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I've been struggling trying to understand how to find the center of a circle given 4 points in the circumference $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$
Please help me. I don't understand when the polygon created is not a rectangle p.ex..
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I've been struggling trying to understand how to find the center of a circle given 4 points in the circumference $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$
Please help me. I don't understand when the polygon created is not a rectangle p.ex..
The question states that all the points are on the circumference. Three non-collinear points already determine a unique circle passing through them, so any three of the four given points may be chosen and the fourth will automatically lie on the circle.
The problem of finding the centre of the circle through three points is well-known. Wikipedia itself gives the following solution $(R_x,R_y)$ for the three points $(A_x,A_y), (B_x,B_y), (C_x,C_y)$:
$$R_x = \left[(A_x^2 + A_y^2)(B_y - C_y) + (B_x^2 + B_y^2)(C_y - A_y) + (C_x^2 + C_y^2)(A_y - B_y)\right] / D$$ $$R_y = \left[(A_x^2 + A_y^2)(C_x - B_x) + (B_x^2 + B_y^2)(A_x - C_x) + (C_x^2 + C_y^2)(B_x - A_x)\right]/ D$$ $$D = 2\left[A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)\right]$$
However, there is no circle passing through four or more points
in general position. For example, if four points are given as $(0,0),(2,0),(0,2)$ and $(-1,-1)$, the first three points determine a circle with centre $(1,1)$ and radius $\sqrt2$, but the fourth point does not lie on this circle. If this arises, the best you can do is minimise the sum-distance of each point to the circle itself, which becomes a least-squares fitting problem. Many resources for this are available too, like this one from Stony Brook.
Fun fact: The type of quadrilateral (for ease of argument, draw the convex one defined by the four points in question) is called cyclic. In Euclid's Elements, Book 3, Proposition 22, it is proven that a quadrilateral is cyclic if and only if its opposite angles are supplementary.
Either way, a convex quadrilateral is cyclic iff its perpendicular bisectors are concurrent, in which case the intersection is the circumcenter (the center of the circle.)
The four points given on the circumference define a convex cyclic quadrilateral uniquely, so just take the perpendicular bisectors and look at the intersection.
Four points constitute a quadrilateral. If these four points lie on the same circle, the quadrilateral thus formed is called cyclic quadrilateral. A cyclic quadrilateral can be but need not be a rectangle. A cyclic quadrilateral has the properties of the sum each pair of interior opposite angles is $180^0$.
For this problem, finding the center does not require the knowledge of whether the 4 points form a rectangle or not. The center is right at the point of intersection of the perpendicular bisectors of any two chords of that circle.
From any two given points (say $A(x_1, y_1)$ and $B(x_2, y_2))$, find the corresponding slope ($m_{AB}$) and midpoint $(M_{AB})$.
Using the above info to get $(L_{AB})$, the equation of the perpendicular bisector (that passes through $M_{AB}$).
Repeat the above process to get $(L_{BC})$.
The co-ordinates of the center is given by solving the simultaneous equations $(L_{AB})$ and $(L_{BC})$.
Note: Three (non-collinear) points uniquely determine the center of the circle passing through them. The 4th point is probably for checking the correctness of the equation of the circle formed.
Assuming that the points are not colinear, for a very simple procedure, consider the general equation of conics the circle $$x^2+y^2+a x+b y +c=0$$ and build, for each data point, the equation $$f_i=x_i^2+y_i^2+a x_i+b y_i+c=0$$ This then reduce to $n$ equations $$a x_i+b y_i+c=-(x_i^2+y_i^2)$$ and you face a simple multilinear regression which will give you parameters $a,b,c$.
Having these numbers, at least as estimates, just complete the square $$x^2+y^2+a x+b y +c=(x+\frac a2)^2-\frac{a^2}4+(y+\frac b2)^2-\frac{b^2}4+c$$ that is to say $$(x+\frac a2)^2+(y+\frac b2)^2=\frac{a^2+b^2-4c}4$$ which makes the center located at $(-\frac a2,-\frac b2)$ with a radius equal to $\frac{\sqrt{a^2+b^2-4c}} 2$.
As said, this is a very simplistic solution but whatever could be the objective function you would want to minimize, it will give you very good starting values for the optimization process.
Quoting NLREG documentation, a more rigourous objective function would be $$F(a,b,r)=\sum_{i=1}^n \left(\sqrt{(x_i-a)^2+(y_i-b)^2}-r\right)^2$$ where $(a,b)$ are the coordinates of the circle and $r$ the radius. As you see, the problem is much more complex and would require optimization.
Unless you
delete/discard one out of the four given points and you do not have a unique circle with a center and radius.This is over-determined but consistent. |
;tldr: You're right about the properties, but the fields don't cause or influence each other in so direct away as a first glimpse of Maxwell's equations might suggest.
Special Relativity is "hidden" in Maxwell's equations. An understanding of their relationship might be useful.
Maxwell's Equations are:
$$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_0}$$$$\nabla\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$$$\nabla \cdot \vec{B}=0$$$$\nabla \times\vec{B}=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$
Take the curl of the second equation and rearrange.
$$\nabla(\nabla\cdot\vec{E})-\nabla^2\vec{E}=-\frac{\partial}{\partial t}(\nabla \times\vec{B})=-\mu_0\frac{\partial\vec{J}}{\partial t}-\frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$
Then rearrange:$$\nabla^2\vec{E}-\frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2}=\mu_0\frac{\partial \vec{J}}{\partial t}+\nabla{(\rho/\epsilon_0)}$$
Taking the curl of the 4th equation and rearranging:
$$\nabla^2\vec{B}-\frac{1}{c^2}\frac{\partial^2\vec{B}}{\partial t^2}=-\mu_0\nabla\times\vec{J}$$
In free space the charge density, $\rho$, and the current density ,$\vec{J}$, are 0
So theses two equations end up with a 0 for the right hand side and become homogeneous wave equations.
Solutions are linear combinations of individual solutions having the form $\vec{E}=\vec{E_0}e^{i(\vec{k}\cdot \vec{x}-\omega t)}$
$k$ is the wave number, $\omega$ is the frequency of the wave.
We can check against Maxwell's equations.
$$\nabla\cdot\vec{E}=i\vec{k}\cdot\vec{E_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}=0$$
$\vec{k}$ is in the direction of propagation, and since its dot product with $E_0$(and any component $E_0$ in a linear combination), then we can conclude that
. the electric field is perpendicular to its direction of propagation
$\nabla \times \vec{E}=i\vec{k}\times\vec{E_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}=-\frac{\partial \vec{B}}{\partial t} \implies \vec{B}=-\frac{\vec{k}}{\omega}\times\vec{E_0}e^{i(\vec{k}\cdot\vec{x}-\omega t)}$
So
It's also worth noting $|\vec{k}/\omega|=1/c$. the magnetic field is perpendicular to both the direction of propagation and the electric field.
The Poynting Vector representing the direction of energy flux of the EM field is : $$\vec{S}=\frac{1}{\mu_0}\vec{E}\times\vec{B}=\frac{-1}{\mu_0}e^{2i(\vec{k}\cdot\vec{x}-\omega t)}\vec{E_0}\times(\frac{\vec{k}}{\omega}\times \vec{E_0})= \frac{-1}{\mu_0}e^{2i(\vec{k}\cdot\vec{x}-\omega t)}(\frac{\vec{k}}{\omega}E_0^2-\vec{E_0}(\frac{\vec{k}}{\omega}\cdot\vec{E_0}))$$
Since $\vec{E}$ is perpendicular to $\vec{k}$:
$$\vec{S}=\frac{-1}{\omega\mu_0}e^{2i(\vec{k}\cdot \vec{x}-\omega t)}E_0^2\vec{k}$$
in agreement with $\vec{k}$ as the direction of propagation.
So Maxwell's equations demonstrate certain properties of electro magnetic radiation without making any explicit reference to special relativity. Some questions arise though.
In the wave equation, the coefficient of the second time derivative is the reciprocal square of the velocity of the wave. So Maxwell's equations imply that electro magnetic waves travel at the speed of light. But with respect to what frame of reference? How can radiation be traveling one speed to a stationary observer on earth and the same speed relative to a spaceship traveling in its direction at half the speed of light? Since it always takes some amount of time for a a cause to manifest an effect, how can the time variation of a magnetic field instantly influence the electric field as Maxwell's Equations seem to imply?
This issue might best be approached through an analysis of electric and magnetic potentials.
Given that $\nabla \cdot \vec{B}=0$, we have $\vec{B}=\nabla \times \vec{A}$ for some vector $\vec{A}$ by a theorem from vector calculus, a vector with zero divergence can be expressed as the curl of some other vector. This Vector is the magnetic potential or the Vector Potential.
So we also have $\nabla \times \vec{E}=-\frac{\partial}{\partial t}(\nabla \times \vec{A})\implies \nabla\times(\vec{E}+\frac{\partial \vec{A}}{\partial t})=0$
Another theorem from vector calculus tells us that if the curl of a vector is zero, it can be expressed as the gradient of some scalar.
