I would be very thankful, but I can also just ask another time. I get the concept of Quaternions, just not how to use them to express
orientation not rotation. |
Is there a known expression for the eigenvalue distribution of a matrix of the form
$$\sum\limits_{i=1}^n k_ia_ia_i^T$$
where $a_i \in \mathcal{R}^m$, with $n > m$, $a_i \sim \mathcal{N}(0,\Sigma)$ and $k_i > 0$? For simplicity we can consider the case $\Sigma = \text{I}_d$. (Equivalently, the expression is $\sum\limits_{i=1}^n a_ia_i^T$ where the $a_i$'s are $\mathcal{N}(0,k_i\text{I}_d)$, or $A\,\text{diag}(\vec{k})\,A^T$.)
In fact, I'm only concerned with the distribution of the determinant of such a matrix. This type of matrix can be thought of as a generalization of the Wishart matrix (which has $k_i$ equal for all $i$), whose eigenvalue distribution is well known, for instance: Determinant of real Wishart matrix.
Most of the generalizations of this result that I have seen deal with when the $a_i$'s are noncentral, i.e. drawn from distributions with differing means but the same covariance. I'm wondering if this result also can be generalized somehow to my case, where they are still mean zero but have differing covariances.
(In fact, even an exact expression is not necessary; an asymptotic upper bound - i.e. where $n$ is very large, and the $k$'s are considered as fixed - on the determinant in terms of the $k_i$'s would also be of interest. Of course one can get such bounds via Hadamard's inequality or applying AM-GM to the trace and determinant, but these bounds are quite weak for matrices whose eigenvalues are not very uniform.)
[Note: I originally posted this on the Statistics StackExchange, but was advised by a commenter to try here instead (and delete the other).] |
I've already posted an answer on this thread, but I found another example I'd like to describe separately. Let $r > 0$ and consider the following problem, coming from compound interest or as one definition of $e^r$:
Show that $f(n) = (1 + \frac{r}{n})^n$ increases with $n$.
One generalize the problem strategy is to allow $n$ to be a continuous variable (probably this trick could have its own article). Now, see if you can prove that $f(n)$ still increases. If you take this mindset, it's natural to use the definition of $n$th power for $n \in {\mathbb R}$ and write
$f(n) = e^{n \log(1 + \frac{r}{n})}$
And the problem has reduced to showing that$x \log(1 + \frac{r}{x}) = \int_0^1 \frac{r}{(1 + \frac{sr}{x})} ds$increases with $x$, which it clearly does. (Here we've used the integral definition of the logarithm, but written in a way typically helpful for analyzing such products.)
Another problem that can be solved through allowing a discrete parameter to be continuous is to prove Stirling's approximation for $n!$ (although to make that proof very clean you can also use other labor saving tricks like Taylor expansion by integration by parts and the dominated convergence theorem).
If you ran into this problem from compound interest, or you were hoping for something more elementary which did not use such a heavy understanding of the exponential function, then you probably want to find a different proof. But finding a different proof still seems to require "generalizing the problem", but in a different way.
Another proof, goes as follows. Imagine that interest at a rate $r$ works as follows: once an amount of money is invested, the value of each unit after a time $t$ is given by $(1 + tr)$. That is, the value of the money grows linearly. Now imagine you had the opportunity to withdraw and immediately reinvest your money at a time of your choice. Having this ability would allow you to raise more money, because it would allow you to accrue interest on the interest you've already earned (hence the name "compound interest"). With this interpretation, the number $(1 + \frac{r}{n})^n$ is the value of each unit of money after time $1$ and $n$ regularly spaced compoundings.
The proof now goes as follows: if you had a choice of when these compoundings would occur, then the more compoundings the better, and the best way to allocate $n$ compoundings is to have them occur at $n$ regularly spaced time intervals. That is, we interpret$(1 + \frac{r}{n})^n = \max \prod_{i=1}^n (1 + a_i r)$under the constraint that $0 \leq a_i \leq 1$ with $\sum a_i = 1$.
For example, it is better to have one compounding than to have none at all, because after withdrawing and reinvesting the money, now not only does the initial investment grow linearly, but also the interest you earned before the withdrawal grows linearly. For the same reason, given $a_1, \ldots, a_n$, the opportunity to compound once more during, say, $0 < t < a_1$, would allow you to increase the amount of money at all later times.
The fact that the best choice of $(a_1, \ldots, a_n)$ is to have $a_1 = a_2 = \ldots = a_n = \frac{1}{n}$ is the principle that the largest product you can obtain when the sum of positive numbers is fixed is to have all the terms equal. This is easy to check with two variables: you can either find the largest rectangle to fit inside an isosceles triangle, or otherwise just note that if $a_1 \neq a_2$, then changing to $a_1' = \frac{(a_1 + a_2)}{2} = a_2'$ gives an improvement for $(1 + a_1 r)(1+ a_2 r) < (1 + a_1' r) (1 + a_2' r)$. The case of $n$ variables actually follows from this observation.
So if you really wanted some elementary solution to the problem, this one would do. It's an interesting example because you can see that either solution involves some kind of generalization, but the two generalizations are unrelated to each other. The first one does not need to / is unable to consider these non-even partitions. The second does not need to / is unable to consider fractional $n$.
By the way, does anyone know how to prove in an elementary way (i.e. expanding) that $\prod_{i=1}^n (1 + a_ir)$ tends to $e^r = \sum \frac{r^k}{k!}$ as $\max |a_i| \to 0$ with $0 \leq a_i \leq 1$ and $\sum a_i = 1$? An easy solution goes by writing the product with the exponential function so that you get the exponential of $\sum \log(1 + a_i r) = \sum \int_0^1 \frac{a_i r}{(1 + s a_i r)} ds$.
You can then integrate by parts (i.e. Taylor expand) to obtain$\sum a_i r - \sum \int_0^1 (1-s) \frac{(a_i r)^2}{(1 + s a_i r)^2} ds$. Now, $\sum a_i r = r$ is the main term. After you take $\max |a_i|$ to be less than $.5 / |r|$, the error term is bounded in absolute value by $C \sum (a_i r)^2 \leq \max \{ |a_i| \} \cdot \sum a_i |r|^2$. I can, of course, move this question to a different thread.
EDIT: I realized later on that there is a completely elementary proof, and it is also completely obvious even though I didn't think of it. Namely, you expand $(1 + \frac{r}{n})^n$ into powers of $r$, and it is easy to see after a little algebra that each coefficient increases with $n$. I still find the other solutions interesting, but this turns out not to be a good demonstration of how generalizing can make a problem easier. By the way, the last question I had asked was answered in this thread:
A limiting product formula for the exponential function |
We know that there are Whitehead theorem for homotopy and homology theory.
Is there the Whitehead theorem for cohomology theory for 1-connected CW complexes?
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We know that there are Whitehead theorem for homotopy and homology theory.
Is there the Whitehead theorem for cohomology theory for 1-connected CW complexes?
Conceptually, the following two theorems (both due to Whitehead) are Eckmann-Hilton duals.
Theorem. A weak homotopy equivalence between CW complexes is a homotopy equivalence.
Theorem. A homology isomorphism between simple spaces is a weak homotopy equivalence.
They don't look dual, but they are. SeeJ.P. May. The dual Whitehead Theorems.
London Math. Soc. Lecture Note Series Vol. 86(1983), 46--54.
The point is that the second statement is really about cohomology, and the standard cellular proof of the first statement dualizes word-for-word to a ``cocellular'' proof of the second. Cocellular constructions are what appear in Postnikov towers, and they can be used more systematically than can be found in the literature. Yet another plug: they are central in the upcoming book "More Concise Algebraic Topology'' by Kate Ponto and myself.
As Sean says, the key point is the Universal Coefficient Theorem, but the details are not completely obvious unless you make some finiteness assumptions.
Suppose that $f:X\to Y$ is such that $H^{\ast}(f;\mathbb{Z})$ is an isomorphism. Let $Z$ be the cofibre of $f$, so $\tilde{H}^{\ast}(Z;\mathbb{Z})=0$. If we can prove that $\tilde{H}_{\ast}(Z;\mathbb{Z})=0$ then we can appeal to the ordinary homological Whitehead theorem. MathJax is mangling my tildes: all (co)homology groups of $Z$ below should be read as reduced.
For any prime $p$, we have a universal coefficient sequence for $H^{\ast}(Z;\mathbb{Z}/p)$ in terms of $H^{\ast}(Z;\mathbb{Z})$, so $H^{\ast}(Z;\mathbb{Z}/p)=0$. As ${\mathbb{Z}/p}$ is a field we also know that $H_{\ast}(Z;\mathbb{Z}/p)$ is a free module with $H^{\ast}(Z;\mathbb{Z}/p)$ as its dual, so we must have $H_{\ast}(Z;\mathbb{Z}/p)=0$. Using the universal coefficient theorem for homology we deduce that $H_{\ast}(Z;\mathbb{Z})/p$ and $\text{ann}(p,H_{\ast}(Z;\mathbb{Z}))$ are zero, so multiplication by $p$ is an isomorphism on $H_{\ast}(Z;\mathbb{Z})$. As this holds for all $p$, we see that $H_{\ast}(Z;\mathbb{Z})$ is a rational vector space. Thus, if it is nontrivial it will contain a copy of $\mathbb{Q}$ so (via universal coefficients again) $H^{\ast}(Z;\mathbb{Z})$ will contain a copy of $\text{Ext}(\mathbb{Q},\mathbb{Z})$. This group is nonzero (in fact, enormous) by a standard calculation, so this contradicts the initial assumption.
Sure.
The basic point is that for simply-connected spaces, you can determine the connectivity of a map by looking at the connectivity of the cofiber instead of the connectivity of the fiber.
In homology, you determine the connectivity of the cofiber by looking at $H_*(C;\mathbb{Z})$, because of the Hurewicz Theorem.
In cohomology, you appeal to: if $X$ is simply-connected, then $X$ is $n$-connected if and only if $[ X, K(G,m)] = *$ for all $m \leq n$ and all abelian groups $G$. This is because (by basic obstruction theory) if $X$ is $(n-1)$-connected, then $[X, K(G,n)] \cong \mathrm{Hom}(\pi_n(X), G )$.
This results in a long list of groups to check, admittedly; but it can be whittled down by Universal Coefficients theorems if you like. |
I am referring to the following three reactions. Two of these reactions are imaginary ($\ce{M}$ and $\ce{X}$ are imaginary).
$$\begin{alignat}{3} \ce{X- + 2e- \;&<=> X^3-} \qquad &&E^\circ_{\ce{X^3-}/\ce{X-}}=0.9\ \mathrm{V}\qquad &&&\text{(a)}\\ \ce{2H+ + MO^2+ + e- \;&<=> M^3+ + H2O} \qquad &&E^\circ_{\ce{MO^2+}/\ce{M^3+}}=0.5\ \mathrm{V} \qquad &&&\text{(b)}\\ \ce{8H+ + MnO4- + 5e- \;&<=> Mn^2+ + 4H2O} \qquad &&E^\circ_{\ce{MnO4-}/\ce{Mn^2+}}=1.5\ \mathrm{V} \qquad &&&\text{(c)} \end{alignat}$$
Now consider the reaction between a solution having equimolar amounts of $\ce{M^3+}$ and $\ce{X^3-}$ and a $\ce{KMnO4}$ solution. (Note that all the species are in their respective standard states, so there is no need for Nernst equation.)
To find out possible reactions, following steps are taken.
Equations $\text{(a)}$ and $\text{(c)}$;
$$ \begin{align} &{-}\text{(a)} \times 5 + \text{(c)} \times 2 : \\ \ce{&16H+ +5X^3- + 2MnO4- -> 2Mn^2+ + 5X- + 8H2O} \qquad E^\circ_1\qquad\text{(1)} \\ \\ &E^\circ_1 = 1.5\ \mathrm{V} + (-0.9\ \mathrm{V}) = 0.6\ \mathrm{V} \\ &\Delta G^\circ_1 = -nE^\circ_1F = -10 \times 0.6 \times F = -6F \end{align} $$
Equations $\text{(b)}$ and $\text{(c)}$;
$$ \begin{align} &{-}\text{(b)} \times 5 + \text{(c)}: \\ \ce{&5M^3+ + H2O + MnO4- -> Mn^2+ + 5MO^2+ + 2H+} \qquad E^\circ_2 \qquad(2)\\ \\ &E^\circ_2 = 1.5\ \mathrm{V} + (-0.5\ \mathrm{V}) = 1.0\ \mathrm{V} \\ &\Delta G^\circ_2 = -nE^\circ_2F = -5 \times 1.0 \times F = -5F \end{align}$$
My questions:
Are there any errors in above calculations? If so, please suggest corrections. According to $E^\circ$ values, reaction $(2)$ is more feasible than reaction $(1)$. Am I correct? According to $\Delta G^\circ$ values, reaction $(1)$ is more feasible than reaction $(2)$. Am I correct? Is it useless to use $E$ values to predict the feasibilities of redox reactions? Should we always resort to $\Delta G$? Is it possible for predictions based on $E$ values and $\Delta G$ values to contradict? If not how can these results be explained? |
F-test is named after the more prominent analyst R.A. Fisher. F-test is utilized to test whether the two autonomous appraisals of populace change contrast altogether or whether the two examples may be viewed as drawn from the typical populace having the same difference. For doing the test, we calculate F-statistic is defined as:
${F} = \frac{Larger\ estimate\ of\ population\ variance}{smaller\ estimate\ of\ population\ variance} = \frac{{S_1}^2}{{S_2}^2}\ where\ {{S_1}^2} \gt {{S_2}^2}$
Its testing procedure is as follows:
Set up null hypothesis that the two population variance are equal. i.e. ${H_0: {\sigma_1}^2 = {\sigma_2}^2}$
The variances of the random samples are calculated by using formula:
${S_1^2} = \frac{\sum(X_1- \bar X_1)^2}{n_1-1}, \\[7pt] \ {S_2^2} = \frac{\sum(X_2- \bar X_2)^2}{n_2-1}$
The variance ratio F is computed as:
${F} = \frac{{S_1}^2}{{S_2}^2}\ where\ {{S_1}^2} \gt {{S_2}^2}$
The degrees of freedom are computed. The degrees of freedom of the larger estimate of the population variance are denoted by v1 and the smaller estimate by v2. That is,
${v_1}$ = degrees of freedom for sample having larger variance = ${n_1-1}$
${v_2}$ = degrees of freedom for sample having smaller variance = ${n_2-1}$
Then from the F-table given at the end of the book, the value of ${F}$ is found for ${v_1}$ and ${v_2}$ with 5% level of significance.
Then we compare the calculated value of ${F}$ with the table value of ${F_.05}$ for ${v_1}$ and ${v_2}$ degrees of freedom. If the calculated value of ${F}$ exceeds the table value of ${F}$, we reject the null hypothesis and conclude that the difference between the two variances is significant. On the other hand, if the calculated value of ${F}$ is less than the table value, the null hypothesis is accepted and concludes that both the samples illustrate the applications of F-test.
Problem Statement:
In a sample of 8 observations, the entirety of squared deviations of things from the mean was 94.5. In another specimen of 10 perceptions, the worth was observed to be 101.7 Test whether the distinction is huge at 5% level. (You are given that at 5% level of centrality, the basic estimation of ${F}$ for ${v_1}$ = 7 and ${v_2}$ = 9, ${F_.05}$ is 3.29).
Solution:
Let us take the hypothesis that the difference in the variances of the two samples is not significant i.e. ${H_0: {\sigma_1}^2 = {\sigma_2}^2}$
We are given the following:
Applying F-Test
${F} = \frac{{S_1}^2}{{S_2}^2} = \frac {13.5}{11.3} = {1.195}$
For ${v_1}$ = 8-1 = 7, ${v_2}$ = 10-1 = 9 and ${F_.05}$ = 3.29. The Calculated value of ${F}$ is less than the table value. Hence, we accept the null hypothesis and conclude that the difference in the variances of two samples is not significant at 5% level. |
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Change to browse by: Bookmark(what is this?) General Relativity and Quantum Cosmology Title: Generators of local gauge transformations in the covariant canonical formalism of fields
(Submitted on 15 Sep 2019)
Abstract: We investigate generators of local gauge transformations in the covariant canonical formalism (CCF) for matter fields, gauge fields and the second order formalism of gravity. The CCF treats space and time on an equal footing regarding the differential forms as the basic variables. The conjugate forms $\pi_A$ are defined as derivatives of the Lagrangian $D$-form $L(\psi^A, d\psi^A)$ with respect to $d\psi^A$, namely $\pi_A := \partial L/\partial d\psi^A$, where $\psi^A $ are $p$-form dynamical fields. The form-canonical equations are derived from the form-Legendre transformation of the Lagrangian form $H:=d\psi^A \wedge \pi_A - L$. We show that the generator of the local gauge transformation in the CCF is given by $\varepsilon^r G_r + d\varepsilon^r \wedge F_r$ where $\varepsilon^r$ are infinitesimal parameters and $G_r$ are the Noether currents which are $(D-1)$-forms. $\{G_r , G_s \} = f^t_{\ rs}G_t$ holds where $\{\bullet, \bullet \}$ is the Poisson bracket of the CCF and $f^t_{\ rs}$ are the structure constants of the gauge group. For the gauge fields and the gravity, $G_r=-\{F_r, H \}$ holds. For the matter fields, $F_r=0$ holds. Additionally, we apply the CCF to the second order formalism of gravity with Dirac fields for the arbitrary dimension ($D \ge 3 $). Submission historyFrom: Satoshi Nakajima [view email] [v1]Sun, 15 Sep 2019 11:07:46 GMT (12kb,D) |
It is often explained, that the rule of thumb for exercising American options is to check when the benefit from the interest rate (sell the stock earlier, get the cash, put in the bank) is higher than the time value of the option. This is all clear in the positive interest rate environment, but the question is then - would we exercise some put options in case $r = 0$, and why would we do that? I thought, that there's no reason for such exercise, however according to this paper it is even sometimes optimal to early exercise American puts when $r<0$. What am I missing?
The classic result is never early exercise an American call if $r \geq 0, d \leq 0.$ If we think in terms of FX, calls and puts are really the same thing and by switching currency, we get never early exercise an American put if $ r \leq 0, d \geq 0.$ If one of these is violated it may be worth early exercising.
Let $r=0$:
The maximum payoff ever from the put is when $S=0$ so $P= K$. So one would always exercise at this maximum because you cant get any better in the future and dont forego any interest. Based on @MarkJoshi 's comment, we have to assume that $S=0$ never recovers, so if $r=0$ you are essentially indifferent between exercising then or later because you gain/loose no interest and the payoff will never decrease. So it holds $P_t=p_t$.
If $S>0$ you can theoretically always have an infinitesimal increase in payoff when $S\to 0$ and wait, however I would expect that there is an exercise boundary at which point you would exercise because the probability of ending OTM vs. being close to maximum payoff already is too high and hence the AM put dominates at this point and $P_t\geq p_t$.
Let $r<0$:
If you exercise at the maximum payoff, your received cash will vanish over time through the negative interest. I expect it is then a question of the expected opportunity cost from not exercising at the maximum vs. the negative interest until maturity. If the volatility of the asset is very high, you have a high probability of moving away from the maximum in the future and would probably rather accept the negative interest. If the volatility is very low or even close to zero, you will likely not move away from the maximum much and rather exercise later to avoid negative interest. Hence the AM put fully dominates and $P_t>p_t$. |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
This problem can be expressed as the original Merton's portfolio problem.
Consider wealth process defined by SDE
$$d X _ { t } = \frac { X _ { t } \alpha _ { t } } { S _ { t } } d S _ { t } + \frac { X _ { t } \left( 1 - \alpha _ { t } \right) } { S _ { t } ^ { 0 } } d S _ { t } ^ { 0 }$$
where $\alpha_t$ is proportion of the investment in the risky asset $S_t$, and $S_t^0$ is the risk-free asset.
Optimality criterion may depend on the risk aversion of the investor, and the problem is to maximize expected utility of the investor for appropriate utility function $U$:
$$E \left[ U \left( X _ { T } \right) \right] \rightarrow \max$$
Classical choice of the utility function is CRRA:
$$u ( x ) = \frac { x ^ { 1 - \gamma } } { 1 - \gamma }$$
where $\gamma$ is constant and corresponds to the risk-aversion of the investor.
If the asset $S_t$ follows Black-Scholes dynamics (in conformance with your assumption of log-normal returns)
$$\begin{aligned} d S _ { t } ^ { 0 } & = r S _ { t } ^ { 0 } d t \\ d S _ { t } & = \mu S _ { t } d t + \sigma S _ { t } d W _ { t } \end{aligned}$$
remarkably there is a closed-form solution which it is to invest a constant proportion of wealth in the risky asset
$$\alpha_t = \frac { \mu - r } { \gamma \sigma ^ { 2 } }$$
Notice that the solution can be interpreted as the mean-variance trade-off. |
Extraordinary claims require extraordinary proofs which really is the reason why this sorts of discussion is important. Similarly, sometimes, you are so blinded to some sorts of a truth and are faced with something so different that you can misread entirely what is being said. If you read this morning's entry, you might get a feel that I am little ambivalent about the true interesting nature of a paper entitled Statistical physics-based reconstruction in compressed sensing by Florent Krzakala, Marc Mézard, François Sausset, Yifan Sun, Lenka Zdeborová. Let's put this in perspective, our current understanding so far is that the universal phase transition observed by Donoho and Tanner seems to be seen with all the solvers featured here, that there are many ensembles for which it fits (not just Gaussian, I remember my jaw dropping when Jared Tanner showed it worked for the ensembles of Piotr Indyk, Radu Berinde et al) and that the only way to break it is to now consider structured sparsity as shown by Phil Schniter at the beginning of the week. In most people's mind, the L_1 solvers are really a good proxy to the L_0 solvers since even greedy solvers (the closest we can find to L_0 solvers) seem to provide similar results. Then there are results like the ones of Shrinivas Kudekar and Henry Pfister. ( Figure 5 of The Effect of Spatial Coupling on Compressive Sensing) that look like some sort of improvement (but not a large one). In all, a slight improvement over that phase transition could, maybe, be attributed to a slightly different solver, or ensemble (measurement matrices). So this morning I made the point that given what I
understoodabout the graphs displayed in the article, it may be at besta small improvement over the Donoho-Tanner phase transition known to hold for not only Gaussian but other types of matrices and for different kinds of solvers, including greedy algorithms and SL0 (that simulate some sorts of L_0 approach). At bestis really an overstatement but I was intrigued mostly because of the use of an AMP solver, so I fired off an inquisitive e-mail on the subject to the corresponding author: Dear Dr. Krzakala, ... I briefly read your recent paper on arxiv with regards to your statistical physics based reconstruction capability and I am wondering if your current results are within the known boundary of what we know of the phase transition found by Donoho and Tanner or if it is an improvement on it. I provided an explanation of what I meant in today's entry (http://nuit-blanche.blogspot.com/2011/09/this-week-in-compressive-sensing.html). If this is an improvement, I'd love to hear about it. If it is is not an improvement, one wonders if some of the deeper geometrical findings featured by the Donoho-Tanner phase transition have a bearing on phase transition on real physical systems.Best regards,Igor.