So:$$\nabla \times (\vec{E}+\frac{\partial \vec{A}}{\partial t})=0 \implies \vec{E}=-\nabla \phi - \frac{\partial \vec{A}}{\partial t}$$
From this we have:
$$-\nabla^2\phi-\frac{\partial(\nabla \cdot \vec{A})}{\partial t}=\frac{\rho}{\epsilon_0}$$
Taking the curl of the expression of $\vec{B}$ in terms of $\vec{A}$ we have
$$\nabla(\nabla \cdot \vec{A})-\nabla^2\vec{A}=\mu_0\vec{J}+\frac{1}{c^2}\frac{\partial}{\partial t} (-\nabla \phi - \frac{\partial \vec{A}}{\partial t})$$
Implying:
$$\nabla^2 \vec{A}-\frac{1}{c^2}\frac{\partial^2 \vec{A}}{\partial t^2}=-\mu_0\vec{J}-\nabla(\nabla \cdot \vec{A}+\frac{1}{c^2}\frac{\partial \phi}{\partial t})$$
Now for any $\vec{v}$, vector $\vec{u}$, and scalar $g$,
$$\nabla \times (\vec{v}+\nabla g)=\nabla \times \vec{v}$$
$$ \nabla \cdot \vec{v}=\nabla \cdot (\vec{v}+\nabla \times \vec{u})$$
Since our fields only depend on the derivatives of potentials, we are allowed to change them so long as the original vector derivatives remain constant.
So:
$$\vec{A'}=\vec{A}+\nabla g$$$$ \phi'=\phi-\frac{\partial g}{\partial t}$$
So modifying with some scalar $g$ will change the potential perahps to a more convenient form without changing the vector derviatives and hence the fields. This
allows us to assert that: $$\nabla \cdot \vec{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0$$ Gauge Freedom
That is called the
. Lorentz Gauge
It simplifies the equations for our potentials. They become:
$$\nabla^2 \vec{A}-\frac{1}{c^2}\frac{\partial ^2 \vec{A}}{\partial t^2}=-\mu_0\vec{J}$$$$\nabla^2\phi-\frac{1}{c^2}\frac{\partial ^2 \phi}{\partial t^2}=\frac{- \rho}{\epsilon_0}$$
We can use
to find integral equation solutions to the potentials from which the fields can be derived. Green's Method
These solutions give us time dependent versions of the Biot-Savart Law and Coulomb's Law. In particular, the solutions necessarily have a dependence not on the charge and current distribution at the present time, but the distributions as they were in the past. In other words we have (time)
. retarded potentials
Their derivatives produce the Jefimenko's Equations for electric and magnetic fields relating the current feild values to past charge and current distribution values. They resolve our questions above by implying that electromagnetic information travels at light speed and the various relations deduced above regarding the electric and magnetic fields do not actually violate causation. They don't influence each other instantaenously, the relationship between their changes is due to changes in the charge and current distributions in the past. |
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2,224 1,255 Summary Wigner's friend seems to lead to certainty in two complimentary contexts Summary:Wigner's friend seems to lead to certainty in two complimentary contexts
This is probably pretty dumb, but I was just thinking about Wigner's friend and wondering about the two contexts involved.
The basic set up I'm wondering about is as follows:
The friend does a spin measurement in the ##\left\{|\uparrow_z\rangle, |\downarrow_z\rangle\right\}## basis, i.e. of ##S_z## at time ##t_1##. And let's say the particle is undisturbed after that.
For experiments outside the lab Wigner considers the lab to be in the basis:
$$\frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle + |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right)$$
He then considers a measurement of the observable ##\mathcal{X}## which has eigenvectors:
$$\left\{\frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle + |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right), \frac{1}{\sqrt{2}}\left(|L_{\uparrow_z}, D_{\uparrow_z}, \uparrow_z \rangle - |L_{\downarrow_z}, D_{\downarrow_z}, \downarrow_z \rangle\right)\right\}$$
with eigenvalues ##\{1,-1\}## respectively.
At time ##t_2## the friend flips a coin and either he does a measurement of ##S_z## or Wigner does a measurement of ##\mathcal{X}##
However if the friend does a measurement of ##S_z## he knows for a fact he will get whatever result he originally got. However he also knows Wigner will obtain the ##1## outcome with certainty.
However ##\left[S_{z},\mathcal{X}\right] \neq 0##. Thus the friend seems to be predicting with certainty observables belonging to two separate contexts. Which is not supposed to be possible in the quantum formalism.
What am I missing? |
I'm focusing on practical control engineering and I love to read articles/papers and books about people how tried to take control theory and apply that onto control engineering.
My biggest interest is adaptive control. Unfortunately, adaptive control is not used as practical control engineering due to the danger of letting the controller estimate a model online. Wrong model can result instability. Also adaptive control requires lots of knowledge about the system before a adaptive controller can be applied. In my opinion, that makes adaptive control theory not adaptive at all.
So I had to take one step back and go back to the offline control tuning, e.g Model Predictive Control, PID control or Linear Quadratic Control. I want to use model based control, so PID control is not an option here for me. I'm going to implement the controller onto a 3-5$ USD microcontroller chip and Model Predictive Control requires a better controller, trust me, I have tried matrix algebra and optimization onto a low RAM micro controller. That does not work. Besides, without quadratic programing and constrains, there is few reason to select a MPC controller over a LQR controller. That leaves me left with the Linear Quadratic Control option.
Linear Quadratic Control minimize a cost function
$$J = \sum^\infty (x^TQx + u^TRu)$$
In this case, I want to use the infinite horizon Linear Quadratic Control. MATLAB/Octave command
lqr(A,B,Q,R) or
dlqr(A,B,Q,R)
Having a LQR controller requires also that you have a model too. For me, I never build a model of a system by using paper and pen. I use measurement data and use subspace identification methods to estimate an over fitted model which means that the models can handle delays internally. My favorite methods are
N4SID for data with much noise,
MOESP for data with less noise,
OKID for small amount of data and less noise and
ERA if you have an impulse response e.g structure element. Those commands can be used from MataveID
The only problem with subspace identification is that you have no information about your state vector $x$. You only know the dimension of it. Not which state represent which velocity or position. To solve this issue, you have to use a linear kalman filter, or called Linear Quadratic Estimator. There is a LQE command in MataveControl that computes he kalman gain matrix $K$.
So that means I'm in need of an observer with my LQR controller. This controller will now be labeled as Linear Quadratic Gaussian controller - LQG, because of the observer.
I have been talked a lot of university doctors, professors and studens and they have always recommended me to include integration into the model, for better tracking of the reference. Even if there is told that integration is only required for slow systems, but in practical point of view integral action is necessary if tracking is important. One scenario where I think integral action is not important at all is flying drones.
Anyway! LQG with integral controller. Is that possible? Assume that we have our state space model, estimated from the subspace identification methods. I put a hat $\hat {(.)}$ over a matrix to mark that the matrix is estimated from data.
$$x(k+1) = \hat Ax(k) + \hat Bu(k)$$ $$y(k) =\hat Cx(k) + \hat Du(k)$$
I begin to estimate my kalman gain matrix $K$ from that model. Now I have my state update observer.
$$\hat x(k+1) = \hat x(k) + K(y_m(k) - \hat Cx(k))$$
In face you don't know what Kalman Filter is. A kalman filter is just a description of the system that you simulate online, meanwhile you control the system. The state update equation above just adjust the error. If output measurement $y_m(k)$ is equal to estimated $\hat Cx(k)$, that means we have a perfect description of your system. If not, then our $K$ gain matrix is going to adjust our state updater equation.
After getting the observer done, I need to put some integral action into my model. Notice that this is the discrete method to include integral action.
$$A^* = \begin{bmatrix} A & 0\\ -C & 1 \end{bmatrix}$$
$$B^* = \begin{bmatrix} B\\ -D \end{bmatrix}$$
From that, I can finally compute my control law $$u(k) = -L\hat x(k) + l_{n+1}x_{n+1}$$ $$x_{n+1} = \sum^k_{k=0} [r(k) - y_m(k)]$$
Where $L$ is the LQR control law and $l_{n+1}$ is the integral control law. The command to compute these control laws is still from the MATLAB/Octave command
lqr. By using the matrices $A^*$ and $B^*$, the output of
lqr is going to get an extra column $l_{n+1}$ or more depending on the dimension $n$ of $\hat C$ matrix.
There is also a command in MataveControl that can compute the LQR contol law and the extended integral control law from
[L, Li] = lqi(sys, Q, R)
Notice that the measured output $y_m(k)$ does not need to be filtered due to the summation.
Question:
By using a reference gain $K_r$ multiplied with the reference vector $r(k)$, this can even more give a better tracking because the reference gain make sure so the steady state follows the reference $r(k)$. I know that the integral action will minimize the steady state errors, but a reference gain $K_r$ will help so the integral don't need to grow so large.
So how can I find the reference gain for a LQG controller with integral action? |
This answer will try and outline all the different possibilities I came across over the last couple of years, including drawbacks. But first, let me outline the problem a little.
To appreciate the problem, a first simplistic starting point is here.What the authors observe is similar to what you observed. "Optimization is Error-Maximization" is an often cited quote.
Why is that?
To build some intuition, look at the unconstrained solution of the MV problem. It is proportional to the inverse of the variance Covariance matrix $\Sigma^{-1}$. You see that by setting the first derivative to zero and solving for the weights.
The role of the spectrum of $\Sigma$
If you look at the Eigenvalues of $\Sigma$, the problem occurs when they are close to $0$. That is, because the inverse $\Sigma^{-1}$ has the inverse eigenvalues, which in turn means that upon solving the problems, certain directions are being extremely magnified. Also, these directions are very unstable, meaning that upon updating $\Sigma$ at a later point in time, directions and thus your solution are likely to show big changes over time and the solution is unstable.
==REMEDIES==
Be aware that all of the following methods will cause a loss of information of some kind. Also, I do not prefer one method over the other.
1. Messing around with the spectrum
One of the more straightforward things: If an Eigenvalue is, say $<10^{-6}$, set it to $10^{-6}$ and recalculate the variance covariance matrix.(If $\Sigma = E^{T} \Lambda E$ with Eigenbasis $E$ and spectrum $\Lambda$, you can define $\hat{\lambda}_i = \text{max}(10^{-6},\lambda_i)$ and then calculate $\hat{\Sigma} = E^{T} \hat{\Lambda} E$.
(Hint: you can force total variance to be the same by rescaling but the difference shouldnt be big).
Bear in mind that there are numerous ways you could do this but this is the most straightforward one I think. More difficult is the question how small do you accept your eigenvalues to be...
2. Factor Models
If we can express $N$ asset classes in terms of $F$ factors, this is effectively a dimension reduction. If you drop the idiosyncratic parts and the variance covariance matrix of the factors is stable (usually, the idea of factors is that their correlation is not too high). You would need to reformulate your problem
3. Shrinkage Estimation
In numerics, there is the common trick to "add a diagonal"($\hat{\Sigma} = \Sigma + \lambda \mathbb{1})$ to get the spectrum away from $0$. Now, if we come from statistics the total error of an estimator can be decomposed in a bias plus a variance component. The idea is to reduce the estimation error further by taking a bias (you can actually use an ansatz like metioned above and calculate the $\lambda$ if I remember correctly). Look at Shrinkage Estimation.
4. Expected Return estimation - Black Litterman method
One finding from employing the BL-method is that the results are more stable. This is because the expected (prior) returns are calculated via the market weights $w_M$ and the variance-covariance matrix: $\mu \approx \Sigma w_M$. Also, you can see that via the very heuristic argument that if your solution $w \approx \Sigma^{-1} \mu$ and $\mu \approx \Sigma w_m$ then this will be stable, as the idea is that both in a way "cancel out". I know this is by no means mathematically correct but just to give you a feeling.
I am sure the list is by no means complete. Take also a look at
robust optimization. I didnt cover this here. |
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∫0π2arctan(1−cos2xsin2x) dx=π24−πarctan2−12. \int\limits_0^{\frac{\pi}{2}}\arctan\left(1-\cos^2 x\sin^2 x\right)\ \mathrm{d}x=\frac{\pi^2}{4}-\pi\arctan\sqrt{\frac{\sqrt{2}-1}{2}}. 0∫2πarctan(1−cos2xsin2x) dx=4π2−πarctan22−1.Can someone prove this ?
Note by Haroun Meghaichi 5 years, 3 months ago
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∫0π/2arctan(1−cos2xsin2x) dx=∫0π/2(π2−arctan(11−sin2xcos2x)) dx=π24−∫0π/2arctan(sin2x) dx−∫0π/2arctan(cos2x) dx=π24−2∫0π/2arctan(cos2x) dx\displaystyle \begin{aligned} \int_0^{\pi/2} \arctan\left(1-\cos^2x\sin^2x\right)\,dx & =\int_0^{\pi/2} \left(\frac{\pi}{2}-\arctan\left(\frac{1}{1-\sin^2x\cos^2x}\right)\right)\,dx \\&=\frac{\pi^2}{4}-\int_0^{\pi/2} \arctan(\sin^2x)\,dx-\int_0^{\pi/2} \arctan(\cos^2x)\,dx \\&=\frac{\pi^2}{4}-2\int_0^{\pi/2}\arctan(\cos^2x)\,dx \\\end{aligned} ∫0π/2arctan(1−cos2xsin2x)dx=∫0π/2(2π−arctan(1−sin2xcos2x1))dx=4π2−∫0π/2arctan(sin2x)dx−∫0π/2arctan(cos2x)dx=4π2−2∫0π/2arctan(cos2x)dx
Consider
I(a)=∫0π/2arctan(a cos2x) dx\displaystyle I(a)=\int_0^{\pi/2} \arctan(a\,\cos^2x)\,dx I(a)=∫0π/2arctan(acos2x)dx
⇒I′(a)=∫0π/2cos2x1+a2cos4x dx=∫0π/2sec2xsec4x+a2 dx=∫0π/2sec2x(1+tan2x)2+a dx=∫0∞dtt4+2t2+a2+1 (tanx=t)\displaystyle \begin{aligned}\Rightarrow I'(a)&=\int_0^{\pi/2} \frac{\cos^2x}{1+a^2\cos^4x}\,dx \\&= \int_0^{\pi/2} \frac{\sec^2x}{\sec^4x+a^2}\,dx\\&= \int_0^{\pi/2} \frac{\sec^2x}{(1+\tan^2x)^2+a}\,dx\\&=\int_0^{\infty} \frac{dt}{t^4+2t^2+a^2+1}\,\,\,\,\,\,\,(\tan x=t) \\\end{aligned} ⇒I′(a)=∫0π/21+a2cos4xcos2xdx=∫0π/2sec4x+a2sec2xdx=∫0π/2(1+tan2x)2+asec2xdx=∫0∞t4+2t2+a2+1dt(tanx=t)
With the substitution t↦a2+1tt\mapsto \dfrac{\sqrt{a^2+1}}{t} t↦ta2+1,
I′(a)=1a2+1∫0∞t2t4+2t2+a2+1 dt\displaystyle I'(a)=\frac{1}{\sqrt{a^2+1}}\int_0^{\infty} \frac{t^2}{t^4+2t^2+a^2+1}\,dt I′(a)=a2+11∫0∞t4+2t2+a2+1t2dt
⇒I′(a)=12a2+1∫0∞a2+1+t2t4+2t2+a2+1 dt=12a2+1∫0∞1+1+a2t2(t−a2+1t)2+2(1+a2+1) dt=12a2+1∫−∞∞dyy2+2(1+a2+1) (t−a2+1t=y)=π2211+a21+a2+1\displaystyle \begin{aligned}\Rightarrow I'(a) &=\frac{1}{2\sqrt{a^2+1}}\int_0^{\infty} \frac{\sqrt{a^2+1}+t^2}{t^4+2t^2+a^2+1}\,dt \\&=\frac{1}{2\sqrt{a^2+1}}\int_0^{\infty} \frac{1+\frac{\sqrt{1+a^2}}{t^2}}{\left(t-\frac{\sqrt{a^2+1}}{t}\right)^2+2(1+\sqrt{a^2+1})}\,dt\\&=\frac{1}{2\sqrt{a^2+1}}\int_{-\infty}^{\infty} \frac{dy}{y^2+2(1+\sqrt{a^2+1})}\,\,\,\,\,\,\,\left(t-\frac{\sqrt{a^2+1}}{t}=y\right) \\&=\frac{\pi}{2\sqrt{2}} \frac{1}{\sqrt{1+a^2}\sqrt{1+\sqrt{a^2+1}}} \\\end{aligned} ⇒I′(a)=2a2+11∫0∞t4+2t2+a2+1a2+1+t2dt=2a2+11∫0∞(t−ta2+1)2+2(1+a2+1)1+t21+a2dt=2a2+11∫−∞∞y2+2(1+a2+1)dy(t−ta2+1=y)=22π1+a21+a2+11
Integrating back,
I(1)−I(0)=I(1)=π22∫01da1+a21+a2+1=π22∫12dtt−1(t+1) (a2+1=t)=π2∫02−1duu2+2 (t−1=u2)=π2(arctanu2∣02−1=π2arctan(2−12)\displaystyle \begin{aligned}I(1)-I(0)=I(1) &=\frac{\pi}{2\sqrt{2}}\int_0^1 \frac{da}{\sqrt{1+a^2}\sqrt{1+\sqrt{a^2+1}}} \\&= \frac{\pi}{2\sqrt{2}}\int_1^{\sqrt{2}} \frac{dt}{\sqrt{t-1}(t+1)}\,\,\,\,\,\,\,(\sqrt{a^2+1}=t) \\&=\frac{\pi}{\sqrt{2}}\int_0^{\sqrt{\sqrt{2}-1}} \frac{du}{u^2+2} \,\,\,\,\,\,\,(t-1=u^2) \\&=\frac{\pi}{2} \left(\arctan\frac{u}{\sqrt{2}}\right|_0^{\sqrt{\sqrt{2}-1}} \\&=\frac{\pi}{2}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\right) \\\end{aligned} I(1)−I(0)=I(1)=22π∫011+a21+a2+1da=22π∫12t−1(t+1)dt(a2+1=t)=2π∫02−1u2+2du(t−1=u2)=2π(arctan2u∣∣∣∣02−1=2πarctan⎝⎛22−1⎠⎞
Hence,
∫0π/2arctan(1−cos2xsin2x) dx=π24−πarctan(2−12)\boxed{\displaystyle \int_0^{\pi/2} \arctan\left(1-\cos^2x\sin^2x\right)\,dx=\dfrac{\pi^2}{4}-\pi\arctan\left(\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)} ∫0π/2arctan(1−cos2xsin2x)dx=4π2−πarctan⎝⎛22−1⎠⎞
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What a great job! Hats off to you @Pranav Arora .
Wonderful solution Pranav!!! How do you think of such things?!⌣¨\ddot\smile⌣¨
Thank you Karthik! :)
I don't think I did anything special in this one, I used only the standard techniques as you can see. I am sure you too would have solved it if you jumped on this one before me. ;)
@Pranav Arora – Oh, no...I tried a lot but did not succeed. My approach was also 'differentiating under the integral' but I was nowhere close to the solution. Great solution once again!⌣¨\ddot\smile⌣¨
Excellent!!!
What level calculus is this..?I am in 12th grade am I supposed to be able to solve this...??
I suppose this is problem solving calculus. You don't regularly learn stuff this hard in school.
@Daniel Liu – Hey are you actually 14 year old...??
Problem Loading...