The authors responded quickly with:
Dear Igor, Thanks for writing about our work in your blog.Please notice, however, that our axes in the figure you show are not the same as those of Donoho and Tanner. For a signal with N components, we define \rho N as the number of non-zeros in the signal, and \alpha N as the number of measurements. In our notation Donoho and Tanner's parameters are rho_DT = rho/alpha and delta_DT = alpha. We are attaching our figure plotted in Donoho and Tanner's way. Our green line is then exactly the DT's red line (since we do not put any restriction of the signal elements), the rest is how much we can improve on it with our method. Asymptotically (N\to \infty) our method can reconstruct exactly till the red line alpha=rho, which is the absolute limit for exact reconstruction (with exhaustive search algorithms).So we indeed improve a lot over the standard L1 reconstruction!We will be of course very happy to discuss/explain/clarify details if you are interested.With best regardsFlorent, Marc, Francois, Yifan, and Lenka
The reason I messed up reading the variables is because I was probably not expecting something that stunning.
Thank you for Florent Krzakala, Marc Mézard, François Sausset, Yifan Sun, Lenka Zdeborová for their rapid feedback.
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Prajapati, Ravindra S and Sirajuddin, Minhajuddin and Durani, Venuka and Sreeramulu, Sridhar and Varadarajan, Raghavan (2006)
Contribution of Cation-\pi Interactions to Protein Stability. In: Biochemistry, 45 (50). pp. 15000-15010.
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Abstract
Calculations predict that cation-\pi interactions make an important contribution to protein stability. While there have been some attempts to experimentally measure strengths of cation-\pi interactions using peptide model systems, much less experimental data are available for globular proteins. We have attempted to determine the magnitude of cation-\pi interactions of Lys with aromatic amino acids in four different proteins (LIVBP, MBP, RBP, and Trx). In each case, Lys was replaced with Gln and Met. In a separate series of experiments, the aromatic amino acid in each cation-\pi pair was replaced by Leu.Stabilities of wild-type (WT) and mutant proteins were characterized by both thermal and chemical denaturation. Gln and aromatic \rightarrow Leu mutants were consistently less stable than corresponding Met mutants, reflecting the nonisosteric nature of these substitutions. The strength of the cation-\pi interaction was assessed by the value of the change in the free energy of unfolding [\Delta \Delta G°=\Delta G°(Met)-\Delta G°- (WT)]. This ranged from +1.1 to -1.9 kcal/mol (average value -0.4 kcal/mol) at 298 K and +0.7 to -2.6 kcal/mol (average value -1.1 kcal/mol) at the $T_m$ of each WT. It therefore appears that the strength of cation-\pi interactions increases with temperature. In addition, the experimentally measured values are appreciably smaller in magnitude than calculated values with an average difference $\vert \Delta G^o_{expt}- \Delta G^o_{calc \vert av}$ of 2.9 kcal/mol. At room temperature, the data indicate that cation-\pi interactions are at best weakly stabilizing and in some cases are clearly destabilizing. However, at elevated temperatures, close to typical $T_{m}\hspace{2mm}'s$, cation-\pi interactions are generally stabilizing.
Item Type: Journal Article Additional Information: Copyright of this article belongs to American Chemical Society. Department/Centre: Division of Biological Sciences > Molecular Biophysics Unit Depositing User: Sumana K Date Deposited: 30 May 2007 Last Modified: 19 Sep 2010 04:34 URI: http://eprints.iisc.ac.in/id/eprint/9549 Actions (login required)
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Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..
There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university
Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.
What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?
Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.
Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P
Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?
Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?
Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"
@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.
Although not the only route, can you tell me something contrary to what I expect?
It's a formula. There's no question of well-definedness.
I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.
It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.
Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.
You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.
@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.
@Eric: If you go eastward, we'll never cook! :(
I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.
@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)
@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.
@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator |
Let the stochastic process $\{X_t \}_{t \in \mathbb{N}}$ be defined as $$X_t = \sum_{i = 0}^{\infty} \phi^i W_{i-t} $$
where $\phi \in \mathbb{R}$ and the $W_i$ are white noise with $ W_i\sim N(0, \sigma) \forall i \in \mathbb{Z}$. It is well known in econometric theory that these kind of processes are solutions to auto-regressive (and ARMA) type equations.
It is quite natural to wonder if the associated distribution function to the process
$$ F_t(x) = P(X_t \le x) \quad x \in \mathbb{R} $$
has a density (this type of question is asked all the time in the theory of stochastic differential equations, that is when we have continuous time $t$). So do we have a density in this case and can it be recovered explicitly? |
Just in view of the
double universal covering provided by $SU(2)$, $SO(3)$ must a quotient of $SU(2)$ with respect to a central discrete normal subgroup with two elements. This is consequence of a general property of universal covering Lie groups:
If $\pi: \tilde{G} \to G$ is the universal covering Lie-group homomorphism, the kernel $H$ of $\pi$ is a discrete normal central subgroup of the universal covering $\tilde{G}$ of $G= \tilde{G}/H$, and $H$ is isomorphic to the fundamental group of $G$, i.e. $\pi_1(G)$ (wich, for Lie groups, is Abelian) .
One element of that subgroup must be $I$ (since a group includes the neutral element). The other, $J$, must verify $JJ=I$ and thus $J=J^{-1}= J^\dagger$. By direct inspection one sees that in $SU(2)$ it is only possible for $J= -I$. So $SO(3) = SU(2)/\{I,-I\}$.
Notice that $\{I,-I\} = \{e^{i4\pi \vec{n}\cdot \vec{\sigma}/2 }, e^{i2\pi \vec{n}\cdot \vec{\sigma}/2 }\}$ stays in the
center of $SU(2)$, namely the elements of this subgroup commute with all of the elements of $SU(2)$. Moreover $\{I,-I\}=: \mathbb Z_2$ is just the first homotopy group of $SO(3)$ as it must be in view of the general statement I quoted above.
A unitary representations of $SO(3)$ is also a representation of $SU(2)$ through the projection Lie group homomorphism $\pi: SU(2) \to SU(2)/\{I,-I\} = SO(3)$. So, studying unitary reps of $SU(2)$ covers the whole class of unitary reps of $SO(3)$. Let us study those reps.
Consider a unitary representation $U$ of $SU(2)$ in the Hilbert space $H$. The central subgroup $\{I,-I\}$ must be represented by $U(I)= I_H$ and $U(-I)= J_H$, but $J_HJ_H= I_H$ so, as before, $J_H= J_H^{-1}= J_H^\dagger$.
As $J_H$ is unitary and self-adjoint simultaneously, its spectrum has to be included in $\mathbb R \cap \{\lambda \in \mathbb C \:|\: |\lambda|=1\}$. So (a)
it is made of $\pm 1$ at most and (b) the spectrum is a pure point spectrum and so only proper eigenspeces arise in its spectral decomposition.
If $-1$ is not present in the spectrum, the only eigenvalue is $1$ and thus $U(-I)= I_H$. If only the eigenvalue $-1$ is present, instead, $U(-I)= -I_H$.
If the representation is
irreducible $\pm 1$ cannot be simultaneously eigenvalues. Otherwise $H$ would be split into the orthogonal direct sum of eigenspaces $H_{+1}\oplus H_{-1}$. As $U(-1)=J_H$ commutes with all $U(g)$ (because $-I$ is in the center of $SU(2)$ and $U$ is a representation), $H_{+1}$ and $H_{-1}$ would be invariant subspaces for all the representation and it is forbidden as $U$ is irreducible.
We conclude that,
if $U$ is an irreducible unitary representation of $SU(2)$, the discrete normal subgroup $\{I,-I\}$ can only be represented by either $\{I_H\}$ or $\{I_H, -I_H\}$.
Moreover:
Since $SO(3) = SU(2)/\{I,-I\}$, in the former case $U$ is also a representation of $SO(3)$. It means that $I = e^{i 4\pi \vec{n}\cdot \vec{\sigma} }$ and $e^{i 2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ are both transformed into $I_H$ by $U$.
In the latter case, instead, $U$ is not a true representation of $SO(3)$, just in view of a sign appearing after $2\pi$, because $e^{i 2\pi \vec{n}\cdot \vec{\sigma}/2 } = -I$ is transformed into $-I_H$ and only $I = e^{i 4\pi \vec{n}\cdot \vec{\sigma}/2 }$ is transformed into $I$ by $U$. |
Quadratic Equations / Functions Learn easily with Video Lessons and Interactive Practice Problems Introduction Quadratic functions are easy to recognize. The polynomial expression known as a quadratic contains a variable that is squared making it a 2nd degree equation, and the graph is U-shaped. Quadratic expressions that are equal to zero are called quadratic equations.
The
standard form of a quadratic equation is: $ax^{2} + bx + c = 0$
The graph of a quadratic equation has a recognizable shape –
a parabola. The parabola may open up or down, and the direction of the opening is determined by the sign of the leading coeeficient.
Solving Quadratic Equations by Factoring
To find the
solutions to quadratic equations, also known as the zeros or roots, set the quadratic expression equal to zero then factor. The values for x identify where the graph touches the x-axis. There are several methods you can use to factor quadratic equations. Greatest Common Factor
Identify the
greatest common factor ( GCF) of all the terms in the quadratic expression and use the reverse of the Distributive Property to factor.
This equation has a GCF equal to 2x.
$\begin{align} 2x^{2} + 2x& = 0\\ 2x(x + 2) &= 0 \end{align}$
$x = -2, 0$
The graph touches the x-axis at -1 and 0.
Square Root Property
Use the
property of square roots to find the zeros of quadratic equations such as this one.
$\begin{align} x^{2} - 36 &= 0\\ x^{2} - 36 + 36 &= 0 +36\\ x^{2} &= 36\\ x&=\sqrt{36}\\ \sqrt{36} &=\pm 6 \end{align}$
The solution to the quadratic equation is -6 and 6.
Foil
The
foil method is used to simplify the product of two binomials, so the reverse of the foil method can be used to factor quadratic equations with trinomial expressions having a equal to 1.
To reverse the foil method, find factors of the c that sum to the b.
$ax^{2} + bx + c = 0$
$x^{2} + 7x + 6 = 0$
For the product of 6, the factors 1 and 6 sum to 7. Inside two sets of parentheses, add the constants of 6 and 1 to x respectively then set each binomial equal to zero and solve to determine the roots of the equations.
$(x + 6)(x+ 1) = 0$
$x = -6, -1$
To check your work, FOIL.
$\begin{align} (x + 6)(x+ 1)&=0\\ x^{2} + 6x + 1x +6&=0\\ x^{2} + 7x +6&=0 \end{align}$
The roots of the equation are -6 and -1.
Difference of Two Squares
A quadratic equation that is the difference of two squares is also known as a
DOTS equation. If you can recognize which quadratic equations are DOTS (difference of two squares), you can save yourself time when factoring quadratic equations.
To identify DOTS, look for a specific pattern in the quadratic equation. Notice there are only two terms and both are perfect squares. The solution is the square root of the constant.
$ax^{2} + bx + c = 0$
$\begin{align} x^{2} - 49 & =0\\ (x +7) (x -7)&=0 \end{align}$
The solution to this DOTS equations is -7 and 7.
Factor by Grouping
When the quadratic equation has a
trinomial expression with $a\neq 1$, you can factor by grouping. There are several steps to this method.
$ax^{2} + bx + c = 0$
$2x^{2} + -6x -8 =0$
Factor by Grouping
Find the product of a and c. Identify two factors that sum to b. Write new values for bx. Group the factors using parentheses Factor out the GCF of each group Set up the binomial factors.
For this equation $a\times c = -16$. -2 and 8 sum to -6. Take a look at the next steps to solve this quadratic equation.
$\begin{align} 2x^{2} -6x -8 & =0\\ 2x^{2} +2x -8x -8 & =0\\ (2x^{2} +2x)+ (-8x - 8)&=0\\ 2x(x +1) -8(x +1)&=0\\ (2x -8)(x +1)&=0 \end{align}$
The solution to the equation is -1 and 4.
Solving Quadratic Equations by Completing the Square
When you are unable to determine factors, you can use the
complete the square method to solve quadratic equations. To determine the roots using this method, there are several steps.
$ax^{2} + bx + c = 0$
Use the opposite operation to move the constant to the other side of the standard form. Take half of b, square it and add to both sides of the equation. Factor the perfect square on left side of the equation. Apply the square root property to solve.
$\begin{align} x^{2} + 2x -7 &= 0\\ x^{2} + 2x -7 +7&= 0 +7\\ x^{2} + 2x &=7\\ x^{2} + 2x + 1 &=7 + 1\\ x^{2} + 2x + 1 &=8\\ (x + 1)^{2}&= 8\\ x + 1 &= \pm\sqrt{8}\\ x + 1 -1&=-1 \pm\sqrt{8}\\ x&= -1\pm\sqrt{8} \end{align}$
The solution is x&= -1\pm\sqrt{8} which is -3.8 and 1.8.
Solving Quadratic Equations with the Quadratic Formula
If there is no way to factor a quadratic equation or you simply prefer, you can always use the
quadratic formula to determine the value(s) of x.
$ax^{2} + bx + c = 0$
Use the quadratic formula to solve this equation but first use the
discriminant to learn about the roots.
$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Using and Understanding the Discriminant
The
discriminant is the value under the radical, and it provides valuable information about the roots of an quadratic equation.
Discriminant ${b^{2}-4ac}$
if > 0 then there are two real roots if = 0 there is one root repeated if < 0 there are two complex roots
For this problem, the discriminant is greater than zero, so there are two real roots.
$x^{2} -6x -4 =0$
$\begin{align} x&=\frac{6\pm\sqrt{36+16}}{2}\\ x&=\frac{6\pm\sqrt{52}}{2}\\ x &= \frac{6}{2}\pm\frac{\sqrt{52}}{2} x&=3\pm 3.6\\ x&= -0.6, 6.6 \end{align}$
The roots for this equation are -0.6 and 6.6. |
I'm right now learning about Monodromy from self-studying Rick Miranda's fantastic book "Algebraic Curves and Riemann surfaces". Today, I read about monodromy, and the monodromy representation of a holomorphic map between compact Riemann surfaces. I understand that we start by having a holomorphic map $F:X \rightarrow Y$, of degree d, where X and Y are Riemann surfaces, and then we remove the branch points from Y, and all the corresponding points in X mapping to them. Let $B=\{b_1,..,b_n\}$ be the branch points, $A=\{a_1,...,a_m\}$ the ramification points. So, fix a point $q \in V = Y-B$. We have that there are d preimages of q in $U=X-A$.
So for a specific branch point b, we choose some small open neighbourhood W of b so that $F^{-1}(W)$ gives a disjoint union $W_i$ of open neighbourhoods of the points mapping to b. Take some path from our basepoint $q$ to $q_0 \in W$, call this path $\alpha$. Choosing some small loop $\beta$ with basepoint q, with winding number 1 around b, and then considering $\alpha^{-1}\circ \beta \alpha$, gives a loop on V, based at q around b. We can now see that this loop only depends on $\beta$, in some sense.
Say that the points that maps to b has multiplicity $n_i,...,n_j$. Then we have that, according to local normal form, there are local coordinates $z_j$ on the open neighbourhoods from above, so that the map takes the form $z=z_j^{n_j}$. Now, we have that the loop around b, when we lift it up here, will simply yield a cyclic permutation of the preimages in the neighbourhood.
Now, my question is mostly:
How do I apply this concretely?Let us take an example (from Miranda's book) :"Let $f(z) = 4z^2(z-1)^2/(2z-1)^2$ define a holomorphic map of degree 4 from $P^1$ to itself. Show that there are three branch points, and that the three permutations in $S_4$ are $\rho_1=(12)(34)$, $\rho_2(13)(24)$ and $\rho_3=(14)(23)$ up to conjugacy."I can find the branch points, and I see that the multiplicity of the two points mapping to it has multiplicity 2, but I don't get how to rigorously show that the above are the associated permutations.
Hope I was clear, and sorry if I wasn't.
UPDATENow, rereading the question properly, maybe he doesn't want me to find the specific permutations, but just simply showing that they have that conjugacy class. I think that is the case. But I would still be curious of how to find the specific permutation that the monodromy induces. |
One way of understanding why the muon decay $$\mu^-\to e^-+\bar{\nu}_e+\nu_\mu\tag{1}$$ is a
Charged Current (CC) process is to write the intermediate step through which the decay (1) proceeds i.e., $$\mu^-\to \nu_\mu+\color{red}{W^-}\to \nu_\mu+\color{red}{e^-+\bar{\nu}_e}.\tag{2}$$ Now the intermediate step explains that since the process is mediated by charged current interaction or the $W$ boson, it is a Charged Current (CC) process. But what makes the process Flavor Changing (FC)? You can find this term here. As far as I can see, the process (1) or (2) conserves the total as well as individual lepton numbers $L_e,L_\mu,L_\tau$. Does one define a flavor number or something?
One way of understanding why the muon decay $$\mu^-\to e^-+\bar{\nu}_e+\nu_\mu\tag{1}$$ is a
When you write the SM Lagrangian in terms of mass eigenstates, charge current couplings (
W) involve generation-mixing matrices, the CKM for quarks and the PMNS for leptons. By sharp contrast, there is no mixing for neutral current couplings ( Z), and generation mixing enters delicately (GIM-suppressed) at the 2- W exchange, so, then loop level—it runs on the CKM matrices of the charge current couplings!
A hadronic decay going through a
W exchange $B\to \psi K$ thus mutates flavor of the same charge, a b to an s, thereby slipping down from the 3rd to the 2nd generation. Or $\Lambda \to p \pi^-$, an s to a d, from the 2nd to the first. Such flavor mutations of fermions of the same charge, all else being unremarkable, are dubbed "flavor changing", FC.
You might, ill-advisedly, marvel at the
μ to e mutation as such a generational slip, and call it lepton-flavor changing, too. (But, neutrino Lepton flavor was invented to provide conservation agreement, as you note, trivially, at the interaction vertex. It has no analogy to quark flavors, and, unlike them, it is not conserved in propagation: the entire point of neutrino oscillations! This is why people repeatedly stress the "nonphysicality" of $\nu_{e,\mu,\tau}$, a shared fiction). Again: The $\nu_{μ,e}$ neutrinos you wrote are not mass eigenstates ($\nu_{3,2,1}$ , real, propagating particles of the fundamental SM); they are "interaction-convenience combination states", specifically invented to conserve lepton flavors in the SM interactions, $L_e,L_\mu,L_\tau$, but not in kinetic terms.
They thus straddle generations (characterized by mass stratification) through the PMNS matrix, mixing generations thoroughly, far more violently than the CKM. One has not even assigned generations to the mass eigenstates, yet, before resolution of the mass hierarchy question. The consensus is that there is
no point in discussing such, yet, unless you were building recondite models. The jumble of neutrino and antineutrino components on the right hand side of your (2), then, involves all 3 generations and makes a mockery out of "flavor" and FC, since it cannot be ignored.
I hope you can see why it hardly makes sense to be discussing FCCC, at all—even though 30 years ago people were actually discussing the $\Delta S =1$ part of the effective weak hamiltonian, subdominant to the strangeness preserving part, just because CKM mixing is small...
I should contrast the above to the substantial utility of the sound FCNC concept, which usefully demarcates small loop corrections of a SM Lagrangian principle—but that would entail mission creep.
A teaching moment reminder, in response to comments:
Much of the popular misconceptions of flavor violations hinge on the misbegotten terms "flavor basis" and "flavor eigenstates" for $\nu_{e,\mu,\tau}$. You may see the peril of the term if you slip back to the 60s and recall the convenience superposition array coupling to the up quark in the charged weak current. That state, $d~'\equiv d \cos\theta_C+s \sin\theta_C$, where the angle is the so-called Cabbibo angle (discovered by Gell-Mann and Levy), is not a mass eigenstate. Therefore it
has no well-defined flavor: it provides access to two flavors, d and s, defined through the quark mass eigenstates. It is called "in the flavor basis", since it couples to the u flavor in the charged weak vertex. (It underlies the FCCC dispatched above). It would be superflous upon sticking to the mass basis and writing down the CKM matrix in the action, which everybody does, today.
Likewise,$\nu_{e,\mu,\tau}$ are convenience superpositions which remind one that in a vast swath of reactions where neutrino oscillations have not changed their identity, $\nu_{e,\mu,\tau}$ are practical reminders that the charged lepton generation they are associated with in production will be mostly the same in their charge current absorption, usually their primary handle of observation. But,
lacking a well-defined mass, they do no possess a well-defined generation, even though it is their mutation in neutrino oscillations (free propagation!) that is informally dubbed "lepton flavor change".
The all-time popular confusion is reflected in widespread poster charts of the standard model generations. These charts tabulate mass eigenstates for quarks, (u,c,t; d,s,b), on the one hand; but, instead of presenting the analog for leptons, real particles (e,μ,τ; ν
1,ν 2,ν 3), assuming the normal hierarchy prevails, they instead stick in the interaction states (ν e,ν μ,ν τ), conferring a misconstrued physical significance to them and all but inviting the logical quandaries dispatched here. |
Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$?
I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery).
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Let $k$ be a field. What is an explicit power series $f \in k[[t]]$ that is transcendental over $k[t]$?
I am looking for elementary example (so there should be a proof of transcendence that does not use any big machinery).
If $k$ has characteristic zero, then $\displaystyle e^t = \sum_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any purported algebraic relation satisfied by $e^t$.
Edit: Let me fill in a few details. Given a polynomial $P$ in $e^t$ of degree $d$ where each coefficient is a polynomial in $k[t]$ of degree at most $m$, the possible terms that appear in any formal derivative of $P$ lie in a vector space of dimension $(m+1)(d+1)$, so by taking at least $(m+1)(d+1)$ formal derivatives we obtain too many linear relationships between the terms $t^k e^{nt}$. The coefficient of $e^{dt}$ in particular eventually dominates all other coefficients.