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2019-10-11 14:20
TURBO stream animation /LHCb Collaboration An animation illustrating the TURBO stream is provided. It shows events discarded by the trigger in quick sequence, followed by an event that is kept but stripped of all data except four tracks [...] LHCB-FIGURE-2019-010.- Geneva : CERN, 2019 - 3. 详细记录 - 相似记录 2019-10-10 15:48 详细记录 - 相似记录 2019-09-12 16:43
Pending/LHCb Collaboration Pending LHCB-FIGURE-2019-008.- Geneva : CERN, 10 详细记录 - 相似记录 2019-09-10 11:06
Smog2 Velo tracking efficiency/LHCb Collaboration LHCb fixed-target programme is facing a major upgrade (Smog2) for Run3 data taking consisting in the installation of a confinement cell for the gas covering $z \in [-500, -300] \, mm $. Such a displacement for the $pgas$ collisions with respect to the nominal $pp$ interaction point requires a detailed study of the reconstruction performances. [...] LHCB-FIGURE-2019-007.- Geneva : CERN, 10 - 4. Fulltext: LHCb-FIGURE-2019-007_2 - PDF; LHCb-FIGURE-2019-007 - PDF; 详细记录 - 相似记录 2019-09-09 14:37
Background rejection study in the search for $\Lambda^0 \rightarrow p^+ \mu^- \overline{\nu}$/LHCb Collaboration A background rejection study has been made using LHCb Simulation in order to investigate the capacity of the experiment to distinguish between $\Lambda^0 \rightarrow p^+ \mu^- \overline{\nu}$ and its main background $\Lambda^0 \rightarrow p^+ \pi^-$. Two variables were explored, and their rejection power was estimated applying a selection criteria. [...] LHCB-FIGURE-2019-006.- Geneva : CERN, 09 - 4. Fulltext: PDF; 详细记录 - 相似记录 2019-09-06 14:56
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First study of the VELO pixel 2 half alignment/LHCb Collaboration A first look into the 2 half alignment for the Run 3 Vertex Locator (VELO) has been made. The alignment procedure has been run on a minimum bias Monte Carlo Run 3 sample in order to investigate its functionality [...] LHCB-FIGURE-2019-003.- Geneva : CERN, 02 - 4. Fulltext: VP_alignment_approval - TAR; VELO_plot_approvals_VPAlignment_v3 - PDF; 详细记录 - 相似记录 2019-07-29 14:20 详细记录 - 相似记录 2019-07-09 09:53
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It is often convenient to parametrize the implied volatility curve to allow easy interpolation of volatility for any strike or maturity. What functional form describes the implied volatility curve for options at varying strikes and fixed maturity?
OptionMetrics uses a kernel smoothing algorithm to interpolate the volatility surface. Their assumptions tend to be based on the academic consensus and have become somewhat industry standard, so the real answer to your question may be that there really is no good functional form.
First, note that there are actually quite a few implied volatility curves...I am afraid there is no "
the" volatility curve. Right off the bat I can think of The put and call bid and offer curves The put and call midmarket price curves The put and call midmarket vol curves The out-of-the-money bid, offer, midmarket price and midmarket vol curves
so that is 12 different curves right there. You can probably already tell that getting a single functional form to fit them all is not going to be easy.
The most common function used is a parabola, though almost always on $\log(K)$ rather than on strike $K$. The second most common choice is cubic splines, either with nodes at every strike or smoothing. It is customary in these cases to specify "cutoffs", which are limiting high and low strikes beyond which volatility is assumed to be constant. That keeps the curve from going negative, or "too" positive.
You will occasionally see implementations based on modifications of the terminal probability distribution, such as Edgeworth expansions.
A Polynomial of degree 2 or 3 ?
But Linear interpolation on a datapoints vector works fine in my experience, let's say you have an index whose options strike :
80/82/84/86/88/90
You usually don't need to calculate vol @ 83. The only case is if you have a different volatility smile (estimated vol. for example) whose data points are 80/85/90 then you can just do linear interp to find your estimated vol @ 82/84.
In the rates world (ie swaptions, caps and floors) I believe most banks are using some form of the
SABR model (Stochastic Alpha Beta Rho) for building the volatility smile.
When we say 'use the SABR model' what we really mean is that the smile shape function is derived from the shape of the smile in a theoretical model of form: $$ \text{d} F_t = \sigma_t F_t^\beta \, \text{d} W^1_t $$ $$ \text{d} \sigma_t = \alpha \sigma_t \, \text{d} W^2_t $$ $$ \text{d}W^1_t \text{d}W^2_t = \rho \, \text{d}t $$ Some clever people found a way to get a good-quality closed-form approximation for the smile function, so effectively you can just plug in the parameters and get your volatity at a given strike for a given value of the forward.
That said, the formula is known to break down at low strikes -- producing negative values for the implied probability distribution. Therefore most houses have put resources into fixing this in one way or another. |
I would like to calculate the polar velocity components given the position $(x,y)$ and velocity $(u_x,u_y)$ in Cartesian coordinates. First of all, $$ r=\sqrt{x^2+y^2}\text{ and }\theta=\tan^{-1}\left(\frac yx\right). $$ By now, I know the angle and radius in the global cylindrical coordinate system. I assume that $\mathbf u=u_r\,\mathrm e_r+u_\theta\,\mathrm e_\theta$. Is it correct to write, \begin{align} u_r&=\cos(\theta)u_x + \sin(\theta)u_y \\ u_\theta&=-\sin(\theta)u_x + \cos(\theta)u_y? \end{align} The problem is that I would like to calculate the components in polar coordinates (not angular velocity though) given that a particle is moving from $(x_1,y_1)$ to $(x_2,y_2)$.
From what I can gather you just want to replace the trigonometric functions with their Cartesian counterparts
$$cos(\theta)=\frac xr=\frac{x}{\sqrt {x^2+y^2}}$$
$$sin(\theta)=\frac yr=\frac{y}{\sqrt {x^2+y^2}}$$
So what you have for $u_r$ becomes
$$u_r=\frac{xu_x}{\sqrt {x^2+y^2}}+\frac{yu_y}{\sqrt {x^2+y^2}}=\frac{xu_x+yu_y}{\sqrt {x^2+y^2}}$$
And $u_\theta$ becomes
$$u_\theta=-\frac{yu_x}{\sqrt {x^2+y^2}}+\frac{xu_y}{\sqrt {x^2+y^2}}=\frac{xu_y-yu_x}{\sqrt {x^2+y^2}}$$
Therefore, the velocity $\vec v$ with polar components expressed in Cartesian coordinates is
$$\mathbf{\vec v=\frac{xu_x+yu_y}{\sqrt {x^2+y^2}}e_r+\frac{xu_y-yu_x}{\sqrt {x^2+y^2}}e_\theta}$$
If you want to explicitly substitute your expressions for $u_x$ and $u_y$ as expressed in your comments, then at any position $(x,y)$ the velocity will be given by
$$\mathbf{\vec v=\frac{x(x_2-x_1)+y(y_2-y_1)}{\delta t\sqrt {x^2+y^2}}e_r+\frac{x(y_2-y_1)-y(x_2-x_1)}{\delta t\sqrt {x^2+y^2}}e_\theta}$$ |
I am trying to solve an equation in Python. Basically what I want to do is to solve the equation:
$$ \frac{1}{x^2}\frac{d}{dx}\left(Gam \frac{dL}{dx}\right)+L\left(\frac{a^2x^2}{Gam}-m^2\right)=0 $$
This is the Klein-Gordon equation for a massive scalar field in a Schwarzschild spacetime. We know $m$ and $Gam=x^2-2x$.
The initial/boundary condition that I know are $L\rvert_{2+\epsilon}=1$ and $L\rvert_{\infty}=0$.
Notice that the asymptotic behavior of the equation is
$$ L(x\to\infty)\to \frac{e^{\pm(m^2-a^2)x}}{x} $$
Then, if $a^2>m^2$ we will have oscillatory solutions, while if $a^2 < m^2$ we will have a divergent and a decay solution. What I am interested in is the decay solution; however when I am trying to solve the above equation transforming it as a system of first-order differential equations and using the shooting method in order to find the $a$ that can give me the behavior that I am interested about, I am always having a divergent solution (for $0<a^2<m^2$). I suppose that it is happening because
odeint is always finding the divergent asymptotic solution.
Is there a way to avoid or tell to
odeint that I am interested in the decay solution? If not, do you know a way that I could solve this problem? Maybe using another method for solving my system of differential equations? If yes, which method?
Basically what I am doing is to add a new system of equations for $a$
$$\frac{d^2a}{dx^2}=0,\\ \frac{da}{dx}(2+\epsilon)=0,\\ a(2+\epsilon)=a_0)$$
in order to have $a$ as a constant. Then I am considering different values for $a_0$ and asking if my boundary conditions are fulfilled.
EDIT 1.