Eisenstein proved (actually, stated) in 1852 that if $f=\sum a_n z^n$ is an algebraic power series with rational coefficients, there exist positive integers $A$ and $B$ such that $A a_n B^n$ are integers for all $n$. In particular, as Eisenstein himself remarks, only finitely many prime numbers appear in the denominators of the coefficients of $f$. For example, $e^z$, $\log(1+z)$, etc., are transcendental.
How about $\sum t^{n!}$? Doesn't a "sea-of-zeroes" argument show it can't be algebraic?
Coming back to the lacunary series, I would prefer the series $f(z)=\sum_{k\ge0}z^{d^k}$, where $d>1$ is an integer, because it is the classical example in Mahler's method; this function satisfies the functional equation $f(z^d)=f(z)-z$. I simply copy Ku.Nishioka's argument from her book "Mahler functions and transcendence" (Theorem 1.1.2). Assume that $f(z)$ is algebraic over $\mathbb C(z)$, hence satisfies an \emph{irreducible} equation $f(z)^n+a_{n-1}(z)f(z)^{n-1}+\dots+a_0(z)=0$ where the coefficients $a_j(z)\in\mathbb C(z)$. Substituting $z^d$ for $z$ and using $f(z^d)=f(z)-z$ we obtain $f(z)^n+(-nz+a_{n-1}(z^d))f(z)^{n-1}+\dots=0$. The left-hand sides of both polynomial relations for $f(z)$ must coincide because of the irreducibility. This in particular implies that $a_{n-1}(z)=-nz+a_{n-1}(z^d)$. Letting $a_{n-1}(z)=a(z)/b(z)$ where $a$ and $b$ are two coprime polynomials we see that $a(z)b(z^d)=-nzb(z)b(z^d)+a(z^d)b(z)$. Since $a(z^d)$ and $b(z^d)$ are coprime, $b(z^d)$ must divide $b(z)$. This is possible if only $\deg b(z)=0$, that is, $b(z)=b$ is a nonzero constant. Then $a(z)=-bnz+a(z^d)$ and comparing the degrees of both sides we see that $a(z)$ is constant as well, hence $nz=0$, a contradiction.
over the rationals every power serie with integer coefficents not periodic is tracendent, over Fp a power serie is algebraic iff the secuence of coeficient is p automatic there is a article od jp allouch tracendence of formal series with it information. |
Get your free trial content now! Video Transcript Transcript Percent Change
Xenia is really into social media. Hoping to increase the number of people following her on VidPicStar, she plans to test out different kinds of content – to learn what people really want to see.
First, she posts a really funny video she took at the zoo. Before the post, she had 240 followers. Her video post is a success! Now she has 312 followers.
The Percentage Change Formula
How successful was her post? Let’s use the
percent change formula to calculate the increase in the number of her followers.
I'll write the formula down for you.
The percent change is equal to the amount of change divided by the original value.
This will give you a decimal answer, so you will need to
multiply the resulting decimal by one hundred to change it to a percent. Use common sense to decide if the percent of change is an increase or decrease. Percentage Change Example 1
Let's put numbers into the formula. Remember, Xenia started out with 240 followers then increased to 312.
312 minus 240 divided by 240. The difference is seventy-two. Now divide by 240. The answer is three-tenths. To change to a percent, multiply three-tenths by one hundred.
The
percent change is a thirty percent increase! Wow! A thirty percent increase is fantastic. Percentage Change Example 2
Xenia wants even more followers. So she posts a picture of a tasty muffin. Will the muffin picture boost the number of followers? Let's see. Oh no! Quite the contrary. Her followers decreased to 265.
What is the percent change? Let's again write out the calculation using the percent change formula.
First, find the difference between the two numbers. Three hundred twelve minus 265 then divide by the orginal value of 312. We get forty-seven. Now divide this by 312 and we get fifteen-hundredths. Fifteen-hundredths times one hundred is equal to fifteen percent. The number of Xenia's followers decreased by 15%.
Her followers really did not like that last posting, but then again, food pics are so passé. Xenia tries again. This time she posts a selfie. Maybe this glamorous pose will give her the boost she wants.
Percent Change Übung Du möchtest dein gelerntes Wissen anwenden? Mit den Aufgaben zum Video Percent Change kannst du es wiederholen und üben. Explain the difference between increase and decrease. Tipps
If the number of followers changes from a bigger to a smaller number, the percent of change is a decrease.
The percent change depends on the original value.
Percent means 'over $100$'.
The decimal number $0.35$ corresponds with the percentage $35~\%$.
Lösung
Xenia likes posting videos and pics on social media. She loves to see the number of her followers increase.
First she posts a video. The number of followers increases from $240$ to $312$.
To calculate the percent change, we calculate the amount of change: $312-240=72$. We divide the result by the original value, $240$. This gives us $\frac{72}{240}=0.3$. By multiplying by $100$, and then adding the percent sign, we get the percent change: $30~\%$. We calculate the amount of change: $312-265=47$. Then we divide the result by the original value, $312$. This gives us $\frac{47}{312}=0.15$. By multiplying by $100$ and adding the percent sign, we get the percent of decrease: $15~\%$.
$\text{percent change}=\frac{\text{amount of change}}{\text{original value}}$
To get the percentage, we multiply the resulting decimal by $100$ then add a percent sign.
Decide which formula you need in order to calculate the percent change. Tipps
Just look at the following example:
Your favorite jeans cost $100~\$$. They are now on sale for a reduced price of $80~\$$.
The price decreased by $20~\%$.
$100~\$$ is the original value (original price).
Lösung
Percent change is the amount of change divided by the original value:
$\text{percent change}=\frac{\text{amount of change}}{\text{original value}}$
Now we have to multiply the resulting decimal by $100$ then add a percent sign.
The amount of change is positive, even if a value decreases. We always subtract the smaller from the bigger value. This amount will be divided by the original value. It won't be divided by the new value. Calculate the percent change. Tipps
The percent of change can be calculated by
$\large \frac{\text{amount of change}}{\text{original value}}$
In order to change a decimal to a percent, we have to multiply the decimal by $100$ the add a percent sign.
Remeber to divide by the original value. This is the value we start with.
Lösung
Let's have a look at the formula for percent of change:
$\text{percent change}=\frac{\text{amount of change}}{\text{original value}}$
What's the amount of change? It's the difference between the new and the original value: $312-240=72$. This amount is divided by the original value. The number of followers before Xenia posted the video: $240$.
This gives us the decimal $\frac{72}{240}=0.3$.
Now we multiply this decimal by $100$, add a percent sign. The posting resulted in an increase of $30~\%$ in the number of her followers.
Determine the percent change for each situation. Tipps
Percent change is always a percentage. Therefore, after solving for a decimal answer, multiply the decimal by $100$ then add a percent sign.
Divide the amount of change by the original value then multiply the decimal by $100$.
An example: the ticket price for four tickets is reduced from $\$120$ to $\$100$.
This leads to: $\frac{120-100}{120}=0.1\bar 6\approx 0.17$.
By multiplying with $100$, we get a percent change of $17~\%$.
Lösung
For each example we will use the following formula:
$\text{percent change}=\frac{\text{amount of change}}{\text{original value}}$
To determine the percent change we have to multiply by $100$ then add a percent sign.
Skis
$\frac{200-175}{200}=0.125=12.5~\%$.
Birthday party
$\frac{120-90}{120}=0.25=25~\%$.
Sister
$\frac{300-250}{250}=0.20=20~\%$.
Rugby club
$\frac{46-40}{40}=0.15=15~\%$.
Identify the most successful posts. Tipps
Use the following formula:
$\frac{\text{part}}{\text{whole}}=\frac{\text{percent}}{100}$
Add the result to $260$ in case of an increase. Subtract it from $260$ in case of a decrease.
You can also compare the percentages:
In case of an increase: the higher the percentage, the higher the new value In case of a decrease: the higher the percentage, the lower the new value Lösung
$260$ is the original value.
The picture of Archimedes in the tub and the picture of Dr.Evil show a decreasing number of followers. How much do the number of followers decrease?
Posting the picture of Archimedes leads to a decrease of $20~\%$ because there are $52$ fewer followers. The number of followers decreases to $208$. The picture of Dr.Evil leads to a decrease of $15~\%$ because there are 39 fewer followers. The number of followers decreases to $221$. The boy sitting on a park bench has $15~\%$ increase, meaning there are $39$ more followers. The new number of followers is $299$. The picture of lion babies on a scale has a $20~\%$ increase, and there are an additional $52$ followers. The new number of followers is $312$. The picture of the beagle has a $40~\%$ change which is an increase of $104$ followers. The new number of followers is $364$. Estimate the percent of change left in each set of blocks. Tipps
Since there are fewer blocks, the percentage will be less than $100$%.
Lösung
Percent change is the the amount of change divided by the original amount. If the new number of toy blocks is less than the original number, the percent change is a decrease.
$ p = \frac{\text{amount of decrease}}{\text{original amount}}$
Example (the red set):
First the teacher calculates the change (original amount $-$ new amount):
$27-24 = 3$.
Then she divides that change by the old amount:
$ p = \frac{3}{27} = \frac19=0, \overline1 \approx 0,11$
In the third step, she converts the decimal to a percentage by multiplying by $100$ then add a percent sign:
$ (p \cdot 100)~ \%= (0,11 \cdot 100) ~ \%= 11~ \%$.
Because we are evaluating for the percentage of blocks that are left and not for the percentage of blocks that have been taken away, we subtract the percent change from $100%$:
$100 ~\% - 11 ~\% = 89 ~\%$ |
CIE 1931 xy chromaticity diagram
showing the gamut
of the sRGB color space and location of the primaries. The D65 white point
is shown in the center. The Planckian locus
is shown with color temperatures labeled in kelvin
. The outer curved boundary is the spectral (or monochromatic) locus, with wavelengths shown in nanometers (labeled in blue). Note that the colors in this displayed file are being specified using sRGB. Areas outside the triangle cannot be accurately colored because they are out of the gamut of sRGB therefore they have been interpreted. Also note how the D65 label is not an ideal 6500-kelvin blackbody
because it is based on atmospheric filtered daylight.
Plot of the sRGB intensities versus sRGB numerical values (red), and this function's slope in log-log space (blue) which is the effective gamma at each point. Below a compressed value of 0.04045 or a linear intensity of 0.00313, the curve is linear so the gamma is 1. Behind the red curve is a dashed black curve showing an exact gamma = 2.2 power law.
sRGB is a standard RGB color space created cooperatively by HP and Microsoft in 1996 for use on monitors, printers and the Internet.
sRGB uses the ITU-R BT.709 primaries, the same as are used in studio monitors and HDTV,
[1] and a transfer function (gamma curve) typical of CRTs. This specification allowed sRGB to be directly displayed on typical CRT monitors of the time, a factor which greatly aided its acceptance.
Unlike most other RGB color spaces, the sRGB gamma cannot be expressed as a single numerical value. The overall gamma is approximately 2.2, consisting of a linear (gamma 1.0) section near black, and a non-linear section elsewhere involving a 2.4 exponent and a gamma (slope of log output versus log input) changing from 1.0 through about 2.3.
Contents Background 1 The sRGB gamut 2 Specification of the transformation 3 The forward transformation (CIE xyY or CIE XYZ to sRGB) 3.1 The reverse transformation 3.2 Theory of the transformation 4 Viewing environment 5 Usage 6 See also 7 References 8 External links 9 Background
The sRGB color space has been endorsed by the W3C, Exif, Intel, Pantone, Corel, and many other industry players. It is used in proprietary and open graphics file formats, such as SVG.
The sRGB color space is well specified and is designed to match typical home and office viewing conditions, rather than the darker environment typically used for commercial color matching.
Much software is now designed with the assumption that an 8-bit-per-channel image file placed unchanged onto an 8-bit-per-channel display will appear much as the sRGB specification recommends. LCDs, digital cameras, printers, and scanners all follow the sRGB standard. Devices which do not naturally follow sRGB (as older CRT monitors did) include compensating circuitry or software so that, in the end, they also obey this standard. For this reason, one can generally assume, in the absence of embedded profiles or any other information, that any 8-bit-per-channel image file or any 8-bit-per-channel image API or device interface can be treated as being in the sRGB color space. However, when the correct displaying of an RGB color space is needed, color management usually must be employed.
The sRGB gamut
sRGB defines the chromaticities of the red, green, and blue primaries, the colors where one of the three channels is nonzero and the other two are zero. The gamut of chromaticities that can be represented in sRGB is the color triangle defined by these primaries. As with any RGB color space, for non-negative values of R, G, and B it is not possible to represent colors outside this triangle, which is well inside the range of colors visible to a human with normal trichromatic vision.
Chromaticity Red Green Blue White point x 0.6400 0.3000 0.1500 0.3127 y 0.3300 0.6000 0.0600 0.3290 Y 0.2126 0.7152 0.0722 1.0000
On an sRGB display, each solid bar should look as bright as the surrounding striped dither. (Note: must be viewed at original, 100% size)
sRGB also defines a nonlinear transformation between the intensity of these primaries and the actual number stored. The curve is similar to the gamma response of a CRT display. It is more important to replicate this curve than the primaries to get correct display of an sRGB image. This nonlinear conversion means that sRGB is a reasonably efficient use of the values in an integer-based image file to display human-discernible light levels.
sRGB is sometimes avoided by high-end print publishing professionals because its color gamut is not big enough, especially in the blue-green colors, to include all the colors that can be reproduced in CMYK printing.
Specification of the transformation The forward transformation (CIE xyY or CIE XYZ to sRGB)
The first step in the calculation of sRGB tristimulus values from the CIE XYZ tristimulus values is a linear transformation, which may be carried out by a matrix multiplication. The numerical values below match those in the official sRGB specification (IEC 61966-2-1:1999) and differ slightly from those in a publication by sRGB's creators.
[2]
It is important to note that these linear RGB values are not the final result. \begin{bmatrix} R_\mathrm{linear}\\G_\mathrm{linear}\\B_\mathrm{linear}\end{bmatrix}= \begin{bmatrix} 3.2406&-1.5372&-0.4986\\ -0.9689&1.8758&0.0415\\ 0.0557&-0.2040&1.0570 \end{bmatrix} \begin{bmatrix} X \\ Y \\ Z \end{bmatrix}
Note also, that if the CIE xyY color space values are given (where
x, y are the chromaticity coordinates and Y is the luminance), they must first be transformed to CIE XYZ tristimulus values by: X = Y x / y,\, Z = Y (1- x - y)/y\,
The intermediate parameters R_\mathrm{linear}, G_\mathrm{linear} and B_\mathrm{linear} for in-gamut colors are defined to be in the range [0,1], which means that the initial
X, Y, and Z values need to be similarly scaled (if you start with XYZ values going to 100 or so, divide them by 100 first, or apply the matrix and then scale by a constant factor to the [0,1] range). The linear RGB values are usually clipped to that range, with display white represented as (1,1,1); the corresponding original XYZ values are such that white is D65 with unit luminance ( X, Y, Z = 0.9505, 1.0000, 1.0890). Calculations assume the 2° standard colorimetric observer. [2]
sRGB was designed to reflect a typical real-world monitor with a gamma of 2.2, and the following formula transforms the linear values into sRGB. Let C_\mathrm{linear} be R_\mathrm{linear}, G_\mathrm{linear}, or B_\mathrm{linear}, and C_\mathrm{srgb} be R_\mathrm{srgb}, G_\mathrm{srgb} or B_\mathrm{srgb}:
C_\mathrm{srgb}=\begin{cases} 12.92C_\mathrm{linear}, & C_\mathrm{linear} \le 0.0031308\\ (1+a)C_\mathrm{linear}^{1/2.4}-a, & C_\mathrm{linear} > 0.0031308 \end{cases}
These gamma-corrected values are in the range 0 to 1. If values in the range 0 to 255 are required, e.g. for video display or 8-bit graphics, the usual technique is to multiply by 255 and round to an integer.
The reverse transformation
Again the sRGB component values R_\mathrm{srgb}, G_\mathrm{srgb}, B_\mathrm{srgb} are in the range 0 to 1. (A range of 0 to 255 can simply be divided by 255).
C_\mathrm{linear}= \begin{cases}\frac{C_\mathrm{srgb}}{12.92}, & C_\mathrm{srgb}\le0.04045\\ \left(\frac{C_\mathrm{srgb}+a}{1+a}\right)^{2.4}, & C_\mathrm{srgb}>0.04045 \end{cases} where a = 0.055 and where C is R, G, or B.
Followed by a matrix multiplication of the linear values to get XYZ:
\begin{bmatrix} X\\Y\\Z\end{bmatrix}= \begin{bmatrix} 0.4124&0.3576&0.1805\\ 0.2126&0.7152&0.0722\\ 0.0193&0.1192&0.9505 \end{bmatrix} \begin{bmatrix} R_\mathrm{linear}\\ G_\mathrm{linear}\\ B_\mathrm{linear}\end{bmatrix} Theory of the transformation
It is often casually stated that the decoding gamma for sRGB data is 2.2, yet the above transform shows an exponent of 2.4. This is because the net effect of the piecewise decomposition is necessarily a changing instantaneous gamma at each point in the range: it goes from gamma = 1 at zero to a gamma of 2.4 at maximum intensity with a median value being close to 2.2. The transformation was designed to approximate a gamma of about 2.2, but with a linear portion near zero to avoid having an infinite slope at
K = 0, which can cause numerical problems. The continuity condition for the curve C_\mathrm{linear} which is defined above as a piecewise function of C_\mathrm{srgb}, is \left(\frac{K_0+a}{1+a}\right)^\gamma=\frac{K_0}{\phi}.
Solving with \gamma = 2.4 and the standard value \phi=12.92 yields two solutions, K_0 ≈ 0.0381548 or K_0 ≈ 0.0404482. The IEC 61966-2-1 standard uses the rounded value K_0=0.04045. However, if we impose the condition that the slopes match as well then we must have
\gamma\left(\frac{K_0+a}{1+a}\right)^{\gamma-1}\left(\frac{1}{1+a}\right)=\frac{1}{\phi}.
We now have two equations. If we take the two unknowns to be K_0 and \phi then we can solve to give
K_0=\frac{a}{\gamma-1},\ \ \ \phi=\frac{(1+a)^\gamma(\gamma-1)^{\gamma-1}}{(a^{\gamma-1})(\gamma^\gamma)}.
Substituting a=0.055 and \gamma=2.4 gives K_0 ≈ 0.0392857 and \phi ≈ 12.9232102, with the corresponding linear-domain threshold at K_0 / \phi ≈ 0.00303993. These values, rounded to K_0=0.03928, \phi=12.92321, and K_0/\phi=0.00304, are sometimes used to describe sRGB conversion.
[3] Publications by sRGB's creators [2] rounded to K_0=0.03928 and \phi=12.92, resulting in a small discontinuity in the curve. Some authors adopted these values in spite of the discontinuity. [4] For the standard, the rounded value \phi=12.92 was kept and the K_0 value was recomputed to make the resulting curve continuous, as described above, resulting in a slope discontinuity from 12.92 below the intersection to 12.70 above. Viewing environment
Parameter Value Luminance level 80 cd/m 2 Illuminant white point x = 0.3127, y = 0.3291 (D65) Image surround reflectance 20% (~medium gray) Encoding ambient illuminance level 64 lux Encoding ambient white point x = 0.3457, y = 0.3585 (D50) Encoding viewing flare 1.0% Typical ambient illuminance level 200 lux Typical ambient white point x = 0.3457, y = 0.3585 (D50) Typical viewing flare 5.0%
The sRGB specification assumes a dimly lit encoding (creation) environment with an ambient correlated color temperature (CCT) of 5000 K. It is interesting to note that this differs from the CCT of the illuminant (D65). Using D50 for both would have made the white point of most photographic paper appear excessively blue.
[5] The other parameters, such as the luminance level, are representative of a typical CRT monitor.
For optimal results, the ICC recommends using the encoding viewing environment (i.e., dim, diffuse lighting) rather than the less-stringent typical viewing environment.
[2]
Usage
Due to the standardization of sRGB on the Internet, on computers, and on printers, many low- to medium-end consumer digital cameras and scanners use sRGB as the default (or only available) working color space. As the sRGB gamut meets or exceeds the gamut of a low-end inkjet printer, an sRGB image is often regarded as satisfactory for home use. However, consumer-level CCDs are typically uncalibrated, meaning that even though the image is being labeled as sRGB, one can't conclude that the image is color-accurate sRGB.
If the color space of an image is unknown and it is an 8- to 16-bit image format, assuming it is in the sRGB color space is a safe choice. This allows a program to identify a color space for all images, which may be much easier and more reliable than trying to track the "unknown" color space. An ICC profile may be used; the ICC distributes three such profiles:
[6] a profile conforming to version 4 of the ICC specification, which they recommend, and two profiles conforming to version 2, which is still commonly used.
Images intended for professional printing via a fully color-managed workflow, e.g. prepress output, sometimes use another color space such as Adobe RGB (1998), which allows for a wider gamut. If such images are to be used on the Internet they may be converted to sRGB using color management tools that are usually included with software that works in these other color spaces.
The two dominant programming interfaces for 3D graphics, OpenGL and Direct3D, have both incorporated half part support for the sRGB color space by using sRGB's gamma curve. OpenGL supports the textures with sRGB gamma encoded color components (first introduced with EXT_texture_sRGB extension, added to the core in OpenGL 2.1) and rendering into sRGB gamma encoded framebuffers (first introduced with EXT_framebuffer_sRGB extension, added to the core in OpenGL 3.0). Direct3D supports sRGB gamma textures and rendering into sRGB gamma surfaces starting with DirectX 9. Correct mipmapping and interpolation of sRGB gamma textures has direct hardware support in texturing units of most modern GPUs (for example nVidia GeForce 8 performs conversion from 8-bit texture to linear values before interpolating those values), and do not have any performance penalty.