Hello, let me try to explain a little more my issue
I am incorporating the value at infinity considering the assimptotic behavior, it means that I will have a relation between the field and its derivative. I will post the code for you if it is helpful:
from IPython import get_ipythonget_ipython().magic('reset -sf')import numpy as npimport matplotlib.pyplot as pltfrom scipy.integrate import odeintfrom math import *from scipy.integrate import ode
These are initial conditions for Schwarzschild. The field is invariant under reescaling, then I can use $L(2+\epsilon)=1$
def init_sch(u_sch): om = u_sch[0] return np.array([1,0,om,0]) #conditions near the horizon, [L_c,dL/dx,a,da/dx]
These are our system of equations
def F_sch(IC,r,rho_c,m,lam,l,j=0,mu=0): L = IC[0] ph = IC[1] om = IC[2] b = IC[3] Gam_sch=r**2.-2.*r dR_dr = ph dph_dr = (1./Gam_sch)*(2.*(1.-r)*ph+L*(l*(l+1.))-om**2.*r**4.*L/Gam_sch+(m**2.+lam*L**2.)*r**2.*L) dom_dr = b db_dr = 0. return [dR_dr,dph_dr,dom_dr,db_dr]
Then I try for different values of "om" and ask if my boundary conditions are fulfilled. p_sch are the parameters of my model. In general what I want to do is a little more complicated and in general I will need more parameters that in the just massive case. Howeve I need to start with the easiest which is what I am asking here
p_sch = (1,1,0,0) #[rho_c,m,lam,l], lam and l are for a more complicated case ep = 0.2ep_r = 0.01r_end = 500n_r = 500000n_omega = 1000omega = np.linspace(p_sch[1]-ep,p_sch[1],n_omega)r = np.linspace(2+ep_r,r_end,n_r)tol = 0.01a = 0for j in range(len(omega)): print('trying with $omega =$',omega[j]) omeg = [omega[j]] ini = init_sch(omeg) Y = odeint(F_sch,ini,r,p_sch,mxstep=50000000) print Y[-1,0] #This is my boundary condition that I am implementing. Basically this should be my conditions at "infinity if abs(Y[-1,0]*((p_sch[1]**2.-Y[-1,2]**2.)**(1/2.)+1./(r[-1]))+Y[-1,1]) < tol: print(j,'times iterations in omega') print("R'(inf)) = ", Y[-1,0]) print("\omega",omega[j]) omega_1 = [omega[j]] a = 10 break if a > 1: break
Basically what I want to do here is to solve the system of equations giving different initial conditions and find a value for "a=" (or "om" in the code) that should be near to my boundary conditions. I need this because after this I can give such initial guest to a secant method and try to fiend a best value for "a". However, always that I am running this code I am having divergent solutions that it is, of course, a behavior that I am not interested. I am trying the same but considering the scipy.integrate.solve_vbp, but when I run the following code:
from IPython import get_ipythonget_ipython().magic('reset -sf')import numpy as npimport matplotlib.pyplot as pltfrom math import *from scipy.integrate import solve_bvpdef bc(ya,yb,p_sch): m = p_sch[1] om = p_sch[4] tol_s = p_sch[5] r_end = p_sch[6] return np.array([ya[0]-1,yb[0]-tol_s,ya[1],yb[1]+((m**2-yb[2]**2)**(1/2)+1/r_end)*yb[0],ya[2]-om,yb[2]-om,ya[3],yb[3]])def fun(r,y,p_sch): rho_c = p_sch[0] m = p_sch[1] lam = p_sch[2] l = p_sch[3] L = y[0] ph = y[1] om = y[2] b = y[3] Gam_sch=r**2.-2.*r dR_dr = ph dph_dr = (1./Gam_sch)*(2.*(1.-r)*ph+L*(l*(l+1.))-om**2.*r**4.*L/Gam_sch+(m**2.+lam*L**2.)*r**2.*L) dom_dr = b db_dr = 0.*y[3] return np.vstack((dR_dr,dph_dr,dom_dr,db_dr))eps_r=0.01r_end = 500n_r = 50000r = np.linspace(2+eps_r,r_end,n_r)y = np.zeros((4,r.size))y[0]=1tol_s = 0.0001p_sch= (1,1,0,0,0.8,tol_s,r_end)sol = solve_bvp(fun,bc, r, y, p_sch)
I am obtaining this error: ValueError:
bc return is expected to have shape (11,), but actually has (8,).ValueError:
bc return is expected to have shape (11,), but actually has (8,). |
2019-09-20 08:41
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Investigation of the $\Delta n = 0$ selection rule in Gamow-Teller transitions: The $\beta$-decay of $^{207}$Hg / Berry, T A (Surrey U.) ; Podolyák, Zs (Surrey U.) ; Carroll, R J (Surrey U.) ; Lică, R (CERN ; Bucharest, IFIN-HH) ; Grawe, H ; Timofeyuk, N K (Surrey U.) ; Alexander, T (Surrey U.) ; Andreyev, A N (York U., England) ; Ansari, S (Cologne U.) ; Borge, M J G (CERN ; Madrid, Inst. Estructura Materia) et al. Gamow-Teller $\beta$ decay is forbidden if the number of nodes in the radial wave functions of the initial and final states is different. This $\Delta n=0$ requirement plays a major role in the $\beta$ decay of heavy neutron-rich nuclei, affecting the nucleosynthesis through the increased half-lives of nuclei on the astrophysical $r$-process pathway below both $Z=50$ (for $N>82$ ) and $Z=82$ (for $N>126$). [...] 2019 - 5 p. - Published in : Phys. Lett. B 793 (2019) 271-275 Article from SCOAP3: PDF; Record dettagliato - Record simili 2019-09-14 06:30
Precision measurements of the charge radii of potassium isotopes / Koszorús, Á (KU Leuven, Dept. Phys. Astron.) ; Yang, X F (KU Leuven, Dept. Phys. Astron. ; Peking U., SKLNPT) ; Billowes, J (Manchester U.) ; Binnersley, C L (Manchester U.) ; Bissell, M L (Manchester U.) ; Cocolios, T E (KU Leuven, Dept. Phys. Astron.) ; Farooq-Smith, G J (KU Leuven, Dept. Phys. Astron.) ; de Groote, R P (KU Leuven, Dept. Phys. Astron. ; Jyvaskyla U.) ; Flanagan, K T (Manchester U.) ; Franchoo, S (Orsay, IPN) et al. Precision nuclear charge radii measurements in the light-mass region are essential for understanding the evolution of nuclear structure, but their measurement represents a great challenge for experimental techniques. At the Collinear Resonance Ionization Spectroscopy (CRIS) setup at ISOLDE-CERN, a laser frequency calibration and monitoring system was installed and commissioned through the hyperfine spectra measurement of $^{38–47}$K. [...] 2019 - 11 p. - Published in : Phys. Rev. C 100 (2019) 034304 Article from SCOAP3: PDF; Record dettagliato - Record simili 2019-09-12 09:23
Evaluation of high-precision atomic masses of A ∼ 50-80 and rare-earth nuclides measured with ISOLTRAP / Huang, W J (CSNSM, Orsay ; Heidelberg, Max Planck Inst.) ; Atanasov, D (CERN) ; Audi, G (CSNSM, Orsay) ; Blaum, K (Heidelberg, Max Planck Inst.) ; Cakirli, R B (Istanbul U.) ; Herlert, A (FAIR, Darmstadt) ; Kowalska, M (CERN) ; Kreim, S (Heidelberg, Max Planck Inst. ; CERN) ; Litvinov, Yu A (Darmstadt, GSI) ; Lunney, D (CSNSM, Orsay) et al. High-precision mass measurements of stable and beta-decaying nuclides $^{52-57}$Cr, $^{55}$Mn, $^{56,59}$Fe, $^{59}$Co, $^{75, 77-79}$Ga, and the lanthanide nuclides $^{140}$Ce, $^{140}$Nd, $^{160}$Yb, $^{168}$Lu, $^{178}$Yb have been performed with the Penning-trap mass spectrometer ISOLTRAP at ISOLDE/CERN. The new data are entered into the Atomic Mass Evaluation and improve the accuracy of masses along the valley of stability, strengthening the so-called backbone. [...] 2019 - 9 p. - Published in : Eur. Phys. J. A 55 (2019) 96 Fulltext from Publisher: PDF; Record dettagliato - Record simili 2019-09-05 06:35
Nuclear charge radii of $^{62−80}$Zn and their dependence on cross-shell proton excitations / Xie, L (Manchester U.) ; Yang, X F (Peking U., SKLNPT ; Leuven U.) ; Wraith, C (Liverpool U.) ; Babcock, C (Liverpool U.) ; Bieroń, J (Jagiellonian U.) ; Billowes, J (Manchester U.) ; Bissell, M L (Manchester U. ; Leuven U.) ; Blaum, K (Heidelberg, Max Planck Inst.) ; Cheal, B (Liverpool U.) ; Filippin, L (U. Brussels (main)) et al. Nuclear charge radii of $^{62−80}$Zn have been determined using collinear laser spectroscopy of bunched ion beams at CERN-ISOLDE. The subtle variations of observed charge radii, both within one isotope and along the full range of neutron numbers, are found to be well described in terms of the proton excitations across the $Z=28$ shell gap, as predicted by large-scale shell model calculations. [...] 2019 - 5 p. - Published in : Phys. Lett. B 797 (2019) 134805 Article from SCOAP3: PDF; Record dettagliato - Record simili 2019-08-14 18:20 Record dettagliato - Record simili |
H(w) =1/ (square root of 2) That sounds a bit confusing, we usually refer to the cutoff point as the -3 dB point.
That is the same though, -3 dB is
half the power.
Let me explain: take your \$H(\omega) = \frac{1}{\sqrt2}\$
That means that at that \$\omega\$ the
voltage is divided by \$\sqrt2\$, if this voltage is applied across a (load) resistor at the output of the filter then the current through that same load resistor will also be divided by \$\sqrt2\$.
What does that mean for the
Power?
I means that the power is
halved.