[7] See also References ^ Charles A. Poynton (2003). Digital Video and HDTV: Algorithms and Interfaces. Morgan Kaufmann. ^ a b c d Michael Stokes, Matthew Anderson, Srinivasan Chandrasekar, Ricardo Motta (November 5, 1996). "A Standard Default Color Space for the Internet – sRGB, Version 1.10". ^ Phil Green and Lindsay W. MacDonald (2002). Colour Engineering: Achieving Device Independent Colour. John Wiley and Sons. ^ Jon Y. Hardeberg (2001). Acquisition and Reproduction of Color Images: Colorimetric and Multispectral Approaches. Universal-Publishers.com. ^ Rodney, Andrew (2005). Color Management for Photographers. Focal Press. p. 121. ^ sRGB profiles, ICC ^ GPU Gems 3, section 24.4.1, http://http.developer.nvidia.com/GPUGems3/gpugems3_ch24.html Standards IEC 61966-2-1:1999 is the official specification of sRGB. It provides viewing environment, encoding, and colorimetric details. Amendment A1:2003 to IEC 61966-2-1:1999 describes an analogous sYCC encoding for YCbCr color spaces, an extended-gamut RGB encoding, and a CIELAB transformation. sRGB on www.color.org The fourth working draft of IEC 61966-2-1 is available online, but is not the complete standard. It can be downloaded from www2.units.it. External links
International Color Consortium Archive copy of http://www.srgb.com, now unavailable, containing much information on the design, principles and use of sRGB A Standard Default Color Space for the Internet – sRGB at w3.org OpenGL extension for sRGB gamma textures at sgi.com Conversion matrices for RGB vs. XYZ conversion Will the Real sRGB Profile Please Stand Up?
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Suppose that we have a time-homogeneous discrete-time Markov chain $(X_n)$. We want to estimate the transition probabilities $p_{ij} = \mathbb{P}[X_{n+1} = j \mid X_n = i]$.
In the case when we have full information, e.g. we observe $(X_1, X_2, \ldots, X_N)$, the maximum likelihood estimator is $$ \hat{p}_{ij} = \frac{ n_{ij} }{ \sum_{k=1}^{m} n_{ik} }, $$ where $n_{ik}$ is the number of times that the process moved from state $i$ to $k$. (I took the expression above from https://stats.stackexchange.com/a/14361/22409)
Unfortunately, in my case we are not able to observe all the transitions. For example, we may observe $X_1, X_6, X_{11}, X_{13}, X_{14}, X_{15}, \ldots$, that is some of the states, but the states that we observe are not regularly spaced.
I have two related questions:
Mathematically, how do we estimate the transition probabilities given such data observations? What statistical software is available for estimating the transition probabilities? |
(1) is true if $char(k)=0$. This follows from a combination of results. First of all, it is true over any field that the spectrum $MGL$ is connective, which means that
$$MGL^{p,q}(X)=0$$
if $p>q+dim(X)$, $X\in Sm/k$ [1, Cor. 2.9]. (Slightly more is true: for any $p\geq q+dim(X)$, the orientation map $MGL\to H\mathbb{Z}$ induces an isomorphism $MGL^{p,q}(X)\cong H\mathbb Z^{p,q}(X)$ [1, Lem 6.4].)
Second, we have the Hopkins-Morel equivalence [1, Thm. 6.11]
$$MGL/(x_1,x_2,\dots) \simeq H\mathbb Z.$$
Assuming this, Spitzweck has shown in [2] that that the slices of MGL are given by
$$s_rMGL\simeq \Sigma^{2r,r}H(MU_{2r}),$$
Moreover, he gives an explicit description of the $r$-effective cover $f_rMGL$ as a homotopy colimit of spectra of the form $\Sigma^{2i,i}MGL$ for $i\geq r$, which shows that $f_rMGL$ is also $r$-connective (being a homotopy colimit of $r$-connective spectra).Because the homotopy $t$-structure is right complete [1, Cor. 1.4], this implies that
$$\mathrm{holim}_{r\to\infty}f_rMGL=0.$$
Now, let $E=\Sigma^{-p,-q}\Sigma^\infty X_+$ with $p>2q$. Take any map $E\to MGL$. Since $H^{p+2r,q+r}(X,A)=0$ for any abelian group $A$ and $r\in\mathbb Z$, this map lifts through all stages of the slice filtration, hence comes from a map $E\to \mathrm{holim}_{r\to\infty}f_rMGL=0$. QED.
(Incidentally, this shows that there is a strongly convergent spectral sequence $H^{\ast\ast}(X,MU_{2\ast})\Rightarrow MGL^{\ast\ast}(X)$.)
If $char(k)>0$ ($k$ need not be perfect), the Hopkins-Morel equivalence is also known if $char(k)$ is inverted [1], so we can at least deduce that $MGL^{p,q}(X)$ is $char(k)$-torsion for all $p>2q$, $X\in Sm/k$ (and it is zero if $p>q+dim(X)$ by connectivity, so for fixed $X$ and $q$ at most finitely many of these groups can be nonzero).
Some comments about (2): if you go through the proof of the characteristic zero case in [3] and try to use Gabber's theorem instead of resolution of singularities, at some point in the proof an isomorphism is replaced by a finite flap map $f: Y\to X$ of degree prime to a given prime $l\neq p$, and the proof will work if that map has a section. Even if you work $\mathbb Z_{(l)}$-locally, you still need a map $g: X\to Y$ such that $fg=deg(f)\cdot\mathrm{id}$, and I don't see why you'd have such a map in $SH\otimes\mathbb Z_{(l)}$. But for $MGL_{(l)}$-modules I guess that the Gysin map [4] should work.
[1] M. Hoyois, From algebraic cobordism to motivic cohomology (pdf)
[2] M. Spitzweck, Relations between slices and quotients of the algebraic cobordism spectrum (pdf)
[3] O. Röndigs, P. Østvær, Modules over motivic cohomology (pdf)
[4] F. Déglise, Around the Gysin triangle II (pdf) |
I have a question concerning fermions in curved space-time.
Please read it to the end before suggesting the spin-connection and vierbein-based approach.
I heard that there is a special way of thinking about spin-1/2 particles (Dirac fermions) in flat space-time: the spinor field $\psi(x)$ is considered a (Grassmanian) scalar multiplet (under the Lorentz transformations), but the matrix-valued 4-vector $\gamma^{\mu}$ transforms as an actual 4-vector.
The value of the $\psi$ field here is in correspondence with the value of the usual spinor-transforming field, but taken at some fixed frame of reference (in which $\gamma^{\mu}$ take the usual fixed values). Quantities like $\bar{\psi} \gamma^{\mu} \psi$ transform like vectors, which is basically why this formalism is equivalent to the standard (with $\psi$ transforming as spinor and constant $\gamma^{\mu}$).
The Dirac action is then just $$ S[\psi] = \int d^4 x \: \bar{\psi} \left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi, $$ which is manifestly Lorentz-invariant in this strange formalism.
My question is about curved space-time of GR. The idea is to write something like $$ S[\psi] = \int d^4 x \: \sqrt{-g} \: \bar{\psi} \left( i \gamma^{\mu} \partial_{\mu} - m \right) \psi, $$ where $\gamma^{\mu}$ transforms as matrix-valued vector under GCTs, $\nabla_{\mu} \psi$ and $\partial_{\mu} \psi$ are equivalent since $\psi$ is basically a (Grassmanian) scalar multiplet. So this new action is manifestly diffeomorphism-invariant, and agrees with the Dirac field in flat space limit. Also (since $\left\{ \gamma^{\mu}, \gamma^{\nu} \right\} = 2 g^{\mu \nu} \cdot 1_{4 \times 4}$) the metric field can be constructed out of (more fundamental?) matrix-valued vector field $\gamma^{\mu}$.
My teacher says it is incorrect, and I am pretty sure it is, but he can't explain why (and that's what really bothers me). One guess is that the interaction between fermions and gravity probably not correct since there is no spin-connection term (like in the standard vierbein-based approach).
So the question then becomes: what should I add in this action to make the fermion-gravity interaction term correct, given that I don't want to abandon this strange formalism and consider the spinor transformation of $\psi$. |
How fast would you throw an object that it never came back down? This is where escape speed comes into the picture. It is the minimum speed required to escape a planet’s gravitational pull.
A spacecraft leaving the surface of the earth should be going at a speed of about 11 kilometres (7 miles) per second to enter the outer orbit. Here, in this article let us dig deeper into the concept of escape speed.
What is Escape Speed?
Escape speed is the minimum speed with which a mass should be projected from the Earth’s surface in order to escape Earth’s gravitation field. Escape speed, also known as escape velocity is defined as:
The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.
For example, if we consider earth as a massive body. The escape velocity is the minimum velocity that an object should acquire to overcome the gravitational field of earth and fly to infinity without ever falling back.
It purely depends on the distance of the object from the massive body and the mass of the massive body. More the mass it will be higher, similarly, the closer distance, higher will be the escape velocity.
For any massive bodies such as planets, stars which are spherically symmetric in nature,
the escape speed for any given distance is mathematically expressed as: Where, is the escape speed v e is the universal gravitational constant (G≅6.67×10 G -11m 3kg -1s -2) is the mass of the massive body(the body from which the object is to be escaped from) M is the distance from the centre of the massive body to the object r
Here we can notice that the above-mentioned relation is independent of the mass of the object which will be escaping from the massive body.
You may also want to check out these topics given below! Derivation of Escape Speed
In general escape, speed is achieved when the object moves with a velocity at which the arithmetic sum of the object’s gravitational potential energy and its Kinetic energy equates to zero. That is, the object should possess greater kinetic energy than the gravitational potential energy to escape to infinity.
The simplest way of deriving the formula is by using the concept of conservation of energy. Let us assume that the object is trying to escape from a planet (which is uniform circular in nature) by moving away from it. The prime force acting on such an object will be the planet’s gravity. As we know, Kinetic energy(K) and the Gravitational Potential Energy(U g) are the only two kinds of energies associated here.
By the principle of conservation of energy, we can write:\(\left ( K+U_{g} \right )_{i}=\left ( K+U_{g} \right )_{f}\)
Where, \(K=\frac{1}{2}mv^{2}\) \(U=\frac{GMm}{r}\)
\(\frac{1}{2}mv_{e}^{2}-\frac{GMm}{r}=0+0\)\(\frac{1}{2}mv_{e}^{2}=\frac{GMm}{r}\)\(\Rightarrow v_{e}=\sqrt{\frac{2GM}{r}}\)
Here U gf is zero as the distance is infinity and Kf will also be zero as final velocity will be zero. Thus, we get:
\(v_{e}=\sqrt{2gr}\)
The minimum velocity required to escape from the gravitational influence of massive body is given by:
\(g=\frac{GM}{r^{2}}\)
Where,
The escape speed of the earth at the surface is approximately 11.186km/s. That means “
an object should have a minimum of 11.186km/s initial velocity to escape from earth’s gravity and fly to infinite space.”
Ideally, If you can jump with initial velocity 11.186km/s you can tour outer space! Isn’t it interesting? For more such brain-twisting concepts do follow the links given below.
Unit Of Escape Speed
Unit of escape speed or the escape velocity is metre per seconds (m.s
-1) which is also the SI unit of escape speed. Dimensional Formula:
Dimensional formula of universal gravitational constant = M -1L 3T -2 Dimensional formula of the mass of the earth = M 1L 0T 0 Dimensional formula of distance to the centre of the earth = M 0L 1T 0 Therefore, the dimensional formula of escape speed after substituting in the equation is = M 0L 1T -1 Frequently Asked Questions 1) Is it better to launch a ship into the orbit from near or away from the equator? Ans: It is better to launch a ship from the equator because the radius is greater at the equator than at the poles. This lowers the escape velocity. 2) Compute the escape velocity for the indicated planet. Use G = 6.67 x 10 -11 N-m2/kg2 a) Mars: Mass 6.46 x 10 23 kg; Radius 3.39 x 106 m Solution:
\(v_{e}=\sqrt{\frac{2GM}{r}}\)
The formula to find the escape speed is as follows:
Substituting the values in the equation, we get\(v_{e}=\sqrt{\frac{2(6.67\times 10^{-11})(6.46\times 10^{23})}{3.39\times 10^6}}\) \(\sqrt{25420766}\) \(\approx 5.04\times 10^3\)
The escape speed for earth is approximately equal to 5.04 x 10
3 m/s. |
Sep 30th 2019, 05:59 AM
# 2
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Join Date: Oct 2017
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Posts: 474
There's no need to introduce work done here. The potential is related to the electric field strength by:
$\displaystyle \vec{E} = - \nabla V$
Assuming the scalar field $\displaystyle V(x,y,z) = c_0 c_1 x^3 + x - 25 yz^4$, we can compute the gradient, $\displaystyle \nabla V$.
$\displaystyle \nabla V = \frac{\partial V}{\partial x} \hat{x} + \frac{\partial V}{\partial y} \hat{y} + \frac{\partial V}{\partial z} \hat{z}$
$\displaystyle = \left(3 c_0 c_1 x^2 +1\right) \hat{x} - 25z^4 \hat{y} - 100 y z^3 \hat{z}$
Assuming that the only force in the problem is the electric force, we have
$\displaystyle \vec{F} = \vec{E} q_0 = m_0 \vec{a}$.
Therefore,
$\displaystyle \vec{a} = \frac{\vec{E} q_0}{m_0} = \frac{-q_0 \nabla V }{m_0}$
$\displaystyle = \frac{q_0}{m_0} \left(-\left(3 c_0 c_1 x^2 +1\right) \hat{x} + 25z^4 \hat{y} + 100 y z^3 \hat{z}\right)$
You might want to double-check that the potential you wrote down is the same as the one I assumed... |
Research Open Access Published: Exponential decay rate for a quasilinear von Karman equation of memory type with acoustic boundary conditions Boundary Value Problems volume 2015, Article number: 122 (2015) Article metrics
1143 Accesses
5 Citations
Abstract
In this paper, we show the exponential decay result of the quasilinear von Karman equation of memory type with acoustic boundary conditions. This work is devoted to investigating the influence of kernel function
g and the effect of the nonlinear term \(|u'|^{\rho}u''\) and to proving exponential decay rates of solutions when g does not necessarily decay exponentially. This result improves on earlier ones concerning the exponential decay. Introduction
Let \(\Omega\subset R^{2} \) be a bounded domain with sufficiently smooth boundary
∂Ω, \(\Gamma_{0} \cup\Gamma_{1} =\partial\Omega\), \(\Gamma_{0} \cap\Gamma_{1} =\emptyset\), \(\Gamma_{0} \) and \(\Gamma_{1} \) have positive measure and \(\nu=(\nu_{1} , \nu _{2} )\) be the outward unit normal vector, and by \(\tau=(-\nu_{2} , \nu_{1} )\) we denote the corresponding unit tangent vector on ∂Ω. We define \(u' = \frac{\partial u}{\partial t}\), \(\Delta u= \sum_{i=1}^{2} \frac{\partial^{2} u }{\partial x_{i}^{2} }\), \(\Delta ^{2} u= \sum_{i=1}^{2} \frac{\partial^{4} u }{\partial x_{i}^{4} }\), where \(x=(x_{1} , x_{2}) \in\Omega\). In this paper, we consider the quasilinear von Karman equation of memory type with acoustic boundary conditions:
where \(\alpha>0 \), \(\rho>0\). The functions
g, p and q satisfy some conditions to be specified later, the von Karman bracket \([\cdot, \cdot ] \) is given by
and
here \(\mu\in(0, \frac{1}{2} )\) is Poisson’s ratio,
The physical applications of the above system are related to the problem of noise control and suppression in practical applications. In this model, problem (1.1)-(1.8) describes small vibrations of a thin homogeneous isotropic plate of uniform thickness of
α with acoustic boundary conditions on a portion of the boundary and the Dirichlet boundary condition on the rest, \(u(x,t)\) denotes the transversal displacement of the plate, the Airy stress function, \(v(x,t)\) a vibrating plate and \(y(x,t)\) the normal displacement to the boundary; and Eq. (1.1) was interpreted by the stresses at any instant dependent on the complete history of strains. In this model, the portion of the boundary denoted by \(\Gamma_{1}\) is a locally reacting plate, with each point on the plate acting like a damped harmonic oscillator in the response to excess stress from the fluid in the interior Eq. (1.7). The coupling between the acoustic stress and the displacement of the boundary is given by Eqs. (1.5)-(1.6). The noise sound propagates through some acoustic medium, for example, through air, in a room which is characterized by a bounded domain Ω and whose walls, ceiling and floor are described by the boundary conditions. This is the description of Wu in [1]. For more physical explanation of wave equations with acoustic boundary, we refer the reader to [2–10]. The acoustic boundary conditions were introduced by Beale and Rosencrans in [3, 11], where the authors proved the global existence and regularity of solutions in a Hilbert space of the linear problem
where
m, p and q are nonnegative functions on the boundary with m and q being strictly positive.
Frota and Larkin [12] eliminated the term \(z_{tt}\) and established global solvability and decay estimates for a linear wave equation with boundary conditions
The decay rate estimated for wave equations of memory type with acoustic boundary conditions was studied by Park and Park [9], and Park
et al. [13] investigated the general decay for a von Karman equation of memory type with acoustic boundary conditions. Park and Ha [14] considered the Klein-Gordon equation with damping \(|u_{t} |^{\rho}u_{t}\) and acoustic boundary conditions. Wu [1] also proved the well-posedness for variable-coefficient wave equation with nonlinear damped acoustic boundary conditions. Recently Boukhatem and Benabderrahmane [15] proved the existence and decay of solutions for a viscoelastic wave equation with acoustic boundary conditions. The semilinear wave equation with porous acoustic boundary conditions was studied by Graber and Said-Houari [16], and Graber [17] investigated the strong stability and uniform decay of solutions to a wave equation with semilinear porous acoustic boundary conditions. The uniform decay for a von Karman plate equation with a boundary memory condition was studied by Park and Park [18]. Park and Kang [19] considered the uniform decay of solutions for von Karman equations of dynamic viscoelasticity with memory. The asymptotic behavior and energy decay of the solutions for a quasilinear viscoelastic problems were studied by many authors [20–22], and Kang [23] proved the exponential decay for quasilinear von Karman equation with memory.
This is done by applying the idea presented in [4, 5] with some necessary modification due to the nature of the problem treated here. To the best of our knowledge, there are no results for a quasilinear von Karman equation of memory type with acoustic boundary conditions. Thus this work is meaningful. In particular, the nonlinear term \(|u'|^{\rho} u''\) is difficult to analyze, and the result of the energy decay is dependent on the kernel
g. So we overcome the issue using the change of Lyapunov functional. The structure of this paper is as follows. In Section 2, we give some notation and material needed for our work. In Section 3, we prove the main results. Preliminaries
In this section, we present some material needed in the proof of our result. Throughout this paper, we define
For a Banach space
X, \(\|\cdot\|_{X} \) denotes the norm of X. For simplicity, we denote \(\|\cdot\|_{L^{2} (\Omega)} \) by norm \(\|\cdot\|\) and \(\|\cdot\|_{L^{1} (\Gamma_{1})}\) by \(\|\cdot\|_{\Gamma_{1}}\), respectively. We define, for all \(1\leq p<\infty\),
For \(0< \mu<\frac{1}{2}\), the bilinear form \(a(\cdot, \cdot)\) is defined by
A simple calculation, based on the integration by part formula, yields
Thus, for \((u, v) \in(H^{4} (\Omega)\cap W)\times W\), it holds
Since \(\Gamma_{0} \neq\emptyset\), we have (see [6]) that \(\sqrt {a(u,u)} \) is equivalent to the \(H^{2} (\Omega)\) norm in
W, that is,
This and the Sobolev imbedding theorem imply that for some positive constants \(C_{p}\), \(\tilde{C}_{p} \) and \(C_{s}\),
And since \(V \hookrightarrow L^{\rho+2}(\Omega)\), \(\rho>0\), there exists a positive constant
K such that
By (2.1) and Young’s inequality, we deduce that
From this and (2.2), it holds that
Now we introduce the relative results of the Airy stress function and von Karman bracket \([\cdot, \cdot]\).
Lemma 2.1
[24]
Let u, w be functions in \(H^{2} (\Omega)\) and v in \(H_{0}^{2} (\Omega )\), where Ω is an open bounded and connected set of \(R^{2} \) with regular boundary. Then Lemma 2.2
[25]
where C and \(C'\) are some positive constants.
and
where \(\xi: R_{+} \to R_{+} \) is a nonincreasing differentiable function. Condition (2.8) was considered by Messaoudi and Mustafa [26] when studying the stability of a memory-type Timoshenko system. For the functions
p and q, we assume that \(p, q \in C (\Gamma_{1} )\) and \(p(x), q(x) >0\) on \(\Gamma_{1} \). This assumption implies that there exist positive constants \(p_{i} \), \(q_{i}\) (\(i=0,1 \)) such that Theorem 2.1 Let the initial data \((u_{0}, u_{1}, y_{0} )\in(H^{4} (\Omega) \cap W) \times (H^{3} (\Omega)\cap W ) \times L^{2} (\Gamma_{1})\) and the conditions above on g, p and q hold. Then problem (1.1)-(1.8) admits a unique solution \((u,y)\) in the class Main result
and
From (2.3), we deduce
where
A direct calculation gives
Moreover, (2.5) gives
Define a modified energy by
This and the assumptions
g and p imply that \(\mathcal{E} (t)\) is nonincreasing and one easily sees that
Therefore, it is enough to obtain the desired energy decay for the modified energy \(\mathcal{E}(t)\), which will be done in what follows. For this object, let us define the functional
where
and
Lemma 3.1 For \(N>0\) large enough, there exist \(\beta_{1} >0 \) and \(\beta_{2} >0 \) such that Proof
From (3.9), we have
and
and
From the results of \(J_{1},J_{2}, \ldots,J_{6}\), we get
Therefore, for
N is sufficiently large,
Here we can take
So that we have
where \(\beta_{1} =N-C\), \(\beta_{2} = N+C\). We complete the proof of Lemma 3.1. □
Lemma 3.2 For any \(t_{0} >0 \) and sufficiently large \(N>0\), there exist positive constants \(\alpha_{0} \) and \(\alpha_{1} \) such that Proof
and
Combining these estimates \(I_{i} \), \(i=1,\ldots,6\) and (3.16), we get
Since
g is continuous and positive for any \(t\geq t_{0} >0 \), we have
We first choose \(\varepsilon>0 \) and \(\delta>0\) so small such that \(g_{0} -\varepsilon>0 \) and \(1-2l-\delta\tilde{C}_{p} >0\), respectively. We also take \(\eta>0 \) sufficiently small and \(N>0\) large enough so that \(\frac{N}{2} g(t) +\varepsilon(1-2l-\delta\tilde{C}_{p} ) -\eta(l+l^{2} +C^{\prime}C_{1}^{-1}4E(0))>0\), \(\alpha g_{0} -\eta(\alpha+\frac {K^{2(\rho+1)}}{\rho+1} (2\alpha^{-1} \mathcal{E}(0))^{\rho})- \varepsilon\alpha>0\), \(\frac{N}{2}-\frac{g(0)C_{s} }{4\eta} ( \alpha+\frac{1}{\rho+1} ) >0 \) and \(N p_{0} -\varepsilon/\delta -\eta>0 \) for any \(t\geq t_{0} \). Then we deduce that desired result. We complete the proof of Lemma 3.2. □
Our main result is the following.
Theorem 3.1 Assume that \(l<1/2 \). Then, for each \(t_{0} > 0 \), there exist positive constants \(C_{0} \) and C such that Proof
This and the fact \(\xi'(t)\leq0\) yield
Now, we define
Since \(\xi(t)\) is a nonincreasing positive function, we can easily observe that \(\mathcal{L}(t)\) is equivalent to \(\mathcal{E}(t)\). Thus (3.20) implies that
Integrating the above inequality, we get
References 1.
Wu, J: Well-posedness for a variable-coefficient wave equation with nonlinear damped acoustic boundary conditions. Nonlinear Anal.
75(18), 6562-6569 (2012) 2.
Avalos, G, Lasiecka, I, Rebarber, R: Uniform decay properties of a model in structural acoustics. J. Math. Pures Appl.
79(10), 1057-1072 (2000) 3.
Beale, JT, Rosencrans, SI: Acoustic boundary conditions. Bull. Am. Math. Soc.
80, 1276-1278 (1974) 4.
Berrimi, S, Messaoudi, SA: Existence and decay of solutions of a viscoelastic equation with a nonlinear source. Nonlinear Anal.
64, 2314-2331 (2006) 5.
Guesmia, A, Messaoudi, SA: General energy decay estimates of Timoshenko systems with frictional versus viscoelastic damping. Math. Methods Appl. Sci.
32, 2102-2122 (2009) 6.
Bucci, F, Lasiecka, I: Exponential decay rates for structural acoustic model with an over damping on the interface and boundary layer dissipation. Appl. Anal.
81(4), 977-999 (2002) 7.
Lasiecka, I: Boundary stabilization of a 3-dimensional structural acoustic model. J. Math. Pures Appl.
78(2), 203-232 (1999) 8.
Mugnolo, D: Abstract wave equation with acoustic boundary conditions. Math. Nachr.
279(3), 299-318 (2006) 9.
Park, JY, Park, SH: Decay rate estimates for wave equation of memory type with acoustic boundary conditions. Nonlinear Anal. TMA
74(3), 993-998 (2011) 10.
Vicente, A: Wave equation with acoustic/memory boundary conditions. Bol. Soc. Parana. Mat.
27(1), 29-39 (2009) 11.
Beale, JT: Spectral properties of an acoustic boundary condition. Indiana Univ. Math. J.
25(9), 895-917 (1976) 12.
Frota, CL, Larkin, NA: Uniform stabilization for a hyperbolic equation with acoustic boundary conditions in simple connected domains. Prog. Nonlinear Differ. Equ. Appl.
66, 297-312 (2005) 13.
Park, SH, Park, JY, Kang, YH: General decay for a von Karman equation or memory type with acoustic boundary conditions. Z. Angew. Math. Phys.
63, 813-823 (2012) 14.
Park, JY, Ha, TG: Well-posedness and uniform decay rates for the Klein-Gordon equations with damping term and acoustic boundary conditions. J. Math. Phys.
50, Article No. 013506 (2009) 15.
Boukhatem, Y, Benabderrahmane, B: Existence and decay of solutions for a viscoelastic wave equation with acoustic boundary conditions. Nonlinear Anal.
97, 191-209 (2014) 16.
Graber, PJ, Said-Houari, B: On the wave equation with semilinear porous acoustic boundary conditions. J. Differ. Equ.
252, 4898-4941 (2012) 17.
Graber, PJ: Strong stability and uniform decay of solutions to a wave equation with semilinear porous acoustic boundary conditions. Nonlinear Anal.
74, 3137-3148 (2011) 18.
Park, PJ, Park, SH: Uniform decay for a von Karman plate equation with a boundary memory condition. Math. Methods Appl. Sci.
28, 2225-2240 (2005) 19.
Park, JY, Kang, JR: Global existence and uniform decay for a nonlinear viscoelastic equation with damping. Acta Appl. Math.
110, 1393-1406 (2010) 20.
Liu, WJ: Uniform decay of solutions for quasilinear system of viscoelastic equations. Nonlinear Anal.
71, 2257-2267 (2009) 21.
Messaoudi, SA, Tatar, NE: Exponential and polynomial decay for a quasilinear viscoelastic equation. Nonlinear Anal.
68, 785-793 (2008) 22.
Messaoudi, SA, Tatar, NE: Exponential decay for a quasilinear viscoelastic equation. Math. Nachr.
282, 1443-1450 (2009) 23.
Kang, JR: Exponential decay for nonlinear von Karman equations with memory. Abstr. Appl. Anal.
2013, Article ID 484596 (2013) 24.
Lagnese, JE: Boundary Stabilization of Thin Plates. SIAM, Philadelphia (1999)
25.
Chueshov, I, Lasiecka, I: Global attractors for von Karman evolutions with a nonlinear boundary dissipation. J. Differ. Equ.
198, 196-231 (2004) 26.
Messaoudi, SA, Mustafa, MI: A stability result in a memory-type Timoshenko system. Dyn. Syst. Appl.
18, 457-468 (2009) 27.
Raposo, CA, Santos, ML: General decay to a von Karman system with memory. Nonlinear Anal.
74, 937-945 (2011) 28.
Rivera, JE, Menzala, GP: Decay rates of solutions of a von Karman system for viscoelastic plates with memory. Q. Appl. Math.
LVII, 181-200 (1999) Acknowledgements
YH Kang was supported by research grants from the Catholic University of Daegu in 2014.
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. |
Lets say you are given a distribution function $f(p)$ and you want to define a temperature, $T_f$, for this distribution. (I assume $\mu = 0$.)
It is then natural to define a temperature the following way: \begin{equation} T_f \equiv \frac{ \int d^3p \ G(p) f(p)}{\int d^3p \ f(p)}, \end{equation} where $G(p)$ is defined by the following equation \begin{equation} T = \frac{ \int d^3p \ G(p) f_{eq}(p,T)}{\int d^3p \ f_{eq}(p,T)}, \end{equation} where $f_{eq}(p,T)$ is the equilibrium thermal distribution function.
I know that if $f_{eq}$ is given by the Maxwell-Boltzmann distribution, then $$G_{MB}(p) = \frac{p^2}{3E},$$ where $E = \sqrt{p^2 + m^2}$.
What I need is to find an expression for $G(p)$ if $f_{eq}$ is the Bose-Einstein or Fermi-Dirac distribution $$ f_{eq} = \frac{1}{e^{E(p)/T} \pm 1}.$$
I do not need an analytic expression for $G(p)$, an integral that I can solve numerically is sufficient. It seems to me that this should be possible to do, but I just can't think of how. |
Suppose we have a $U$ shaped hollow tube consisting of three parts as : $T_{v1}$, $T_{v2}$, and $T_{h}$ , representing two vertical tubes and one horizontal tube respectively. The configuration is kept fixed as such so that the horizontal tube acts as a base. Let $T_{v1}$ have a cross-sectional area $A$ and $T_{v2}$ have cross-sectional area $2A$
(The horizontal tube may have any small finite cross-sectional area). An ideal fluid is placed in the tube which starts executing a periodic motion (I'm not sure if this would be simple harmonic motion or not). Assume at a certain instant $h_1 >h_2$ and that the liquid is flowing from $T_{v1}$ to $T_{v2}$. Now I apply Bernoulli's equation at the two liquid surfaces in $T_{v1}$ and $T_{v2}$. $$P_{atm} + 1/2{\rho}v_1^2 + {\rho}gh_1=P_{atm} +1/2{\rho} v_2^2 +{\rho}gh_2$$Using equation of continuity $$v_1=2v_2$$ This would lead to $$3/2{\rho}v_2^2= {\rho}g(h_2 - h_1)$$This gives me a complex solution for velocity. How is this possible? Please help me with this. Thanks in advance.
Suppose we have a $U$ shaped hollow tube consisting of three parts as : $T_{v1}$, $T_{v2}$, and $T_{h}$ , representing two vertical tubes and one horizontal tube respectively. The configuration is kept fixed as such so that the horizontal tube acts as a base. Let $T_{v1}$ have a cross-sectional area $A$ and $T_{v2}$ have cross-sectional area $2A$
I think that the problem here is that you cannot use Bernoulli's principle in this problem because the fluid motion is not steady. I think this document will help you. You could try to do the calculations as shown there, by using Newton's second law.
The problem is you cannot assume at a certain time $h_{1} > h_{2}$! How did you assume this condition?! I mean you don't know the height of fluid column in $T_{v1}$ and $T_{v2}$. In fact, the only thing you know certainly is that $v_{1} = 2v_{2}$ then we have:
$$\frac{1}{2} \rho v_{1}^{2} + \rho g h_{1} = \frac{1}{2} \rho v_{2}^{2} + \rho g h_{2}$$
or:
$$\frac{1}{2} \rho v_{2}^{2} + \rho g h_{2} = \frac{1}{2} \rho 4v_{2}^{2} + \rho g h_{1}$$
or again:
$$\frac{3}{2} \rho v_{2}^{2} = \rho g (h_{2} - h_{1})$$
or furthermore:
$$h_{2} - h_{1} = \frac{3}{2}\frac{v_{2}^{2}}{g}$$
So the only thing you could say that $h_{2} > h_{1}$ and that's the reason why your solution was incorrect. Note that you can always calculate the height difference $h_{2} - h_{1}$ not their absolute values because height also needs reference like pressure.
Also for those who suggested to use unsteady Bernoulli equation! Such a nice invention!? I mean there is not time-dependent or steady Bernoulli equation. Bernoulli equation is just energy balance and as long as there are no non-conservative forces should work no matter the flow is steady or unsteady!
If you are looking for when there are acceleration, we could solve this problem like this:
$$\int_{1}^{2} \rho \frac{d v}{d t} dl + \frac{1}{2}\rho v_{2}^{2} + \rho g h_{2} = \frac{1}{2} \rho v_{1}^{2} + \rho g h_{1}$$
Let's say the path between 1 and 2 includes three parts as $1 \rightarrow 3$, $3 \rightarrow 4$, and $4 \rightarrow 2$. In part $1 \rightarrow 3$ the velocity is $v_{1}$, in the path $3 \rightarrow 4$ if the cross section is $A_{34}$ and its length is $L_{34}$, then velocity is $v_{34} = \frac{A}{A_{34}}v_{1}$. In the path $4 \rightarrow 2$ the velocity is $v_{2}$. So the integral would be expanded as:
$$\int_{1}^{2} \rho \frac{d v}{d t} dl = \int_{1}^{3} \rho \frac{d v}{d t} dl + \int_{3}^{4} \rho \frac{d v}{d t} dl + \int_{4}^{2} \rho \frac{d v}{d t} dl = \rho \frac{d v_{1}}{d t} h_{1} + \rho \frac{d v_{1}}{d t}\frac{A}{A_{34}}L_{34} + \rho \frac{d v_{2}}{d t}h_{2}$$
Also, we know that $\frac{1}{2}v_{1} = v_{2}$, so:
$$\int_{1}^{2} \rho \frac{d v}{d t} dl = \rho \frac{d v_{1}}{d t} h_{1} + \rho \frac{d v_{1}}{d t}\frac{A}{A_{34}}L_{34} + \frac{1}{2}\rho \frac{d v_{1}}{d t}h_{2}$$
Finally:
$$\rho \frac{d v_{1}}{d t} h_{1} + \rho \frac{d v_{1}}{d t}\frac{A}{A_{34}}L_{34} + \frac{1}{2}\rho \frac{d v_{1}}{d t}h_{2} + \frac{1}{4} \rho v_{1}^{2} + \rho g h_{2} = \frac{1}{2} \rho v_{1}^{2} + \rho g h_{1} $$
If we take: $a = h_{1} + \frac{A}{A_{34}} L_{34} + \frac{1}{2}h_{2}$, then:
$$a \frac{d v_{1}}{d t} = \frac{1}{4} v_{1}^{2} + g (h_{1}-h_{2})$$
Then by solving this ODE, we have:
$$v_{1}(t) = 2\sqrt{g (h_{1}-h_{2})} \tanh\Big(\frac{t}{\tau}\Big)$$
Where the relaxation time is defined as: $$\tau = \frac{a}{\sqrt{g (h_{1}-h_{2})}}$$ |
Annals of Mathematical Statistics
Опубликовано на портале: 31-03-2004
Theodore W. Anderson, Donald A. DarlingAnnals of Mathematical Statistics. 1952. Vol. 23. No. 2. P. 193-212.
The statistical problem treated is that of testing the hypothesis that $n$ independent, identically distributed random variables have a specified continuous distribution function F(x). If F_n(x) is the empirical cumulative distribution function and \psi(t) is some nonnegative weight function (0 \leqq t \leqq 1), we consider n^{\frac{1}{2}} \sup_{-\infty and n\int^\infty_{-\infty}\lbrack F(x) - F_n(x) \rbrack^2 \psi\lbrack F(x)\rbrack dF(x). A general method for calculating the limiting distributions of these criteria is developed by reducing them to corresponding problems in stochastic processes, which in turn lead to more or less classical eigenvalue and boundary value problems for special classes of differential equations. For certain weight functions including \psi = 1 and \psi = 1/\lbrack t(1 - t) \rbrack we give explicit limiting distributions. A table of the asymptotic distribution of the von Mises \omega^2 criterion is given.
Опубликовано на портале: 01-07-2004
Peter A.W. LewisAnnals of Mathematical Statistics. 1961. Vol. 32. P. 1118-1124.
Anderson and Darling proposed the use of the statistic for testing the hypothesis that a sample of size n has been drawn from a population with a specified continuous cumulative distribution function Gn(x) is the empirical distribution function defined on the sample on the size n Author consider here the problem of determining and tabulating the distribution function of this statistics. |
Hex 1.0
Hydrogen-electron collision solver
#include <tuple>
#include "arrays.h"
#include "input.h"
#include "matrix.h"
#include "radial.h"
#include "parallel.h"
#include "opencl.h"
class PreconditionerBase Preconditioner template. More... class NoPreconditioner Solution driver without actual preconditioner. More... class CGPreconditioner CG iteration-based preconditioner. More... class GPUCGPreconditioner CG iteration-based preconditioner (GPU variant). More... class JacobiCGPreconditioner Jacobi-preconditioned CG-based preconditioner. More... class SSORCGPreconditioner SSOR-preconditioned CG-based preconditioner. More... class ILUCGPreconditioner ILU-preconditioned CG-based preconditioner. More... class DICCGPreconditioner DIC-preconditioned CG-based preconditioner. More... class SPAICGPreconditioner SPAI-preconditioned CG-based preconditioner. More... class TwoLevelPreconditioner Two-resolution preconditioner. More... class MultiresPreconditioner Multi-resolution (V-cycle) preconditioner. More... class Preconditioners Preconditioner traits. More...
cArray iChol (cArrayView const &A, lArrayView const &I, lArrayView const &P) Sparse incomplete Cholesky decomposition. More... SymDiaMatrix DIC (SymDiaMatrix const &A) DIC preconditioner. More... SymDiaMatrix SSOR (SymDiaMatrix const &A) SSOR preconditioner. More... CooMatrix SPAI (SymDiaMatrix const &A, const iArrayView diagonals) SPAI preconditioner. More...
Setup the diagonal incomplete Cholesky preconditioner. It is a essentially the original matrix with a preconditioned diagonal and the strict upper and lower triangles normalized by the preconditioned diagonal, i.e. a matrix
\[ \mathbf{P} = \mathbf{\tilde{L}}_\mathbf{A} + \mathbf{D}^{-1} + \mathbf{\tilde{L}}_\mathbf{A}^T \]
for the preconditioner
\[ \mathbf{M} = (\mathbf{D} + \mathbf{L}_\mathbf{A}) \mathbf{D}^{-1} (\mathbf{D} + \mathbf{U}_\mathbf{A}) = (1 + \mathbf{\tilde{U}}_\mathbf{A}^T) \mathbf{D} (1 + \mathbf{\tilde{U}}_\mathbf{A}) \]
The formula for the elements of \( \mathbf{D} \) is
\[ d_i = a_{ii} - \sum_{k < i} a_{ik} d_{k}^{-1} a_{ki} \ , \]
and is to be evaluated along the diagonal, re-using the just computed values \( d_i \). Hence, the access pattern in dense matrix would be
\[ \pmatrix { \ast & & & \ast & & \cr & \ast & & \ast & & \cr & & \ast & \ast & & \cr \ast & \ast & \ast & ? & & \cr & & & & & \cr & & & & & \cr } \]
In the case of the sparse SymDiaMatrix, the asterisks will occur only on the nonzero diagonals.
A Matrix in SymDiaMatrix format that is to be preconditioned.
This routine computes the LDL-decomposition of a symmetric matrix,
\[ A = L D L^T \ , \]
where \( L \) is a lower triangular matrix normalized so that it has units on the diagonal and \( D \) is a diagonal matrix.
A Matrix elements in the form of a consecutive array \( \left\{a_i\right\}_{i=1}^N \) as in
\[ \pmatrix { a_1 & & & & \cr a_2 & a_3 & & & \cr a_4 & a_5 & a_6 & & \cr a_7 & a_8 & a_9 & a_{10} & \cr \vdots & & & & \ddots \cr } \]Whenever \( a_k \) is equal to zero, it is (or can be) omitted from the input array.
I Array of column indices (one for every element of A). P Array of row pointers, i.e. starting positions of rows of A. For dense matrix it would be 0, 1, 3, 6, 10, ... The last element must be equal to the length of both A and I.
Compute sparse aproximate inverse of a given symmetrix diagonal matrix A. The sparse structure of the SPAI is set by the second parameter that contains list of non-lower diagonal indices (greater than or equal to zero).
This function uses Lapack routine ZGELSD.
Symmetric successive over-relaxation preconditioner for \( \omega = 1 \). (Essentially symmetrized Gauss-Seidel). The resulting matrix contains normalized lower (and upper) triangle and in the place of the unit diagonal is the inverse diagonal of \( \mathbf{A} \). So, having the preconditioner |
I'm currently taking a convex optimization class and we're using the textbook by Boyd & Vandenberghe. I was solving an exercise problem when I came across a question (for anybody curious the problem is Exercise 9.1). I have two questions in total regarding the same problem. One about a detail within the problem and one regarding the solving itself. Here's the problem:
Consider the problem of minimizing a quadratic function:
$$\text{minimize}\quad f(x) = \frac{1}{2}x^T P x + q^T x + r$$
where $P \in S^n$ (i.e. an $n \times n$ symmetric matrix) but we do not assume $P$ is positive semidefinite.
Show that if $P$ is
notpositive semidefinite (and therefore objective function $f$ is not convex) then the problem is unbounded below. My approach
By definition if a matrix is not positive semidefinite then that means there exists some $z$ such that
$$z^TPz \lt 0$$
Following this logic we could say that $x = \alpha z$ with $x, z \in \Bbb{R}^n$ and $\alpha \in \Bbb{R}$. The objective function changes to
$$f(x) = \frac{1}{2}\alpha^2 (z^TPz) + q^T(\alpha z) + r$$
This is where I get stuck. According to the solution manual, it states that if $\alpha$ becomes large then $f \rightarrow -\infty$ and we can therefore conclude that the function is unbounded below.
My two questions are as follows:
Why does the matrix not being positive semidefinite imply the function is not convex?
How was the conclusion drawn that if $\alpha$ becomes large then the function diverges to $-\infty$?
Thank you. |
Simplify result of this definite integral
It is well known that for $n\in\mathbb{N}$ and $n>0$ (an maybe even for more than these restrictions): $$I_n = \int_0^\infty\frac{x^n}{e^x-1}dx = \zeta(n+1)n!$$ which, analytically can be shown easily by expanding $1/(1-e^{-x})$ into a geometric series, which leads to trivial integrals, and by using $\zeta(n+1)=\sum_{l=1}^\infty l^{-(n+1)}$. So, eg.: $$I_1 = \pi^2/6$$ $$I_2 = 2\zeta(3)$$ a.s.o...
Now, if I try even the simplest case with sage, I get this 'nifty' little results
sage: integrate(x/(exp(x)-1),x,0,oo)-1/6*pi^2 + limit(-1/2*x^2 + x*log(-e^x + 1) + polylog(2, e^x), x,+Infinity, minus)
Is there any trick to simplify this down to the final result, or is this about as far as I can get with sage alone?
PS.: it is probably needless to say that (once again ... :( ...)
In[1]:= Integrate[x/(Exp[x] - 1), {x, 0, Infinity}]Out[1]:= Pi^2/6 |
I just started having a look at 2 Higgs Doublet Models (2HDM) and came across this:
The general idea seems pretty simple. Introduce a second Higgs doublet that acquires a vev.
The general Yukawa interaction term (ignoring leptons) for both doublets $\Phi^1$ and $\Phi^2$ then looks like: $$ \mathcal{L}_Y = Y_{ij}^{U1}\overline{Q}_{L,i}\tilde{\Phi}^1U_{R,j} + Y_{ij}^{D1}\overline{Q}_{L,i}\Phi^1D_{R,j} + Y_{ij}^{U2}\overline{Q}_{L,i}\tilde{\Phi}^2U_{R,j} + Y_{ij}^{D2}\overline{Q}_{L,i}\Phi^2D_{R,j} $$ Now there are several types of 2HDMs two of which go by the simple names Type I and Type II:
If I got it right the short summary of both is the following:
Type I: One doublet is just like the standard model Higgs i.e. it is responsible for both fermion and vector boson masses while the other doublet only interacts with the gauge bosons and does not have any Yukawa couplings
i.e. we set $Y^{U1}$ and $Y^{D1}$ $\rightarrow0$.
Type II: Again both doublets are responsible for the gauge boson masses while in the fermionic sector they split up such that one doublet is responsible for the up-type masses while the other one takes care of the down-type masses (including charged leptons).
Or in short we set $Y^{U1}$ and $Y^{D2}$ $\rightarrow0$.
My question is now:
Which mechanism drives the corresponding yukawa couplings to zero such that the 2 doublets interact in the type I or II way.