On a dB Power scale that means -3 dB
Using that "half of the power" as a reference point is useful because if we divide a wideband signal into a low frequency part and a high frequency part then we can do that using a low pass filter and a highpass filter with the same cutoff frequency. At that cutoff frequency half of the power ends up at the output of the lowpass filter and the other half ends up at the output of the highpass filter. |
Suppose you have some linear algebra background. The most importantthing you need to know is that the inner product has the same meaningof what you have learnt in linear algebra class. The inner product$$\left\langle \phi|\psi\right\rangle =\int\phi^{*}(x)\psi(x)dx$$
has the meaning related to a projection of one vector ontoanother vector (for true projection, the wavefunctions needed to be normalized). It is similar to the projection of a three dimensionalvector $\mathbf{\vec{v}}=a\hat{\mathbf{x}}+b\hat{\mathbf{y}}+c\hat{\mathbf{z}}$ onto another unit vector $\mathbf{\hat{x}}$ which gives you the results $\mathbf{\vec{v}}\cdot\mathbf{\hat{x}}=a$.
First, the inner product can give you the "length square" of the wavefunction:$$\left\langle \psi|\psi\right\rangle =\int\psi^{*}(x)\psi(x)dx =\int|\psi(x)|^2dx$$similar to the $\mathbf{\vec{v}}\cdot\mathbf{\vec{v}}=a^{2}+b^{2}+c^{2}$, so you can normalize your wavefunction by the condition $\left\langle \psi|\psi\right\rangle =1$.
Second, it allows you to show that two wavefunctions are orthogonalto each other, given by the condition that the inner product evaluatedto zero $\left\langle \phi|\psi\right\rangle =0$ which is the analogto the $\mathbf{\hat{x}}\cdot\mathbf{\hat{y}}=0$.
Third, if we write the wavefunction $\psi(x)$ as a linear combinationof orthonormal wavefunctions $\psi_{n}(x)$: $$\psi(x)=\sum_{n}c_{n}\psi_{n}(x)$$similar to a general vector in linear algebra, then we will have the innerproduct $\left\langle \psi_{n}|\psi\right\rangle =c_{n}$. The meaningof $c_{n}$ is the probability amplitude and it is a complex number in general.So the probability $p_{n}$ of the wavefunction $\psi$ having the component $\psi_{n}$ is given by $p_{n}=|c_{n}|^{2}=|\left\langle \psi_{n}|\psi\right\rangle |^{2}$. The meaning here is very important when you learn how to preform measurement.
Lastly, you should add two wavefunctions amplitude together before you take the square, similar to adding the amplitude of two water waves. More precisely, if the new wavefunction is $\psi(x) = A[\psi_{a}(x)+\psi_{b}(x)]$, then the probability density at position $x$ is $A^2|\psi_{a}(x)+\psi_{b}(x)|^{2}$. Note that $A$ is the normalization constant given by the condition $\left\langle \psi|\psi\right\rangle=1$. It is where the quantum effect arise. Dont take the square and then add them together. |
To evaluate detection performance, we plot the
miss-rate $mr(c) = \frac{fn(c)}{tp(c) + fn(c)}$ against the number of false positives per image $fppi(c)=\frac{fp(c)}{\text{#img}}$ in log-log plots. $tp(c)$ is the number of true positives, $fp(c)$ is the number of false positives, and $fn(c)$ is the number of false negatives, all for a given confidence value $c$ such that only detections are taken into account with a confidence value greater or equal than $c$. As commonly applied in object detection evaluation the confidence threshold $c$ is used as a control variable. By decreasing $c$, more detections are taken into account for evaluation resulting in more possible true or false positives, and possible less false negatives. We define the log average miss-rate (LAMR) as shown, where the 9 fppi reference points are equally spaced in the log space: $\DeclareMathOperator*{\argmax}{argmax}LAMR = \exp\left(\frac{1}{9}\sum\limits_f \log\left(mr(\argmax\limits_{fppi\left(c\right)\leq f} fppi\left(c\right))\right)\right)$
For each fppi reference point the corresponding mr value is used. In the absence of a miss-rate value for a given f the highest existent
fppi value is used as new reference point. This definition enables LAMR to be applied as a single detection performance indicator at image level. At each image the set of all detections is compared to the groundtruth annotations by utilizing a greedy matching algorithm. An object is considered as detected (true positive) if the Intersection over Union (IoU) of the detection and groundtruth bounding box exceeds a pre-defined threshold. Due to the high non-rigidness of pedestrians we follow the common choice of an IoU threshold of 0.5. Since no multiple matches are allowed for one ground-truth annotation, in the case of multiple matches the detection with the largest score is selected, whereas all other matching detections are considered false positives. After the matching is performed, all non matched ground-truth annotations and detections, count as false negatives and false positives, respectively.
Neighboring classes and ignore regions are used during evaluation. Neighboring classes involve entities that are semantically similar, for example bicycle and moped riders. Some applications might require their precise distinction (
enforce) whereas others might not ( ignore). In the latter case, during matching correct/false detections are not credited/penalized. If not stated otherwise, neighboring classes are ignored in the evaluation. In addition to ignored neighboring classes all persons annotations with the tags behind glass or sitting-lying are treated as ignore regions. Further, as mentioned in Section 3.2, EuroCity Persons Dataset Publication, ignore regions are used for cases where no precise bounding box annotation is possible (either because the objects are too small or because there are too many objects in close proximity which renders the instance based labeling infeasible). Since there is no precise information about the number or the location of objects in the ignore region, all unmatched detections which share an intersection of more than $0.5$ with these regions are not considered as false positives.
Note that submissions with provided publication link and/or code will get priorized in below list (COMING SOON).
Method User LAMR (reasonable)▼ LAMR (small) LAMR (occluded) LAMR (all) External data used Publication URL Publication code Submitted on YOLOv3_640 HUI_Tsinghua-Daim... 0.273 0.564 0.623 0.456 no no 2019-05-17 04:56:27 View R-FCN (with OHEM) ECP Team 0.163 0.245 0.507 0.330 ImageNet yes no 2019-04-01 17:10:03 View SSD ECP Team 0.131 0.235 0.460 0.296 ImageNet yes no 2019-04-02 13:56:14 View Faster R-CNN ECP Team 0.101 0.196 0.381 0.251 ImageNet yes no 2019-04-01 17:06:33 View YOLOv3 ECP Team 0.097 0.186 0.401 0.242 ImageNet yes no 2019-04-01 17:08:05 View HRNet Hongsong Wang 0.061 0.138 0.287 0.183 ImageNet no no 2019-08-05 17:11:04 View
Method User LAMR (reasonable)▼ LAMR (small) LAMR (occluded) LAMR (all) External data used Publication URL Publication code Submitted on Faster R-CNN ECP Team 0.201 0.359 0.701 0.358 ImageNet yes no 2019-05-02 10:10:01 View FasterRCNN with M... Qihua Cheng 0.150 0.253 0.653 0.295 ImageNet no no 2019-07-08 08:48:13 View HRNet Hongsong Wang 0.079 0.156 0.265 0.153 ImageNet no no 2019-08-05 17:11:04 View |
Only the first two sections of this long question are essential. The others are just for illustration. Background
Advanced quadratures such as higher-degree composite Newton–Cotes, Gauß–Legendre, and Romberg seem to be mainly intended for cases where one can finely sample the function but not integrate analytically. However, for functions with structures finer than the sampling interval (see Appendix A for an example) or measurement noise, they cannot compete with simple approaches such as the midpoint or trapezoid rule (see Appendix B for a demonstration).
This is somewhat intuitive as, e.g., the composite Simpson rule essentially “discards” one quarter of the information by assigning it a lower weight. The only reason such quadratures are better for sufficiently boring functions is that properly handling border effects outweighs the effect of discarded information. From another point of view, it is intuitively clear to me that for functions with a fine structure or noise, samples that are remote from the borders of the integration domain must be almost equidistant and have almost the same weight (for a high number of samples). On the other hand, quadrature of such functions may benefit from a better handling of border effects (than for the midpoint method).
Question
Assume that I wish to numerically integrate noisy or fine-structured one-dimensional data.
The number of sampling points is fixed (due to function evaluation being costly), but I can freely place them. However, I (or the method) cannot place sampling points interactively, i.e., based on results from other sampling points. I also do not know potential problem regions beforehand. So, something like Gauß–Legendre (non-equidistant sampling points) is okay; adaptive quadrature is not since it requires interactively placed sampling points.
Have any methods going beyond the midpoint method been suggested for such a case?
Or: Is there any proof that the midpoint method is best under such conditions?
More generally: Is there any existing work on this problem?
Appendix A: Specific example of a fine-structured function
I wish to estimate $\int_0^1f(t)\, \mathrm{d}t$ for: $$ f(t) = \sum_{i=1}^{k} \frac{\sin(ω_i t-φ_i)}{ω_i},$$ with $φ_i∈ [0,2π]$ and $\log{ω_i} ∈ [1,1000]$. A typical function looks like this:
I chose this function for the following properties:
It can be integrated analytically for a control result. It has fine structure on a level that makes it impossible to capture all of it with the number of samples I am using ($<10^2$). It is not dominated by its fine structure. Appendix B: Benchmark
For completeness, here is a benchmark in Python:
import numpy as npfrom numpy.random import uniformfrom scipy.integrate import simps, trapz, romb, fixed_quadbegin = 0end = 1def generate_f(k,low_freq,high_freq): ω = 2**uniform(np.log2(low_freq),np.log2(high_freq),k) φ = uniform(0,2*np.pi,k) g = lambda t,ω,φ: np.sin(ω*t-φ)/ω G = lambda t,ω,φ: np.cos(ω*t-φ)/ω**2 f = lambda t: sum( g(t,ω[i],φ[i]) for i in range(k) ) control = sum( G(begin,ω[i],φ[i])-G(end,ω[i],φ[i]) for i in range(k) ) return control,fdef midpoint(f,n): midpoints = np.linspace(begin,end,2*n+1)[1::2] assert len(midpoints)==n return np.mean(f(midpoints))*(n-1)def evaluate(n,control,f): """ returns the relative errors when integrating f with n evaluations for several numerical integration methods. """ times = np.linspace(begin,end,n) values = f(times) results = [ midpoint(f,n), trapz(values), simps(values), romb (values), fixed_quad(f,begin,end,n=n)[0]*(n-1), ] return [ abs((result/(n-1)-control)/control) for result in results ]method_names = ["midpoint","trapezoid","Simpson","Romberg","Gauß–Legendre"]def med(data): medians = np.median(np.vstack(data),axis=0) for median,name in zip(medians,method_names): print(f"{median:.3e} {name}")print("superimposed sines")med(evaluate(33,*generate_f(10,1,1000)) for _ in range(100000))print("superimposed low-frequency sines (control)")med(evaluate(33,*generate_f(10,0.5,1.5)) for _ in range(100000))
(I here use the median to reduce the influence of outliers due to functions that have only high-frequency content. For the mean, the results are similar.)