For the $\rho$-Parameter to be 1 at least at tree level both doublets need to have the Standard Models Higgs hypercharge of 1. Therefore naturally both doublets should act like the SM Higgs i.e. have Yukawa interactions with up- and down-type fermions.
Is this just some ad hoc Ansatz assuming some more fundamental model will explain it or did I miss something? |
Lets start by writing the definition of the equilibrium constant, for some general reaction:
$$\sum_i^{n_\text{reac}} a_iA_i = \sum_j^{n_\text{prod}}b_jB_j$$
The above is a general reaction, where $a_i$ and $b_j$ is the stoichiometry and $A_i$ and $B_i$ is the
molecules. For this reaction we can write the equilibrium constant:
$$K=\frac{\prod_j^{n_\text{prod}}\frac{B_j^{b_j}}{U_j^{\circ}}}{\prod_i^{n_\text{reac}}\frac{A_i^{a_i}}{U_i^{\circ}}}$$
Here $\prod$ is the symbol that denotes we take the product over a range. The $U_i^\circ$ are a standard unit, to ensure that our equilibrium constant is
well defined.
Now if we look at your reaction:
$$\mathrm{H_2O\left( l \right)} \rightleftharpoons \mathrm{H_2O\left( g \right)}$$
We can from the above equations identify that; $a_1 = 1$, $b_1=1$, $A_1=\left[ \mathrm{H_2O} \right]$ and $B_1=p_\mathrm{H_2O}$.
We can thus write our equilibrium constant as:
$$K=\frac{\frac{p_\mathrm{H_2O}}{1\text{ bar}}}{\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}}$$
As you might note, I have also inserted the $U$s. For pressure $U=1\text{ bar}$ and for concentrations $U=1\text{ M}$. Now we need one last ingredient. You might have learned that the activity of liquids is $1$, i.e. the means that we have to set $\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}=1$, this is due to the definition of activity. If interested I would suggest to take a look at this. As our final result we can see that:
$$K=\frac{p_\mathrm{H_2O}}{1\text{ bar}}$$
I.e. we have that the equilibrium constant equals the partial pressure of water in standard units. |
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If the sum of squares of two numbers is 97, then which one of the foll[#permalink]
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Updated on: 28 Feb 2019, 15:12
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I see this as a special quadratics problem.
\(A^2+B^2=97\), and we're asked about \(AB\). This should ring your special quadratics alarm bells for sure.
Note that \((A \pm B)^2 = A^2 \pm 2AB+B^2 \geq 0\). (Squares are never negative.) This means that \(A^2+B^2 \geq \mp 2AB\), so \(\frac{97}{2} \geq |AB|\). And since this means that \(AB\) definitely cannot be \(64\), we're done.
(For those of you citing the AM-GM inequality, yes, AM-GM is out of GMAT scope, and yes, I basically just re-derived the two-unknown case of it. However, this doesn't mean that one needs to know AM-GM, per se, to solve this problem, so I'm not entirely convinced that this problem is out of GMAT scope.)
Originally posted by AnthonyRitz on 28 Feb 2019, 01:55.Last edited by AnthonyRitz on 28 Feb 2019, 15:12, edited 3 times in total.
Re: If the sum of squares of two numbers is 97, then which one of the foll[#permalink]
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28 Feb 2019, 02:47
AnthonyRitz wrote:
I see this as a special quadratics problem.
\(A^2+B^2=97\), and we're asked about \(AB\). This should ring your special quadratics alarm bells for sure.
Note that \((A \pm B)^2 = A^2 \pm 2AB+B^2 \geq 0\). (Squares are never negative.) This means that \(A^2+B^2 \geq \mp 2AB\), so \(\frac{97}{2} \geq |AB|\). And since \(AB\) definitely cannot be \(64\), we're done.
(Yes, for those of you citing the AM-GM inequality, it is out of GMAT scope, and I basically just re-derived the two-unknown case of it. However, this doesn't mean that one needs to know AM-GM, per se, to solve this problem, so I'm not entirely convinced that this problem is out of GMAT scope.) |
In this article on recurrent neural networks by Razvan Pascanu, $\mathbf x_t$ is the state at time $t;$ $\mathbf u_t$ the input at time $t$; and $\mathcal E$ is the cost function:
A proof is given of the fact that if the absolute value of the main eigenvalue of the matrix of recurrent weights is $\rho < 1$ then the vanishing gradient problem would occur:
If we consider a linear version of the model (i.e. set $\sigma$ to the identity function in eq. (2)) we can use the
power iteration methodto formally analyze this product of Jacobian matrices and obtain tight conditions for when the gradients explode or vanish. It is sufficientfor $\rho < 1$, where $\rho$ is the spectral radius of the recurrent weight matrix $\mathbf W_{rec}$ for long term components to vanish (as $t \to\infty$) and necessaryfor $\rho >1$ for them to explode.
We generalize this result for nonliner functions $\sigma$ where $\lvert \sigma'(x)\rvert$ is bounded, $\lVert diag(\sigma'(\mathbf x_k )) \rVert \leq \gamma \in \mathcal R$, by relying on singular values.
We first prove that it is sufficient for $\lambda_1 <\frac{1}{\gamma}$, where $\lambda_1$ is the largest singular value of $\mathbf W_{rec}$, for the vanishing gradient problem to occur. Note that we assume the parametrization given by eq. (2). The Jacobian matrix $\frac{\partial \mathbf x_{k+1}}{\partial \mathbf x_k}$ is given by $\mathbf W_{rec}^T\, diag(\sigma'(\mathbf x_k)).$ The 2-norm of this Jacobian is bounded by the product of the norms of the the two matrices (see eq. (6)). Due to our assumption, this implies that it is smaller than $1.$
$$\forall k, \; \left\lVert \frac{\partial \mathbf x_{k+1}}{\partial \mathbf x_k} \right\rVert \leq \left \lVert \mathbf W_{rec}^T \right \rVert \, \left \lVert diag(\sigma'(\mathbf x_k)) \right \rVert < \frac{1}{\gamma}\gamma <1 \tag{6}$$
Let $\eta \in \mathrm R$ be such that $\forall k, \, \left\lVert \frac{\partial \mathbf x_{k+1}}{\partial \mathbf x_k} \right \rVert \leq \eta < 1.$ The existence of $\eta$ is given by eq. (6). By induction over $i$, we can now show that
$$\left \lVert \frac{\partial \mathcal{E}_t}{\partial \mathbf x_t} \left( \prod_{i=k}^{t-1} \frac{\partial \mathbf x_{i+1}}{\partial \mathbf x_i} \right) \right \rVert \leq \eta^{t-k}\; \left\lVert \frac{\partial \mathcal{E}_t}{\partial \mathbf x_t}\right \rVert \tag 7$$
As $\eta < 1, $ it follows that, according to eq. (7), long term contributions (for which $t-k$ is large) go to $0$ exponentially fast with $t-k.$ $\tag*{$\square$}$
Can I get some help explaining Eq.6?
In particular, I understand that if $\lambda_1 $ is the largest singular value of $\mathbf W_{\text{rec}}$, and $\lambda_1 < \frac{1}{\gamma} $, the absolute value of the largest eigenvalue $\lambda_1^2 < \frac{1}{\gamma^2}$. This latter value (spectral radius) corresponding to the norm of $\mathbf W_{\text{rec}}$, i.e. $\left\lVert \mathbf{W}_{\text{rec}} \right\rVert$.
Therefore
$$\left\lVert \frac{\partial \mathbf x_{k+1}}{\partial \mathbf x_k} \right\rVert \leq \left \lVert \mathbf W_{\text{rec}}^\top \right \rVert \, \left \lVert \text{diag}(\sigma'(\mathbf x_k)) \right \rVert < \frac{1}{\gamma^2}\gamma = \frac{1}{\gamma}$$
If $\gamma>1 \implies \lambda_1^2 < 1$ (and $\lambda_1 < 1)$, exponentiating this $\frac{1}{\gamma}$ will result in vanishing gradients. If $\gamma < 1$ the gradients will explode.
It makes sense, but it is different from eq. (6). |
Often big datasets and high dimensionality go hand in hand. Sometimes the dimensionality is so high that storage of the full MCMC chain in memory becomes an issue. There are a number of ways around this, including: calculating the Monte Carlo estimate on the fly; reducing the dimensionality of the chain using a test function; or just periodically saving a the chain to the hard disk and starting from scratch. To give you more flexibility we allow an SGMCMC algorithm to be run step by step. This allows you to do what you want with the output of the chain. This guide goes into more detail about how to do this, but it needs more TensorFlow knowledge, such as knowledge of TensorFlow sessions and how to build your own placeholders. For more details on these see the TensorFlow for R documentation.
To demonstrate this concept we fit a two layer Bayesian neural network to the MNIST dataset. The MNIST dataset consists of \(28 \times 28\) pixel images of handwritten digits from zero to nine. The images are flattened to be a vector of length 784. The dataset is available as a standard dataset from the TensorFlow library, with a matrix of 55000 training vectors and 10000 test vectors, each with their corresponding labels. First, let’s construct the dataset and a testset. We assume you’ve read some of the earlier vignettes, so are familiar with how to do this. The MNIST dataset can be downloaded using the
sgmcmc function
getDataset as follows:
library(sgmcmc)# Download and load MNIST datasetmnist = getDataset("mnist")# Build dataset list and testset listdataset = list("X" = mnist$train$images, "y" = mnist$train$labels)testset = list("X" = mnist$test$images, "y" = mnist$test$labels)
We’ll build the same neural network model as in the original SGHMC paper (Chen et. al 2014). Suppose \(Y_i\) takes values in \(\{0,\dots,9\}\), so is the output label of a digit, and \(\mathbf x_i\) is the input vector, with \(\mathbf X\) the full \(N \times 784\) dataset, where \(N\) is the number of observations. Then we model as follows \[ Y_i | \theta, \mathbf x_i \sim \text{Categorical}( \beta(\theta, \mathbf x_i) ), \\ \beta(\theta, \mathbf x_i) = \sigma \left( \sigma \left( \mathbf x_i^T B + b \right) A + a \right). \] Here \(A\), \(B\), \(a\), \(b\) are parameters to be inferred with \(\theta = (A, B, a, b)\); \(\sigma(.)\) is the softmax function. \(A\), \(B\), \(a\) and \(b\) are matrices with dimensions: \(100 \times 10\), \(784 \times 100\), \(1 \times 10\) and \(1 \times 100\) respectively. Each element of these parameters is given a Normal prior to give \[ A_{kl} | \lambda_A \sim N(0, \lambda_A^{-1}), \quad B_{jk} | \lambda_B \sim N(0, \lambda_B^{-1}), \\ a_l | \lambda_a \sim N(0, \lambda_a^{-1}), \quad b_k | \lambda_b \sim N(0, \lambda_b^{-1}), \\ j = 1,\dots,784; \quad k = 1,\dots,100; \quad l = 1,\dots,10; \] where \(\lambda_A\), \(\lambda_B\), \(\lambda_a\) and \(\lambda_b\) are hyperparameters. Finally we assume \[ \lambda_A, \lambda_B, \lambda_a, \lambda_b \sim \text{Gamma}(1, 1). \]
As you can see this is a lot of high dimensional parameters, and unless you have a lot of RAM to hand, a standard chain of length \(10^4\) will not fit into memory. First let’s create the
params dictionary, and then we can code the
logLik and
logPrior functions. We’ll sample initial \(\lambda\) parameters from a standard Gamma, and the rest from a standard Normal as follows
# Sample initial weights from standard Normald = ncol(dataset$X) # dimension of chainparams = list()params$A = matrix( rnorm(10*100), ncol = 10 )params$B = matrix(rnorm(d*100), ncol = 100)# Sample initial bias parameters from standard Normalparams$a = rnorm(10)params$b = rnorm(100)# Sample initial precision parameters from standard Gammaparams$lambdaA = rgamma(1, 1)params$lambdaB = rgamma(1, 1)params$lambdaa = rgamma(1, 1)params$lambdab = rgamma(1, 1)
Now let’s declare the
logLik and
logPrior functions. Remember that for ease of use, all distribution functions implemented in the
TensorFlow Probability package are located at
tf$distributions (for more details see the Get Started page).
logLik = function(params, dataset) { # Calculate estimated probabilities beta = tf$nn$softmax(tf$matmul(dataset$X, params$B) + params$b) beta = tf$nn$softmax(tf$matmul(beta, params$A) + params$a) # Calculate log likelihood of categorical distn with probabilities beta logLik = tf$reduce_sum(dataset$y * tf$log(beta)) return(logLik)}logPrior = function(params) { distLambda = tf$distributions$Gamma(1, 1) distA = tf$distributions$Normal(0, tf$rsqrt(params$lambdaA)) logPriorA = tf$reduce_sum(distA$log_prob(params$A)) + distLambda$log_prob(params$lambdaA) distB = tf$distributions$Normal(0, tf$rsqrt(params$lambdaB)) logPriorB = tf$reduce_sum(distB$log_prob(params$B)) + distLambda$log_prob(params$lambdaB) dista = tf$distributions$Normal(0, tf$rsqrt(params$lambdaa)) logPriora = tf$reduce_sum(dista$log_prob(params$a)) + distLambda$log_prob(params$lambdaa) distb = tf$distributions$Normal(0, tf$rsqrt(params$lambdab)) logPriorb = tf$reduce_sum(distb$log_prob(params$b)) + distLambda$log_prob(params$lambdab) logPrior = logPriorA + logPriorB + logPriora + logPriorb return(logPrior)}
Now suppose we want to make inference using stochastic gradient Langevin dynamics (SGLD). If we do this in the normal way then we will most likely run out of memory when the function builds the array to store the output. So instead we just initialize an
sgld object using
sgldSetup. Similarly we could build an
sgldcv object using
sgldcvSetup or an
sgnht object using
sgnhtSetup. Then we can run the SGLD algorithm one step at a time and decide what to do with the output at each iteration ourselves. We’ll just set our stepsize to
1e-4 for this example. To make the results reproducible we’ll set the seed to 13.
stepsize = 1e-4sgld = sgldSetup(logLik, dataset, params, stepsize, logPrior = logPrior, minibatchSize = 500, seed = 13)
This
sgld object is a type of
sgmcmc object, which is basically just a list with a number of entries. The most important of these entries to us is called
params, which holds a list, with the same names as you had in the
params you fed to
sgld, but this list contains
tf$Variable objects. This is how you access the tensors which hold your current parameter values in the chain. For more details on the attributes of these objects, see the documentation for
sgldSetup,
sgldcvSetup etc.
Now that we have created the
sgld object, you want to initialise the TensorFlow graph and the
sgmcmc algorithm you’ve chosen. If you are using a standard algorthm, this will just initialise the TensorFlow graph and all the tensors that were created. If you’re using an algorithm with control variates (e.g.
sgldcv), then this will also find the MAP estimates of the parameter and calculate the full log posterior gradient at that point. The function we use to do this is
initSess as follows
sess = initSess(sgld)
The
sess returned by
initSess is the current TensorFlow session, which is needed to run the SGMCMC algorithm of choice, and to access any of the tensors you need, such as
sgld$params.
Now we have everything to run an SGLD algorithm step by step as follows
Here the function
sgmcmcStep will update
sgld$params using a single update of SGLD, or whichever SGMCMC algorithm you chose. The function
getParams will return a list of parameters as
R objects rather than as tensors to make life easier for you.
Our simple example is fine, but we really would like to calculate a Monte Carlo average of the parameters on the fly. Also with these large examples, they take a long time to run, so it’s useful to check how the algorithm is doing every once in a while. This is especially useful when tuning by trial and error as you can stop an algorithm early if it’s doing badly. This is why we let you declare the TensorFlow session yourself: it lets you create your custom tensors to print algorithm progress, or to create your own test functions to reduce the chain dimensionality (they have to be declared before the TensorFlow session starts).
Let’s delete everything after we created our
sgld object. Now we’re going to demonstrate a more complicated step by step example where we print performance and calulate the Monte Carlo estimate on the fly. Suppose we have test data \(X^*\), and test labels \(y^*\), and at some iteration \(i\) our SGMCMC algorithm outputs values for all the parameters \(\theta_t\). Then the probability that our neural network model will classify a given test observation to class \(k\) is given by \(\beta_k(\theta_t, \mathbf x_i^*)\); i.e. the \(k^{th}\) element of \(\beta(\theta_t, \mathbf x_i^*)\), which was defined earlier. A common performance measure for a classifier is the log loss, defined by \[ ll = - \frac{1}{N} \sum_{i=1}^{N_{\text{test}}} \sum_{k=1}^K y^*_{i,k} \log \beta( \theta_t, \mathbf x_i^* ).\]
This is also \(-\frac{1}{N}\) times the log likelihood at the current parameter, given the test set. So it’s very easy for us to calculate this in practice and output it. We’ll do this every 100 iterations to check the algorithm’s performance and check for convergence. To do this, we need to create a new placeholder to hold the test set, and then create a tensor that will calculate the log loss, which can easily be done using the
logLik function already declared. First we’ll create a placeholder for both
X and
y in the test set, and make sure these have the same dimensions so can hold the full test set.
testPlaceholder = list()testPlaceholder[["X"]] = tf$placeholder(tf$float32, dim(testset[["X"]]))testPlaceholder[["y"]] = tf$placeholder(tf$float32, dim(testset[["y"]]))
Now we can create a tensor that calculates the log loss. We’ll link this to our
testPlaceholder and the current parameter values, located at
sgld$params.
# Get number of observations in test set, ensuring it's a double (R equivalent of float)Ntest = as.double(nrow(testset[["X"]]))logLoss = - logLik(sgld$params, testPlaceholder) / Ntest
Now we’ll declare the TensorFlow session, and run the chain step by step, calculating an average parameter estimate and printing the log loss of the current state every 100 iterations
sess = initSess(sgld)# Fill a feed dict with full test set (used to calculate log loss)feedDict = dict()feedDict[[testPlaceholder[["X"]]]] = testset[["X"]]feedDict[[testPlaceholder[["y"]]]] = testset[["y"]]# Burn-in chainmessage("Burning-in chain...")message("iteration\tlog loss")for (i in 1:10^3) { # Print progress if (i %% 100 == 0) { progress = sess$run(logLoss, feed_dict = feedDict) message(paste0(i, "\t", progress)) } sgmcmcStep(sgld, sess)}# Initialise Monte Carlo estimate using value after burn-inavParams = getParams(sgld, sess)# Run chainmessage("Running SGMCMC...")for (i in 1:10^4) { sgmcmcStep(sgld, sess) # Update av Params currentState = getParams(sgld, sess) for (paramName in names(avParams)) { avParams[[paramName]] = (avParams[[paramName]] * i + currentState[[paramName]]) / (i + 1) } # Print progress if (i %% 100 == 0) { progress = sess$run(logLoss, feed_dict = feedDict) message(paste0(i, "\t", progress)) }}
Obviously calculating the log loss is costly to do every 100 iterations as the test set has \(10^4\) observations itself, this was just for demonstration purposes. In practice we’d recommend subsampling this test set when calculating the log loss, or leaving more iterations until it’s calculated. |
I am looking for a full account of the relationship between the various versions of Floer theory on a symplectic manifold $M$. If we take the usual Floer equation (Hamiltonian version) \begin{equation*} \frac{\partial u}{\partial s} + J \left( \frac{\partial u}{\partial t} - X_H(u) \right) = 0 \ , \end{equation*} there are two "natural" types of boundary conditions: periodic conditions $u(s,t) = u(s,t+1)$ on the cylinder, or "fixed ends" $u(s,0) \in L_0$ and $u(s,1) \in L_1$ on the strip where the $L_i$ are Lagrangian submanifolds of $M$. One can always find a new complex structure $\tilde{J}$ with map $v : \mathbb{R}^2 \to M$ and transform the Floer equation into the unperturbed Cauchy-Riemann equation \begin{equation*} \frac{\partial v}{\partial s} + \tilde{J} \frac{\partial v}{\partial t} = 0 \ . \end{equation*} The two sets of boundary conditions then become: periodic $v(s,t) = \phi_H(v(s,t+1))$, or "fixed ends" $v(s,0) \in L_0$ and $v(s,1) \in \phi_H(L_1)$ where $\phi_H$ is the time-1 symplectomorphism generated by $X_H$.
I can roughly see how this infers several relationships between both types of Floer theories, but my understanding of all the possible correspondences is far from complete. In particular, fixed-end boundary conditions identify solutions of Hamiltion's equations with intersections of $L_0$ and $\phi_H(L_1)$. Fine, but what can be said about periodic boundary conditions? Is Lagrangian Floer theory of the intersection of the diagonal with the graph of a Hamiltonian diffeomorphism the only version of Lagrangian Floer theory that has a correspondence with periodic boundary conditions? There is also apparently a way to show that, on a compact manifold, intersections of a Lagrangian with are in one-to-one with critical points of Morse function on the Lagrangian and moreover, that Lagrangian Floer theory of the zero section of $T^*M$ is identified with Morse theory on $M$ (or something along those lines).
I have spent quite a bit of time going around in circles with the literature and so if someone could provide me with a nice reference with where these things are explained in detail (or even perhaps willing to describe some themselves) it would be very useful.
Thanks. |
The issue stems to an unfortunate coincidence for small values of $n$, namely that the Rankin-Selberg $L$-function is, in some sense, "close to" being the Dirichlet series you write when both $F$ and $G$ have degree $1$ or $2$ Euler factors. But for $n \geq 3$, this is no longer the case.
The Rankin-Selberg product $\pi \otimes \pi'$ of two automorphic representations $\pi,\pi'$ of $\mathrm{GL}_n(\mathbb{A})$ and $\mathrm{GL}_m(\mathbb{A})$ is something that is defined locally as follows.
N.B. The $L$-function of an automorphic representation is the canonical example of an element of the Selberg class, and it is conjectured that there is a bijection between these two sets of $L$-functions.
Write $L(s,\pi) = \prod_p L(s,\pi_p)$ and $L(s,\pi') = \prod_p L(s,\pi_p')$, where the Euler factors are of the form\[L(s,\pi_p) = \prod_{j = 1}^{n} \frac{1}{1 - \alpha_{\pi,j}(p) p^{-s}}, \quad L(s,\pi_p') = \prod_{k = 1}^{m} \frac{1}{1 - \alpha_{\pi',k}(p) p^{-s}}.\]We can write\[L(s,\pi) = \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n)}{n^s}\]where $\lambda_{\pi}(n)$ is the multiplicative function satisfying\[\lambda_{\pi}(p^r) = \sum_{\substack{r_1,\ldots,r_n = 0 \\ r_1 + \cdots + r_n = r}}^{r} \alpha_{\pi,1}(p)^{r_1} \cdots \alpha_{\pi,n}(p)^{r_n},\]and similarly for $L(s,\pi')$.
Then the Rankin-Selberg $L$-function is essentially equal to $L(s,\pi \otimes \pi') = \prod_p L(s,\pi_p \otimes \pi_p')$ with\[L(s,\pi_p \otimes \pi_p') = \prod_{j = 1}^{n} \prod_{k = 1}^{m} \frac{1}{1 - \alpha_{\pi,j}(p) \alpha_{\pi',k}(p) p^{-s}}.\](This is not quite correct at primes dividing the levels of $\pi$ and $\pi'$.)
For $n = m = 1$, $\pi$ and $\pi'$ are just Dirichlet characters $\chi$ and $\psi$, so that $\alpha_{\pi,1}(p) = \chi(p)$, $\alpha_{\pi',1}(p) = \psi(p)$, $\lambda_{\pi}(n) = \chi(n)$, and $\lambda_{\pi'}(n) = \psi(n)$. In this case, it is indeed the case that\[L(s,\pi \otimes \pi') = \sum_{n = 1}^{\infty} \frac{\chi(n) \psi(n)}{n^s} = \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}.\]
For $n = m = 2$, things are a little more complicated. In this case, $\alpha_{\pi,1}(p) \alpha_{\pi,2}(p)$ is equal to $\chi(p)$ for some Dirichlet character $\chi$, and similarly $\alpha_{\pi',1}(p) \alpha_{\pi',2}(p) = \psi(p)$ for some Dirichlet character $\psi$. (More precisely, $\pi$ and $\pi'$ are associated to modular forms, and $\chi$ and $\psi$ are their nebentypen.) Then $L(s,\pi \otimes \pi')$ is
not equal to $\sum_{n = 1}^{\infty} \lambda_{\pi}(n) \lambda_{\pi'}(n) n^{-s}$, but it is surprisingly close to being so:\[L(s,\pi \otimes \pi') = L(2s,\chi \psi) \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}.\](Again, this is not quite true; one has to correct the Euler factors at the bad primes.)
However, if $n$ or $m$ are at least $3$, then $\sum_{n = 1}^{\infty} \lambda_{\pi}(n) \lambda_{\pi'}(n) n^{-s}$ is, in some sense, very far from being equal to $L(s,\pi \otimes \pi')$. You can check this manually by comparing the Euler products of each of these two Dirichlet series.
I hope this answers your question somewhat. I have no idea what you actually mean by "Is it only due to the fact that the operation of mapping $(a_n,b_n)$ to $a_n b_n$ doesn't commute with the normalization operation that is required to define an L-function or are there other reasons ?" |
The computation requires two processes:
row operations of the type used in Gaussian elimination (with some restrictions because we require that the integer equivalence class be preserved) and the corresponding column operations, the Euclidean algorithm for finding the greatest common divisor of two integers.
Specifically, you are allowed to
interchange two rows or two columns, multiply a row or column by $\pm1$ (which are the invertible elements in $\mathbf{Z}$), add an integer multiple of row to another row (or an integer multiple of a column to another column).
The first goal is to reach diagonal form. Let's first work on column 1: using operation 3 for rows repeatedly, you can, by following the Euclidean algorithm, form a row whose first element is the GCD of the elements in column 1. You can then obtain a matrix with the GCD in the $(1,1)$ position and zeroes in the rest of column 1.
Now work on row 1: do the same thing, but using column operations; eventually you will have the GCD of row 1 in the $(1,1)$ position, and zeroes elsewhere in row 1. You will most likely have messed up column 1, but that's OK. Go back and redo column 1, then redo row 1, and repeat until all elements in row and column 1 are 0 except for the $(1,1)$ element. This process is guaranteed to terminate because the GCD gets smaller each time.
Now you can move on to row/column 2, and repeat the process. Continue for row/column 3, and so on, until you have reached diagonal form.
You may not be done at this point, because the diagonal elements may not satisfy the divisibility requirement of the Smith normal form. You can, however, enforce this by some additional moves as in the following example:$$\begin{bmatrix}8 & 0\\0 & 12\end{bmatrix}\longrightarrow\begin{bmatrix}8 & 8\\0 & 12\end{bmatrix}\longrightarrow\begin{bmatrix}8 & 8\\-8 & 4\end{bmatrix}\longrightarrow\begin{bmatrix}24 & 8\\0 & 4\end{bmatrix}\longrightarrow\begin{bmatrix}24 & 0\\0 & 4\end{bmatrix}\longrightarrow\begin{bmatrix}4 & 0\\0 & 24\end{bmatrix}.$$The idea is again to use the Eulidean algorithm. After adding column 1 to column 2, you may have to do several row operations to obtain the GCD of the two original diagonal elements. In the example above, only one row operation was needed for this.
Addendum: Details for the particular example in the OP's question.If you add row 2 to row 1, and then multiply row 1 by $-1$, you get a pivot of 1. You can then subtract suitable multiples of row 1 from rows 2 and 4, and end up with$$\begin{bmatrix}1 & 561 & -174 & -80 \\ 0 & -3477 & 1080 & 474 \\ 0 & -255 & 81 & 24 \\ 0 & -4182 & 1299 & 570\end{bmatrix},$$which, by subtracting multiples of column 1 from columns 2, 3, and 4, leads to$$\begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -3477 & 1080 & 474 \\ 0 & -255 & 81 & 24 \\ 0 & -4182 & 1299 & 570\end{bmatrix}.$$We next need to do a series of row operations involving rows 2, 3, and 4 that results in the GCD of 3477, 255, and 4182. If we always choose the pivot of smallest nonzero magnitude, the steps would be Subtract 14 times row 3 from row 2, and 17 times row 3 from row 4. Add 2 times row 2 to row 3, and subtract row 2 from row 4. Subtract row 4 from row 2, and add row 4 to row 3. Add 3 times row 3 to row 2, and 6 times row 3 to row 4. Add row 2 to row 3, and subtract row 2 from row 4. Add 2 times row 3 to row 2. Swap rows 2 and 3. Negate row 2.
You should now have$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 3 & 234 & -1410 \\ 0 & 0 & -651 & 3906 \\ 0 & 0 & -147 & 882\end{bmatrix}.$$You can now subtract suitable multiples of column 2 from columns 3 and 4. Then you just have to deal with the $2\times2$ block in the lower right. You should be able to get it from here. |
Let $A$ be a finite dimensional $\mathrm{C}^*$-algebra.
We have the usual absolute value:
$$|a|:=\sqrt{a^*a},$$
and the cone of positive elements:
$$A^+:=\{a\in A:\exists \,b\in A, a=b^*b\}.$$
An operator $T\in B(A)$ is said to be positive if $T(A^+)\subset A^+$.
Does it hold that $|Tf|=T|f|$ for positive $T$?.
Context:
I don't have any reason to believe this but I have a $T$-invariant finite trace $h\in A^{'}$ ($h\circ T=h$) and if I have the above I have
$$\|Tf\|_{\mathcal{L}^1}=h(|Tf|)=h(T|f|)=h(|f|)=\|f\|_{\mathcal{L}^1},$$ which would be useful to me. |
The Hamiltonian for a relativistic charged particle moving in a static electromagnetic field is the well known: $$H=c\sqrt{\left(\mathbf{P}-q\mathbf{A}\right)^{2}+m^{2}c^{2}}+q\phi$$ where,\begin{align*} \mathbf{B} & =\nabla\times\mathbf{A},\\ \mathbf{E} & =-\nabla\phi. \end{align*} Now let's suppose that one wants to write the magnetic field in terms of the magnetic scalar potential, $\phi_M$, rather than of $\mathbf{A}$, that is for a magnetic field written as: $$\mathbf{B}=-\nabla\phi_{M}.$$ What would the Hamiltonian look like, in terms of $\phi_M$ ?
One way would be to express the vector potential as function[al] of the scalar potential and substitute in the Hamiltonian.
Magnetic field (induction) can be expressed as gradient of a potential function only in a limited region of space where the integral
$$ \int_1^2 \mathbf B\cdot d\mathbf s $$ between two points does not depend on the path; it may depend only on the endpoints.
Suppose we have such a region and such a potential function so magnetic field is a gradient.
Then, we can ( in principle ) solve the equation
$$ - \nabla \phi = \nabla \times \mathbf A $$
for unknown function $\mathbf A(\mathbf x)$. Suppose we found a solution which is expressed as function[al] of $\phi$ and space coordinates. Then we can substitute for $\mathbf A$ in the standard Hamiltonian you mentioned and thus obtain Hamiltonian that refers to $\phi$ and space coordinates only.
For example, if the magnetic field is uniform, so we can express it as $\mathbf B=[0,0,B_0]$, the potentials that would give the magnetic field correctly could be defined as
$$ \phi = -B_0z $$
$$ \mathbf A = \bigg[-\frac{1}{2}B_0 y,\frac{1}{2}B_0 x,0\bigg] $$
However, it is easy to see that vector potential at some point is not simply a function of scalar potential at the same point; if we are to relate the two, we need to involve also the spatial coordinates explicitly:
$$ \mathbf A = \bigg[-\frac{\phi}{2} \frac{y}{z},\frac{\phi}{2}\frac{x}{z},0\bigg ] $$
Now we can substitute for $\mathbf A$ in the Hamiltonian.
Finding $\mathbf A$ was simple here but for general magnetic field, the relation would be more complicated, probably integral: $\mathbf A$ would be a functional of $\phi$. Such complicated expression would be hard to work with in the Hamiltonian. Better use directly the vector potential.
As far as I know, each vector field can be unambiguously split into a gradient and a rotational part:
$$\mathbf{F} = \nabla \phi + \nabla \times \mathbf A$$
This is called the Helmholtz decomposition: https://en.wikipedia.org/wiki/Helmholtz_decomposition
Due to your assumption $\mathbf{B} = \nabla \times \mathbf A$ it follows that the gradient part is identically zero. |
The model here is the binomial option pricing model, so the second term in the brackets represents the expected future value of the option (under riskneutral probabilities).The aim of the option holder is always to maximize the value of his option. He can at any point sell the option at the fair market price $E(V_{n+1})$ or exercise it to get $G_n$. So if ...
you don't need $ud=1.$ In fact, there are now about 30 binomial trees which converge to Black--Scholes in the large step limit. Most of them do not have $ud=1.$ All you need is$$d < e^{r \Delta t} < u$$The tree recombines provided $u$ and $d$ don't change from step to step.See my book More Mathematical Finance for a comprehensive review and ...
You don't mention if the puts in question are exotic or vanilla, but assuming they are vanilla, you should read this paper by Chen and Joshi. In it, they find optimal performance by using smoothed, truncated Tian-parameter binomial lattices with Richardson extrapolation -- where the idea is to run one extra low-cost (long $\Delta T$) tree in order to ...
In binomial tree models, there is no such a thing as a path. The binomial tree represents information about the distribution of the zero-curve at a given time and preserve enough information between different times to let you compute conditional expectations. Generally, you can not price path-dependant instruments in a model based on trees—because there is ...
Not all binomial trees take $u=e^{\sigma\sqrt{\Delta t}}$. Thinking of the binomial tree as a discrete approximation (on a grid) to a continuous process, it makes sense that a variety of choices for where to place grid points will work.For a listing of a few different choices of $u$, see the Tian Tree settings and others. From this Sitmo page you can see,...
From the gentleman and scholar Emanuel Derman. Emanuel states "the last two pages answer the question asked".https://www.dropbox.com/s/cg299qsbquuqdru/TwitterNotesOnBDT.2017.pdf?dl=0&m=Please thank him directly on Twitter.
Assuming continuously compounded returns for a multi-period model with $N$ being the number of periods:\begin{cases}&\log u \quad \text{with probability q}\\&\log d \quad \text{with probability 1-q}\end{cases}given the stock price at maturity$$\log\left(\frac{S_T}{S_0}\right)=i\log u+(N−i)\log d=i\log\left(\frac{u}{d}\right)+N\log d$$where $...
one of the most fundamental results states that the binomial model converges towards the Black Scholes model if the step size $\Delta t$ converges to zero.The Black Scholes model is an option pricing model where the underlying is given by$$S_T = S_0 \cdot \exp \Bigl(\sigma W_T - \frac 12 \sigma^2 T \Bigr).$$By choosing$$u = \exp(\sigma \sqrt{\...
Note that the tree is recombining. You have $u=1.2$ and $d=0.8$ with $ud=0.96$. Your tree for the asset price reads asAt time zero: 100At time one: 80 or 120At time two: 64 or 96 or 144The transition probabilities are $q_u=0.55$ and $q_d=0.45$. For your put option with strike price $K=104$, you thus obtain by backward inductionAt time two: 40 or 8 ...
You have to look at the terms and conditions on your individual bond. The way the specifications usually work is that a call will result in accrued interest being paid, effectively making up for the lost coupon. Sometimes there's even an extra penalty. A put will result in a loss of coupon in almost all cases, and so is almost always done just after a ...
There is a good quick well-known approximation for at-the-money options:$$\textrm{Call,Put} = 0.4 S \sigma \sqrt{T}.$$See further discussion atWhat are some useful approximations to the Black-Scholes formula?.
you have to be careful to distinguish between trinomial trees in a theoretical sense which do not give unique prices, and trinomial trees chosen as an approximation to the risk-neutral measure of the BS model. In the second case, they are an effective numerical method as are binomial trees.Trinomial trees are more useful when you want to ensure nodes lie ...
The condition$$ud=1\text{, or equivalently }u=1/d$$is necessary to ensure convergence of the Binomial tree's mean $\mu$ and standard deviation $\sigma$ to nonfinite values when $n$ (number of steps) goes to infinity.Cox-Rubinstein-Ross showed in their famous paper, that to achieve this, we must have:$$u=e^{\sigma\sqrt{t/n}}\text{, }d=e^{-\sigma\sqrt{...
For a martingale $dX=a(X,t)\,dt+b(X,t) dW(t)$ where $a$ and $b$ are not constant, your tree will not recombine in general [edit]. This is the main issue. See for instance: Florescu, I. and F. G. Viens (2008, March). Stochastic volatility: Option pricing using a multinomial recombining tree. Applied Mathematical Finance 15 (2), 151-181. It deals with the case ...
It's a pitty that you don't show in your question how you get to your value for $c_0$ but the idea is that you build a portfolio $X_0 = \Delta S_0 - \lambda$ and you infer the values for $\Delta$ and $\lambda$ so that $X_1 = c_1$ both in the up and down scenario. Then, because of the law of one price, $X_0 = c_0$.So for us $X_1 = \Delta S_1 + (1+r) \lambda$...
You have forgotten the combinatorial factors for binomial probabilities on your terms. You need $$ {n\choose k} p^n(1-p)^{n-k},$$ not just $$ p^n(1-p)^{n-k}.$$ The second term should have a factor of $6$ and the third should have a factor of $15,$ etc.
In R you can use fOptions package to draw Binomial Tree graphs.Here is a simple code snippet#Install the package and load itinstall.packages('fOptions')library(fOptions)#Calculate the value of the option and plotoptionVals<-BinomialTreeOption(TypeFlag="ce",S=100,X=100,Time=3,r=0.05,b=0,sigma=0.2,n=3,title="example binomial tree")BinomialTreePlot(...
A condition for correct calibration of the short rate model is that it exactly reproduce the present values of fixed (option-free) cashflows - that is, that it give the same answer as ordinary discounting at the spot rate. If it doesn't, you've done something wrong - sort of like using a model that violates put-call parity. (Actually, it's exactly like that.)...
You can calibrate the model by discretizing in time, and using a forward induction method as originally proposed by Jamishidian in 1991:F.Jamshidian, Forward Induction and Construction of Yield Curve Diffusion Models, J.Fixed Income 6, 62-74 (1991).Although he formulated this induction in the language of the binomial tree, the method is more general, and ...
All you need is to use the discretization to implement the MC approach. The following links should get you started:http://www.lcy.net/files/BDT_Seminar_Paper.pdfhttp://www-2.rotman.utoronto.ca/~hull/TechnicalNotes/TechnicalNote23.pdfhttp://www.iorcf.unisg.ch/Forschung/~/media/Internet/Content/Dateien/InstituteUndCenters/IORCF/Abschlussarbeiten/Frey%...
The argument that the American and European call are worth the same is model independent. So it holds for the binomial model. So there is no need to check tosee if the early exercise occurs because it won't.Of course, if you have written general purpose code, it is much easier to testfor early exercise and always have the test fail than to try and deal ...
To rule out arbitrage in the one-period model, we must assume$$0 < d < 1+r < u,$$where $u$ is the up-factor, $d$ is the down-factor and $r$ is the risk-free interest rate. This chain of inequalities is the no-arbitrage condition.To see what happens if it doesn't hold, consider the case in which$$0 < 1+r < d < u.$$Let $S$ denote ...
"But just for fun, let's say Pr(S1=Su)=1% and Pr(S1=Sd)=99%, in which case, on average, the call at time 1 would be worth 0.01*10 = 0.1$.How would anyone be willing to pay 9.28$ for that ?I'm pretty sure I'm missing something very basic, I hope someone can explain what it is."How would anyone pay 100 for the stock given these probabilities? You don't ...
Your formula for $p$ is $$p = \frac{e^{{(\alpha - \delta})h} - d}{u - d},$$where $\alpha$ is not expected return on stock but continuous risk free rate, i.e. 1%.If you use $\alpha$ as 1%, you will get $p=0.009125828 $ which is within $[0,1]$EDIT: With the information given in the question, it must satisfy following equality:$$S_0e^{\alpha - \delta}=...
Quick answerThe payoff you mention is that of a call spread, i.e. long a call $C_1$ struck at $K_1$ and short a call $C_2$ struck at $K_2$, with $K_2>K_1$. The price of the instrument is therefore: $V = C_1 - C_2$.[First way] If you are stuck because this payout seems 'unsual' to you, an easy way to reach your goal (assuming you know how to use ...
there are many different trees. The first one, the CRR tree, used$$u = e^{\sigma\sqrt{h}}$$and $d = 1/u.$ However, you can take any real-world drift and still get the sameprices in the limit so you can put$$u = e^{\mu h +\sigma\sqrt{h}}, \text{ and } d = e^{\mu h -\sigma\sqrt{h}}$$for any fixed $\mu.$$\mu = 0$ is a poor choice for convergence. ...
Thanks to P.Windridge's comment, I can now answer my own question.Indeed the convergence to standard normal in question can follow from a triangular array version of CLT called the Lindeberg-Feller CLT. Proof can be found on Durrett's Probability: Theory and Examples (freely available online).I reference the statement of the theorem from Durrett:... |
Title Renormalization for Autonomous Nearly Incompressible BV Vector Fields in Two Dimensions Publication Type Journal Article Year of Publication 2016 Authors Bianchini, S, Bonicatto, P, Gusev, NA Journal SIAM Journal on Mathematical Analysis Volume 48 Pagination 1-33 Abstract
Given a bounded autonomous vector field $b \colon \mathbb{R}^d \to \mathbb{R}^d$, we study the uniqueness of bounded solutions to the initial value problem for the related transport equation \begin{equation*} \partial_t u + b \cdot \nabla u= 0. \end{equation*} We are interested in the case where $b$ is of class BV and it is nearly incompressible. Assuming that the ambient space has dimension $d=2$, we prove uniqueness of weak solutions to the transport equation. The starting point of the present work is the result which has been obtained in [7] (where the steady case is treated). Our proof is based on splitting the equation onto a suitable partition of the plane: this technique was introduced in [3], using the results on the structure of level sets of Lipschitz maps obtained in [1]. Furthermore, in order to construct the partition, we use Ambrosio's superposition principle [4].