The medians of the relative integration errors are:
superimposed sines6.301e-04 midpoint8.984e-04 trapezoid1.158e-03 Simpson1.537e-03 Romberg1.862e-03 Gauß–Legendresuperimposed low-frequency sines (control)2.790e-05 midpoint5.933e-05 trapezoid5.107e-09 Simpson3.573e-16 Romberg3.659e-16 Gauß–Legendre
Note: After two months and one bounty without result, I posted this on MathOverflow. |
Given the standard Ising partition function: $$Z(\theta ,h) = \sum\limits_{\bf{x}} {\exp \left\{ {\theta \sum\limits_{(i,j) \in E} {{x_i}{x_j}} + h\sum\limits_{i \in V} {{x_i}} } \right\}}, $$
is there a closed form expression for the lower bound of the free-energy (Pressure) per-site, defined as,
$$\psi (\theta ,h) = \mathop {\lim }\limits_{n \to \infty } \frac{{\log \left( {Z(\theta ,h)} \right)}}{n}.$$
My hunch is that is $2\theta + h$ based on taking limits of an approximation suggested in this article
Any ideas on its suitability as a lower bound?This post imported from StackExchange Physics at 2014-06-21 09:02 (UCT), posted by SE-user user48476 |
Evaluation of Binary Classifiers
Evaluation is important:
Baseline
So for evaluating a classifier we need to set some baseline
base rate accuracy of a trivial classifier the one that always predicts the majority class random rate accuracy of random guess need to have some domain knowledge to assign Random Distribution
Skewed Classes Suppose we have a binary classifier, e.g. cancer prediction. We built some classification model $h_{\theta}(x)$ if we have $h_{\theta}(x) = 1$, we predict cancer, and if $h_{\theta}(x) = 0$, we predict no cancer. Then we find out that we have 1% errors for our classifier on test set, and 99% were correctly diagnosed But now suppose only 0.5% of patients have cancer This is a skewed class - it's a tiny portion of another class We would predict better by always returning 0 (by using the trivial classifier) (we'll have 0.5% error which is better than 1%) $\Rightarrow$ We need different evaluation metrics, not just error rate
Confusion Matrix
Confusion matrix is a $2 \times 2$ Contingency Table
We divide our predictions and mis-predictions into this matrix
Diagnostic Testing Measures [1]
Actual Class $y$ Positive Negative $h_{\theta}(x)$ Test outcome Test outcome positive True positive ($\text{TP}$) False positive ($\text{FP}$, Type I error) Precision = $\cfrac{\# \text{TP}}{\# \text{TP} + \# \text{FP}}$ Test outcome negative False negative ($\text{FN}$, Type II error) True negative ($\text{TN}$) Negative predictive value = $\cfrac{\# \text{TN}}{\# \text{FN} + \# \text{TN}}$ Sensitivity = $\cfrac{\# \text{TP}}{\# \text{TP} + \# \text{FN}}$ Specificity = $\cfrac{\# \text{TN}}{\# \text{FP} + \# \text{TN}}$ Accuracy = $\cfrac{\# \text{TP} + \# \text{TN}}{\# \text{TOTAL}}$
Main values of this matrix: True Positive - we predicted "+" and the true class is "+" True Negative - we predicted "-" and the true class is "-" False Positive - we predicted "+" and the true class is "-" (Type I error) False Negative - we predicted "-" and the true class is "+" (Type II error) (see also Statistical Tests of Significance#Type I and Type II Errors)
The following measures can be calculated: Accuracy Misclassification Error (or Error Rate) Positive predictive value (or precision) $P = \cfrac{\text{TP}}{\text{TP} + \text{FP}}$ Negative predictive value True Positive Rate (also Sensitivity or Recall) Fraction of positive examples correctly classified $\text{tpr} = \cfrac{\text{TP}}{\text{TP} + \text{FN}}$ False Positive Rate (also Fall-Out) Fraction of negative examples incorrectly classified $\text{fpr} = \cfrac{\text{FP}}{\text{FP} + \text{TN}}$ Specificity Support - fraction of positively classified examples $\text{sup} = \cfrac{\text{TP} + \text{FP}}{N} = \cfrac{\text{predicted pos}}{\text{total}}$
Accuracy and Error Rate
In practice, these are the most widely used metrics
Accuracy: $\text{acc} = \cfrac{TP + TN}{N}$ fraction of correctly classified examples Error Rate: $\text{error} = \cfrac{FN + FP}{N} = 1 - \text{acc}$ Fraction of misclassified examples
Precision
For all input data that we predicted $h_{\theta}(x) = 1$ what fraction actually have $y = 1$?
$P = \text{Precision} = \cfrac{\text{# TP}}{\text{# predicted positives}} = \cfrac{\text{# TP}}{\text{# TP} + \text{# FP}}$
Out of all the people we thought have cancer, how many actually had it? High precision is good we don't tell many people that they have cancer when they actually don't
Recall
For all input data that actually have $y = 1$, what fraction did we correctly detect as $h_{\theta}(x) = 1$?
$R = \text{Recall} = \cfrac{\text{# TP}}{\text{# actual positives}} = \cfrac{\text{# TP}}{\text{# TP + # FN}}$
Out of all the people that do actually have cancer, how much we identified? The higher the better: We don't fail to spot many people that actually have cancer
For a classifier that always returns zero (i.e. $h_{\theta}(x) = 0$) the Recall would be zero That gives us more useful evaluation metric And we're much more sure The F Measure is a combination of Precision and Recall
Example
Diagnostic Testing Wikipedia Example [2]
Patients with bowel cancer (as confirmed on endoscopy) Positive Negative Fecal Occult Blood Screen Test Outcome Test Outcome Positive True Positive (TP) = 20 False Positive (FP) = 180 Positive predictive value
= TP / (TP + FP)
= 20 / (20 + 180) = 10% Test Outcome Negative False Negative (FN) = 10 True Negative (TN) = 1820 Negative predictive value
= TN / (FN + TN)
= 1820 / (10 + 1820) ≈ 99.5% Sensitivity
= TP / (TP + FN)
= 20 / (20 + 10) ≈ 67% Specificity
= TN / (FP + TN)
= 1820 / (180 + 1820) = 91%
Visual Analysis
Visual ways of evaluating the performance of a classifier
Not Binary Classifiers
When we have multi-class classifiers we can use:
Contingency Table just show misclassified examples side-by-side Cost Matrix we define the cost for each misclassification and calculate the total cost some measures can be extended to multiclass classifiers:
See Also
Sources |
LaTeX uses internal counters that provide numbering of pages, sections, tables, figures, etc. This article explains how to access and modify those counters and how to create new ones.
Contents
A counter can be easily set to any arbitrary value with
\setcounter. See the example below:
\section{Introduction} This document will present several counting examples, how to reset and access them. For instance, if you want to change the numbers in a list. \begin{enumerate} \setcounter{enumi}{3} \item Something. \item Something else. \item Another element. \item The last item in the list. \end{enumerate}
In this example
\setcounter{enumi}{3} sets the value of the item counter in the list to 3. This is the general syntax to manually set the value of any counter. See the reference guide for a complete list of counters.
All commands changing a counter's state in this section are changing it globally.
Counters in a document can be incremented, reset, accessed and referenced. Let's see an example:
\section{Another section} This is a dummy section with no purpose whatsoever but to contain text. This section has assigned the number \thesection. \stepcounter{equation} \begin{equation} \label{1stequation} \int_{0}^{\infty} \frac{x}{\sin(x)} \end{equation}
In this example, two counters are used:
\thesection
section at this point. For further methods to print a counter take a look on how to print counters.
\stepcounter{equation}
equation. Other similar commands are
\addtocounter and
\refstepcounter, see the reference guide.
Further commands to manipulate counters include:
\counterwithin<*>{<ctr1>}{<ctr2>}
<ctr2> to the counters that reset
<ctr1> when they're incremented. If you don't provide the
*,
\the<ctr1> will be redefined to
\the<ctr2>.\arabic{<ctr1>}. This macro is included in the LaTeX format since April 2018, if you're using an older version, you'll have to use the
chngctr package.
\counterwithout<*>{<ctr1>}{<ctr2>}
<ctr2> from the counters that reset
<ctr1> when they're incremented. If you don't provide the
*,
\the<ctr1> will be redefined to
\arabic{<ctr1>}. This macro is included in the LaTeX format since April 2018, if you're using an older version, you'll have to use the
chngctr package.
\addtocounter{<ctr>}{<num>}
<num> to the value of the counter
<ctr>.
\setcounter{<ctr>}{<num>}
<ctr>'s value to
<num>.
\refstepcounter{<ctr>}
\stepcounter but you can use LaTeX's referencing system to add a
\label and later
\ref the counter. The printed reference will be the current expansion of
\the<ctr>.