URL https://doi.org/10.1137/15M1007380 DOI 10.1137/15M1007380 Renormalization for Autonomous Nearly Incompressible BV Vector Fields in Two Dimensions
Research Group: |
2019-09-27 09:59
Higgs boson pair production at colliders: status and perspectives / Di Micco, Biagio (Universita e INFN Roma Tre (IT)) ; Gouzevitch, Maxime (Centre National de la Recherche Scientifique (FR)) ; Mazzitelli, Javier (University of Zurich) ; Vernieri, Caterina (SLAC National Accelerator Laboratory (US)) ; Alison, John (Carnegie-Mellon University (US)) ; Androsov, Konstantin (INFN Sezione di Pisa, Universita' e Scuola Normale Superiore, Pisa (IT)) ; Baglio, Julien Lorenzo (CERN) ; Bagnaschi, Emanuele Angelo (Paul Scherrer Institut (CH)) ; Banerjee, Shankha (University of Durham (GB)) ; Basler, P (Karlsruhe Institute of Technology) et al. This document summarises the current theoretical and experimental status of the di-Higgs boson production searches, and of the direct and indirect constraints on the Higgs boson self-coupling, with the wish to serve as a useful guide for the next years. The document discusses the theoretical status, including state-of-the-art predictions for di-Higgs cross sections, developments on the effective field theory approach, and studies on specific new physics scenarios that can show up in the di-Higgs final state. [...] LHCHXSWG-2019-005.- Geneva : CERN, 2019 - 274. Registo detalhado - Registos similares 2019-05-10 11:18 Registo detalhado - Registos similares 2019-04-02 20:51
Simplified Template Cross Sections – Stage 1.1 / Delmastro, Marco (Centre National de la Recherche Scientifique (FR)) ; Berger, Nicolas (Centre National de la Recherche Scientifique (FR)) ; Bertella, Claudia (Chinese Academy of Sciences (CN)) ; Duehrssen-Debling, Michael (CERN) ; Kivernyk, Oleh (Centre National de la Recherche Scientifique (FR)) ; Langford, Jonathon Mark (Imperial College (GB)) ; Milenovic, Predrag (University of Belgrade (RS)) ; Pandini, Carlo Enrico (CERN) ; Tackmann, Frank (Deutsches Elektronen-Synchrotron (DE)) ; Tackmann, Kerstin (Deutsches Elektronen-Synchrotron (DE)) et al. Simplified Template Cross Sections (STXS) have been adopted by the LHC experiments as a common framework for Higgs measurements. Their purpose is to reduce the theoretical uncertainties that are directly folded into the measurements as much as possible, while at the same time allowing for the combination of the measurements between different decay channels as well as between experiments. [...] arXiv:1906.02754; LHCHXSWG-2019-003; DESY-19-070.- Geneva : CERN, 2019 - 14 p. Fulltext: LHCHXSWG-2019-003 - PDF; 1906.02754 - PDF; Registo detalhado - Registos similares 2019-03-27 12:46
Recommended predictions for the boosted-Higgs cross section / Becker, Kathrin (Albert Ludwigs Universitaet Freiburg (DE)) ; Caola, Fabrizio (University of Durham (GB)) ; Massironi, Andrea (CERN) ; Mistlberger, Bernhard (Massachusetts Inst. of Technology (US)) ; Monni, Pier (CERN) ; Chen, Xuan (Zurich U.) ; Frixione, Stefano (INFN e Universita Genova (IT)) ; Gehrmann, Thomas Kurt (Universitaet Zuerich (CH)) ; Glover, Nigel (IPPP Durham) ; Hamilton, Keith Murray (University of London (GB)) et al. In this note we study the inclusive production of a Higgs boson with large transverse momentum. We provide a recommendation for the inclusive cross section based on a combination of state of the art QCD predictions for the gluon-fusion and vector-boson-fusion channels. [...] LHCHXSWG-2019-002.- Geneva : CERN, 2019 - 14. Fulltext: PDF; Registo detalhado - Registos similares 2019-03-01 22:49
Higgs boson cross sections for the high-energy and high-luminosity LHC: cross-section predictions and theoretical uncertainty projections / Calderon Tazon, Alicia (Universidad de Cantabria and CSIC (ES)) ; Caola, Fabrizio (University of Durham (GB)) ; Campbell, John (Fermilab (US)) ; Francavilla, Paolo (Universita & INFN Pisa (IT)) ; Marchiori, Giovanni (Centre National de la Recherche Scientifique (FR)) ; Becker, Kathrin (Albert Ludwigs Universitaet Freiburg (DE)) ; Bertella, Claudia (Chinese Academy of Sciences (CN)) ; Bonvini, Marco (Sapienza Universita e INFN, Roma I (IT)) ; Chen, Xuan (Zuerich University (CH)) ; Frederix, Rikkert (Technische Universität Muenchen (DE)) et al. This note summarizes the state-of-the-art predictions for the cross sections expected for Higgs boson production in the 27 TeV proton-proton collisions of a high-energy LHC, including a full theoretical uncertainty analysis. It also provides projections for the progress that may be expected on the timescale of the high-luminosity LHC and an assessment of the main limiting factors to further reduction of the remaining theoretical uncertainties.. LHCHXSWG-2019-001.- Geneva : CERN, 01 - 17. Fulltext: PDF; Registo detalhado - Registos similares 2016-07-15 07:28
Analytical parametrization and shape classification of anomalous HH production in EFT approach / Carvalho Antunes De Oliveira, Alexandra (Universita e INFN, Padova (IT)) ; Dall'Osso, Martino (Universita e INFN, Padova (IT)) ; De Castro Manzano, Pablo (Universita e INFN, Padova (IT)) ; Dorigo, Tommaso (Universita e INFN, Padova (IT)) ; Goertz, Florian (CERN) ; Gouzevitch, Maxime (Universite Claude Bernard-Lyon I (FR)) ; Tosi, Mia (CERN) In this document we study the effect of anomalous Higgs boson couplings on non-resonant pair production of Higgs bosons (HH) at the LHC. We explore the space of the five parameters $\kappa_\lambda$, $\kappa_t$, $c_2$, $c_{g}$, and $c_{2g}$ in terms of the corresponding kinematics of the final state, and describe a suggested partition of the space into a limited number of regions featuring similar phenomenology in the kinematics of HH final state, along with a corresponding set of representative benchmark points. [...] LHCHXSWG-2016-001.- Geneva : CERN, 2016 Fulltext: PDF; Registo detalhado - Registos similares 2015-08-03 09:58
Benchmark scenarios for low $\tan \beta$ in the MSSM / Bagnaschi, Emanuele (DESY) ; Frensch, Felix (Karlsruhe, Inst. Technol.) ; Heinemeyer, Sven (Cantabria Inst. of Phys.) ; Lee, Gabriel (Technion) ; Liebler, Stefan Rainer (DESY) ; Muhlleitner, Milada (Karlsruhe, Inst. Technol.) ; Mc Carn, Allison Renae (Michigan U.) ; Quevillon, Jeremie (King's Coll. London) ; Rompotis, Nikolaos (Seattle U.) ; Slavich, Pietro (Paris, LPTHE) et al. The run-1 data taken at the LHC in 2011 and 2012 have led to strong constraints on the allowed parameter space of the MSSM. These are imposed by the discovery of an approximately SM-like Higgs boson with a mass of $125.09\pm0.24$~GeV and by the non-observation of SUSY particles or of additional (neutral or charged) Higgs bosons. [...] LHCHXSWG-2015-002.- Geneva : CERN, 2015 - 24. Fulltext: PDF; Registo detalhado - Registos similares 2015-03-20 14:24
Recommendations for the interpretation of LHC searches for $H_5^0$, $H_5^{\pm}$, and $H_5^{\pm\pm}$ in vector boson fusion with decays to vector boson pairs / Zaro, Marco (Paris U., IV ; Paris, LPTHE) ; Logan, Heather (Ottawa Carleton Inst. Phys.) We provide theory input for the interpretation of the LHC searches for the production of Higgs bosons $H_5^0$, $H_5^{\pm}$, and $H_5^{\pm\pm}$ that transform as a fiveplet under the custodial symmetry. We choose as a benchmark the Georgi-Machacek model, in which isospin-triplet scalars are added to the Standard Model Higgs sector in such a way as to preserve custodial SU(2) symmetry. [...] LHCHXSWG-2015-001.- Geneva : CERN, 30 - 19p. Fulltext: PDF; Registo detalhado - Registos similares |
What is the motivation for including the compactness and semi-simplicity assumptions on the groups that one gauges to obtain Yang-Mills theories? I'd think that these hypotheses lead to physically "nice" theories in some way, but I've never, even from a computational perspective. really given these assumptions much thought.
As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group $G$ with corresponding Lie algebra $g$ should be real and have a positive (semi)definite associative/invariant bilinear form $\kappa: g\times g \to \mathbb{R}$, cf. the kinetic part of the Yang Mills action. The bilinear form $\kappa$ is often chosen to be (proportional to) the Killing form, but that need not be the case.
If $\kappa$ is degenerate, this will induce additional zeromodes/gauge-symmetries, which will have to be gauge-fixed, thereby effectively diminishing the gauge group $G$ to a smaller subgroup, where the corresponding (restriction of) $\kappa$ is non-degenerate.
When $G$ is semi-simple, the corresponding Killing form is non-degenerate.But $G$ does
not have to be semi-simple. Recall e.g. that $U(1)$ by definition is not a simple Lie group. Its Killing form is identically zero. Nevertheless, we have the following YM-type theories:
QED with $G=U(1)$.
the Glashow-Weinberg-Salam model for electroweak interaction with $G=U(1)\times SU(2)$.
I recommend that you to read the chapter 15.2 in "The Quantum Theory of Fields" Volume 2 by Steven Weinberg, he answers precisely your question.
Here a short summary In a gauge theory with algebra generators satisfying $$ [t_\alpha,t_\beta]=iC^\gamma_{\alpha\beta}t_\gamma $$ it can be checked that the field strength tensor $F^\beta_{\mu\nu}$ transforms as follows $$ \delta F^\beta_{\mu\nu}=i\epsilon^\alpha C^\beta_{\gamma\alpha} F^\gamma_{\mu\nu} $$ We want to construct Lagrangians. A free-particle kinetic term must be a quadratic combination of $F^\beta_{\mu\nu}$ and Lorentz invariance and parity conservation restrict its form to $$ \mathcal{L}=-\frac{1}{4}g_{\alpha\beta}F^\alpha_{\mu\nu}F^{\beta\mu\nu} $$ where $g_{\alpha\beta}$ may be taken symmetric and must be taken real for the Lagrange density to be real as well. The Lagrangian above must be gauge-invariant thus it must satisfy $$ \delta\mathcal{L}=\epsilon^\delta g_{\alpha\beta}F^\alpha_{\mu\nu}C^\beta_{\gamma\delta}F^{\gamma\mu\nu}=0 $$ for all $\epsilon^\delta$. In order not to impose any functional restrictions for the field strengths $F$ the matrix $g_{\alpha\beta}$ must satisfy the following condition $$ g_{\alpha\beta}C^\beta_{\gamma\delta}=-g_{\gamma\beta}C^\beta_{\alpha\delta} $$ In short, the product $g_{\alpha\beta}C^\beta_{\gamma\delta}$ is anti-symmetric in $\alpha$ and $\gamma$. Furthermor the rules of canonical quantization and the positivity properties of the quantum mechanical scalar product require that the matrix $g_{\alpha\beta}$ must be positive-definite. Finally one can prove that the following statements are equivalent There exists a real symmetric positive-definite matrix $g_{\alpha\beta}$ that satisfies the invariance condition above. There is a basis for the Lie algebra for which the structure constants $C^\alpha_{\beta\gamma}$ are anti-symmetric not only in the lower indices $\beta$ and $\gamma$ but in all three indices $\alpha$, $\beta$ and $\gamma$. The Lie algebra is the direct sum of commuting compact simple and $U(1)$ subalgebras.
The proof for the equivalence of these statements as well as a more in-detail presentation of the material can be found in the aforementioned book by S. Weinberg.
A proof for the equivalence for $g_{\alpha\beta}=\delta_{\alpha\beta}$ (actually the most common form) was given by M. Gell-Mann and S. L. Glashow in Ann. Phys. (N.Y.)
15, 437 (1961)
It's because you want the kinetic part of the Yang Mills action $$ \int Tr({\bf{F^2}}) dV$$ to be positive definite. To guarantee this, the Lie algebra inner product you're using (Killing form) needs to be positive definite. This is guaranteed if the gauge group is compact and semi-simple. (I'm not sure if it's
only if G is compact and semi simple though. Maybe someone else could fill in this detail). |
In probability theory and statistics, the coefficient of variation (CV), also known as relative standard deviation (RSD), is a standardized measure of dispersion of a probability distribution or frequency distribution.
Relative Standard Deviation, RSD is defined and given by the following probability function:
${100 \times \frac{s}{\bar x}}$
Where −
${s}$ = the sample standard deviation
${\bar x}$ = sample mean
Problem Statement:
Find the RSD for the following set of numbers: 49, 51.3, 52.7, 55.8 and the standard deviation are 2.8437065.
Solution: Step 1 - Standard deviation of sample: 2.8437065 (or 2.84 rounded to 2 decimal places). Step 2 - Multiply Step 1 by 100. Set this number aside for a moment.
${2.84 \times 100 = 284}$
Step 3 - Find the sample mean, ${\bar x}$. The sample mean is:
${\frac{(49 + 51.3 + 52.7 + 55.8)}{4} = \frac{208.8}{4} = 52.2.}$
Step 4Divide Step 2 by the absolute value of Step 3.
${\frac{284}{|52.2|} = 5.44.}$
The RSD is:
${52.2 \pm 5.4}$%
Note that the RSD is expressed as a percentage. |
Standard busy beaver function draws attention to final count of nonzero symbols on tape. We could instead look at largest amount of nonzero symbols appearing on tape at
any point of computation. This function's lower bound would be $\Sigma(n)$ and upper bound would be $S(n)$ (max shifts function). Was there made any research on such functions? If so, are there any known values for this?
Standard busy beaver function draws attention to final count of nonzero symbols on tape. We could instead look at largest amount of nonzero symbols appearing on tape at
Suppose there is a machine using $n$ states which, at some point, has $X$ nonzero symbols on the tape. One can build a machine with $O(n)$ states which simulates a run of the original machine with a tape alphabet of three symbols $\{0,1,2\}$, with the following small change: whenever the original machine changes $1$ to $0$, the new machine changes it to $2$; whenever the tape is read, $2$ is interpreted as $0$. The new machine ends up with $Y \in [X,2X]$ nonzero symbols on the tape (we get $\Theta(X)$ instead of $X$ since the simulation requires using two symbols for each original symbol). So your new function, let's call it $F(n)$, satisfies $B(n) \leq F(n) \leq B(O(n))$.
Your function (call it $F$) obviously satisfies $$\Sigma(n) \le F(n) \le space(n)$$ where $space(n)$ is the maximum number of tape squares visited by any $n$-state halting TM in the same class as those considered in the definition of $\Sigma$. Furthermore, $$space(n) \le \Sigma(3n-1)$$ as proved in the following paper:
Ben-Amram, A. M., B. A. Julstrom, and K. Zwick. "A Note on Busy Beavers and Other Creatures." Mathematical Systems Theory 29.4 (1996).
Therefore $$\Sigma(n) \le F(n) \le \Sigma(3n-1)$$ |
This depends, of course, on the underlying probability space, most notably the filtration. In general, there are plenty of progressive processes which are not predictable, and this should be fairly standard material. If one works with the raw (un-augmented) filtration on the canonical path space of continuous functions, then the notions of predictability and progressive measurability coincide. I don't have the reference accessible at the moment, but I would bet any amount of money that it's in the first volume of Dellacherie-Meyer.
On the other hand, if you augment your filtration to make it complete and right-continuous, as one so often does, there are plenty of easy examples of progressive processes which are not predictable. (Actually, right-continuity is all that matters in the following.) For example, define $(X_t)_{t \ge 0}$ by $X_t=0$ for $t < 1$ and $X_t=\pm 1$ with some nontrivial (say, equal) probabilities. Consider the augmented filtration generated by $X$,
$\mathcal{F}_t = \cap_{s > t} \sigma(X_s) \vee \mathcal{N}$,
where $\mathcal{N}$ is the set of null sets of the ambient probability space. Then the right-limit process $X_+=(X_{t+})_{t \ge 0}$ is progressive but not predictable, with respect to this filtration. To see this, recall that the predictable processes are (by definition) those generated by the left-continuous adapted processes. Because $X$ is constant strictly before time $1$, every adapted process must be a.s. constant and deterministic on $[0,1)$. A left-continuous process is then also a.s. constant and deterministic on $[0,1]$, and we conclude that the same is true of predictable processes. As $X_{1+}$ is random, it follows that $X_+$ is not predictable. To see that $X_+$ is progressive, just note that it is right-continuous and adapted, because $X_{1+}$ is $\mathcal{F}_1$-measurable due to the right-continuity imposed on the filtration. |
There is no shortage of idiots and assholes in this world. I see them everyday in my mail inbox. The rich Nigerian widow, Idi Amin’s grandson, Gaddafi’s daughter-in-law, Saddam Husain’s son-in-law, are all looking for a “reliable” partner in their business adventures and seek my help in this matter. Of course, they promise to share a fortune with me. The crooks think that I will bite the bait. I have my way of dealing with such idiots.. In addition, I use many other tools to discourage such shameless intruders. I avoid scum pits like Whatsapp, where anyone is free to share and forward any muck which they have received from other idiots. We all have to join hands and stop the stink. Zero-tolerance is the key to a peaceful world.
In case you did’nt know this. It is not at all difficult to share news/pics without Facebook, Instagram, or Whatsapp. Linux/FOSS gurus are considered as social misfits and condemned, since they never use these hitech tools. No wonder, no one wants to talk to them.
Proof of the pudding, lies in the eating. Proof of the eating, lies in the feedback. Proof of the feedback does not exist yet.
Do you trust your child with this school ? What quality of education are you expecting to give your child ?
Vedic maths as a business model ? Why not try witchcraft ? I received today an invitation for a weird event to discuss ways of making money using Vedic mathematics.
To make their point more tempting and attractive, they have also added a suggestive picture in the invite. The point is, many gullible and greedy people, particularly parents, are expected to bite this bait and fall into this trap.
“Look before you leap”, is a very popular idiom. I must add — “Look, before you make your children leap”. Before you sink your money and drown your children, look at some sane advise from a mature teacher of mathematics:
https://drpartha.wordpress.com/2013/07/07/15-2013-and-you-still-believe-vedic-maths-is-good-for-your-children/
If making money is your aim, why not try witchcraft ? It can fetch you much more money than mathematics and make you rich, faster.
Use your commonsense, and you be the judge.
Adding a plugin to show maths using LaTeX in you wordpress posts is a pain.
The worst plugin experience was with WP KaTeX.
WP KaTeX : absolutely no documentation to help usage, after plugin installation. No mention of limitations or restrictions. No description for the recommended jsDelivr CDN addon. No examples or sample sites or tutorials or support forums.
$ latex
\frac{1}{\pi} = \frac{2\sqrt{2}}{9801}\sum\limits_{k=0}^\infty\frac{(4k)!(1103+26390k)}{(k!)^4 (396^{4k})} $ வர்ட் பிரெஸில் தமிழ் எழுத முடியுமா ? என்று என்னை கேட்டார்கள் ….இதொ என் பதில்: தமிழன் நினைத்தால் எதையும் செய்வான் .
This is a sequel to the opinion-survey/feedback form given in this blog. We would love to get some feedback on the feedback form itself !
All items marked with a star (*) are essential, and cannot be skipped.
You can also send your opinion and suggestions directly by mail to drpartha@gmail.com
Thanks for giving us your opinion and suggestions using the form below:
This new contact form was made using
WPForms plugin. You can use this form to send us your comments, issues, queries and suggestions.
You can also send all your comments, issues, queries and suggestions, directly by email to : drpartha@gmail.com
Use this opportunity responsibly, and remember to follow these
basic rules of decency. |
I know that a linear operator is bounded if and only if it is continuous. But what about a bounded linear functional? Is this just a special case of a linear operator, and hence it is also bounded if and only if it is continuous?
But if that is the case, it would seem to make the definition of the weak topology redundant:
The weak topology on $X$ is the weakest topology $U\subset2^X$ with respect to which every bounded linear functional $\Lambda:X\to \mathbb{R}$ is continuous.
If every bounded linear functional is continuous due to being bounded, then why would continuity even be required in the definition of the weak topology? |
To give a little bit more detail on 0-0-0's answer: Let $V$ be an $\mathbb{F}$-vector space of finite dimension.
We first prove the following theorem:
Theorem: Let $\phi\in\mathsf{Hom}(V,V)$, $U\subseteq V$ invariant under $\phi$. Let $\phi|_U:U\to U$ be the restriction of $\phi$ to $U$. Then $q_{\phi|_U}$ divides $q_\phi$ where $q_\phi$ is the minimal polynomial of $\phi$, similar for $\phi|_U$.
Proof: Let $q_\phi$ be the minimal polynomial, i.e. $q_\phi(\phi)=\mathbf{0}$ where $\mathbf{0}$ is the null-endomorphism. Now still $q_\phi(\phi)|_U=\mathbf{0}$. It is straightforward to check, as $U$ is invariant, that $q_\phi(\phi)|_U=q_\phi(\phi|_U)=\mathbf{0}$. Now, per definition $q_{\phi|_U}(\phi|_U)=\mathbf{0}$ and as this minimal polynomial generates the ideal$$I(\phi|_U)=\{p\in\mathbb{F}[X]\mid p(\phi|_U)=\mathbf{0}\}$$we have $q_{\phi|_U}\mid q_\phi$. $\Box$
As a corollary, we obtain the following:
Corollary: Let $\phi\in\mathsf{Hom}(V,V)$. If $\phi$ is diagonalizable and $U\subseteq V$ is invariant under $\phi$, then $\phi|_U:U\to U$ is diagonalizable.
Proof: $\phi$ is diagonalizable iff $$q_\phi=\prod_{i=1}^m(X-\lambda_i)$$ for pairwise distinct eigenvalues $\lambda_i$ of $\phi$. Now, by the above theorem, we have that $q_{\phi|_U}$ divides $q_\phi$, i.e. also $q_{\phi|_U}$ splits into distinct linear factors with algebraic multiplicity 1. Thus $\phi|_U$ is diagonalizable. $\Box$
Now let $U\subseteq V$ be a $\phi$-invariant subspace for $\phi\in\mathsf{Hom}(V,V)$ and $\mathrm{dim}(U)=k$. By the corollary, we have that $\phi|_U$ is diagonalizable, i.e. there exists an eigenbasis $B=(v_1,\dots,v_k)$ for $U$ s.t. $\phi|_U$ is represented by a diagonal matrix. Now, the vectors in this eigenbasis for $U$ are obviously also eigenvector w.r.t. $\phi$. Thus $U=\mathrm{span}(v_1,\dots,v_k)$ for these basis vectors as desired.
We can summarize this good with the following theorem:
Theorem: Let $\phi\in\mathsf{Hom}(V,V)$ be diagonalizable. Then a subspace $U\subseteq V$ with $\mathrm{dim}(U)=k$ is $\phi$-invariant iff $U=\mathrm{span}(v_1,\dots,v_k)$ for eigenvectors $v_i$.
The one direction we just proved, the other is a simple observation. |
Let $F$ be a Maass cusp form for $\mathrm{SL}(3,\mathbb{Z})$ (level 1 trivial character).
Let $g$ be a Maass cusp form for $\Gamma_0(N)$ with character $\chi$ mod $N$. For convenience, you may assume $\chi$ is primitive mod $N$. You may also take $g$ to be an Eisenstein series $E(z,s,\chi)=\sum_{\gamma\in \Gamma_\infty\backslash \Gamma_0(N)}\overline{\chi}(\gamma) \Im(\gamma z)^s$.
If $g$ were level 1 (full modular group $\mathrm{SL}(2,\mathbb{Z})$), then we know that the Rankin-Selberg $L$-function of $F\times g$ is a nice integral \[\text{gamma factors}\cdot L(s,F\times g)=\int_{\mathrm{SL}(2,\mathbb{Z})\backslash \mathbb{H}} F\left(\begin{pmatrix}z&\\&1\end{pmatrix}\right)g(z)\det(z)^{s-1}\, \frac{dx \,dy}{y^2}.\]
The standard unfolding technique applies nicely to the Fourier-Whittaker expansion of $F$. See page 372 of Goldfeld, Dorian. Automorphic forms and L-functions for the group $\mathrm{GL}(n,\mathbb{R})$. Vol. 99. Cambridge University Press, 2006.
However, when $g$ has higher level with character, I don't know how to define the integral. That is my question.
Also, remind that if $f$ and $g$ are automorphic forms on $\mathrm{GL}(2)$ with characters $\chi_f$ and $\chi_g$ we can define their Rankin-Selberg $f\times \bar g$ by using some Eisenstein series $E(z,s, \chi_f\bar{\chi_g})$ to balance the characters of $f\bar g$.
And for my question, adelically, there is definite answer. But I am looking for something computable in classic settings. |
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