The basic syntax to create a new counter is by
\newcounter. Below an example that defines a numbered environment called
example:
\documentclass{article} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \newcounter{example}[section] \newenvironment{example}[1][]{\refstepcounter{example}\par\medskip \textbf{Example~\theexample. #1} \rmfamily}{\medskip} \begin{document} This document will present... \begin{example} This is the first example. The counter will be reset at each section. \end{example} Below is a second example \begin{example} And here's another numbered example. \end{example} \section{Another section} This is a dummy section with no purpose whatsoever but to contain text. This section has assigned the number \thesection. \stepcounter{equation} \begin{equation} \label{1stequation} \int_{0}^{\infty} \frac{x}{\sin(x)} \end{equation} \begin{example} This is the first example in this section. \end{example} \end{document}
In this LaTeX snippet the new environment
example is defined, this environment has 3 counting-specific commands.
\newcounter{example}[section]
section or omit the parameter if you don't want your defined counter to be automatically reset.
\refstepcounter{example}
\label afterwards.
\theexample
For further information on user-defined environments see the article about defining new environments
You can print the current value of a counter in different ways:
\theCounterName
2.1 for the first subsection in the second section.
\arabic
2.
\value
\setcounter{section}{\value{subsection}}).
\alph
b.
\Alph
B.
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ii.
\Roman
II.
\fnsymbol
†.
\theCounterName is the macro responsible to print
CounterName's value in a formatted manner. For new counters created by
\newcounter it gets initialized as an Arabic number. You can change this by using
\renewcommand. For example if you want to change the way a subsection counter is printed to include the current section in italics and the current subsection in uppercase Roman numbers, you could do the following:
\renewcommand\thesubsection{\textit{\thesection}.\Roman{subsection}} \section{Example} \subsection{Example}\label{sec:example:ssec:example} This is the subsection \ref{sec:example:ssec:example}. Default counters in LaTeX
Usage Name For document structure For floats For footnotes For the enumerate environment Counter manipulation commands \addtocounter{CounterName}{number} \stepcounter{CounterName} \refstepcounter{CounterName}
It works like
\stepcounter, but makes the counter visible to the referencing mechanism (
\ref{label} returns counter value)
\setcounter{CounterName}{number} \newcounter{NewCounterName}
If you want the
NewCounterName counter to be reset to zero every time that another OtherCounterName counter is increased, use: \newcounter{NewCounterName}[OtherCounterName]
\setcounter{section}{\value{subsection}}.
\value{CounterName} \theCounterName
for example:
\thechapter,
\thesection, etc. Note that this might result in more than just the counter, for example with the standard definitions of the
article class
\thesubsection will print
Section. Subsection (e.g.
2.1).
For more information see: |
HAL8999 – 3/100 Chapter 2 of Hands on ML Cost functions virtualenv setup code to get the dataset
The chapter follows a rudimentary machine learning project from business case to final product. California census data is analyzed to build a model which will predict media housing price in a district based on other factors using a linear regression model with a Root Mean Square Error (RMSE) function to measure performance i.e. as a cost function.\(\displaystyle RMSE(X, h) = \sqrt{\frac{1}{m}\sum_{i=1}^{m}(h(x^{(i)}) – y^{(i)})^{2}}\)
The function h is the “hypothesis” function which operates on the feature vector \(x^{(i)}\). RMSE isn’t the only cost function by any stretch of the imagination but it seems to get a lot of use.
From this point the author goes through the dev environment setup process I went through a few days ago and it’s pretty clear from the instructions that the work is being done on a Mac.
The code to download the housing tarball is a little sloppy and would have downloaded it every time I ran the cell so I added a simple test to only do the download if the file didn’t already exist.
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[crayon-5da50c411dfc9396710679 ]import os
import tarfile
import pandas as pd
from six.moves import urllib
DL_ROOT = "https://raw.githubusercontent.com/ageron/handson-ml/master/"
HOUSING_PATH = os.path.join("datasets", "housing")
HOUSING_URL = DL_ROOT + "datasets/housing/housing.tgz"
def fetch_housing_data(housing_url=HOUSING_URL, housing_path=HOUSING_PATH):
"""If the housing data isn't on disk, fetch it"""
if not os.path.isdir(housing_path):
os.makedirs(housing_path)
tgz_path = os.path.join(housing_path, "housing.tgz")
if not os.path.isfile(tgz_path):
print("Downloading housing data…")
urllib.request.urlretrieve(housing_url, tgz_path)
housing_tgz = tarfile.open(tgz_path)
housing_tgz.extractall(path=housing_path)
housing_tgz.close()
def load_housing_data(housing_path=HOUSING_PATH):
"""Load the housing data into a dataframe"""
csv_path = os.path.join(housing_path, "housing.csv")
return pd.read_csv(csv_path)
fetch_housing_data()
housing = load_housing_data()
housing.head()
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<class 'pandas.core.frame.DataFrame'>
RangeIndex: 20640 entries, 0 to 20639
Data columns (total 10 columns):
longitude 20640 non-null float64
latitude 20640 non-null float64
housing_median_age 20640 non-null float64
total_rooms 20640 non-null float64
total_bedrooms 20433 non-null float64
population 20640 non-null float64
households 20640 non-null float64
median_income 20640 non-null float64
median_house_value 20640 non-null float64
ocean_proximity 20640 non-null object
dtypes: float64(9), object(1)
memory usage: 1.6+ MB |
ABCD is a parallelogram.
Prove the following:
$\frac{BF}{FA} = \frac{AD}{AE}$
$\frac{S_{ADF}}{S_{AEF}} = \frac{AD}{AE}$
$S_{EBF} = S_{ADF}$
$S_{BCE} = \frac{1}{2}S_{ABCD}$
I solved the first 3, but could not solve the 4th:
1.
$$\text{Thale's theorm:}$$ $$\frac{AE}{CB} = \frac{AF}{FB} = \frac{EF}{FC}$$ $$\frac{AE}{AD} = \frac{EF}{FC}$$ $$\downarrow$$ $$\frac{AE}{CB} = \frac{AF}{FB} = \frac{EF}{FC} = \frac{AE}{AD}$$ $$\frac{AF}{FB} = \frac{AE}{AD}$$ $$\downarrow$$ $$\boxed{\frac{FB}{AF} = \frac{AD}{AE}}$$
2.
$$\text{Let P be a point on ED such that FP will be perpendicular to ED.}$$ $$\div\begin{cases} S_{AEF} = \frac{AE\cdot FP}{2} \\ S_{ADF} = \frac{AD\cdot FP}{2}\end{cases}$$ $$\frac{S_{AEF}}{S_{ADF}} =\frac{AE}{AD}$$ $$\downarrow$$ $$\boxed{\frac{S_{ADF}}{S_{AEF}} =\frac{AD}{AE}}$$
3. $$\text{Let G be a point on AB such that EG will be perpendicular to AB. Then:}$$ $$\frac{S_{AEF}}{S_{EFB}} = \frac{\frac{AF\cdot EG}{2}}{\frac{FB\cdot EG}{2}} = \frac{AF}{FB} = \frac{AE}{AD} = \frac{S_{AEF}}{S_{ADF}}$$ $$\frac{S_{AEF}}{S_{EFB}} = \frac{S_{AEF}}{S_{ADF}}$$ $$\downarrow$$ $$\frac{S_{EFB}}{S_{AEF}} = \frac{S_{ADF}}{S_{AEF}}$$ $$\boxed{S_{EFB} = S_{ADF}}$$
I have absolutely no clue. I see no way of creating a relation between the areas of the triangle and the parallelogram. I thought of trying to somehow prove that the area of BCE is identical to that of BCD or BAD, but couldn't find a way to connect those either, as they don't have a shared perpendicular. |
Here is the URL: https://github.com/avaneev/biteopt
I've tested it on numerous global optimization benchmarking functions (included), and on real-world hyperparameter optimization problems I have. Seems to be working quite well except in comparison to deterministic methods it's necessary to make several attempts at different random seeds, so the iteration budget may be high. But stochasticity of this method gives a chance to solve a problem which can't be sufficiently solved by deterministic methods. Anyway, most benchmarking functions solve in 1 attempt given enough iteration budget.
Works best for non-convex problems, and can also solve convex problems, but of course slower than deterministic methods. Can also solve non-linear constrained problems, but such constraints increase convergence time considerably (though, this application was not tested thoroughly).
Still working on the method.
I would like to hear comments from users that have some practical models (e.g. black-box hyperparameter optimization) which are still needed to be solved acceptably - whether this method works or not for their models, possibly with the description of the model.
Here is the description of the method. The algorithm consists of the following elements:
A cost-ordered population of previous solutions is maintained. A solution is an independent parameter vector which is evolved towards a better solution. On every iteration, the best solution is evolved. $$x_\text{new}=x_\text{best}$$ Below,
iis either equal to rand(1, N) or in the range [1; N], depending on the
AllpProbprobability. Probabilities are defined in the range [0; 1] and in many instances in the code were replaced with simple resetting counters for more efficiency. Parameter values are internally normalized to [0; 1] range and, to stay in this range, are wrapped in a special manner before each function evaluation. Algorithm's hyper-parameters (probabilities) were pre-selected and should not be changed.
Depending on the
RandProbprobability, a single (or all) parameter value randomization is performed using "bitmask inversion" operation. $$mask= 2^{1+\lfloor(0.999999997-rand(0\ldots1)^4 )\cdot MantSize\rfloor}-1$$ $$MantMult=2^{MantSize}$$ $$x_\text{new}[i] = \frac{\lfloor x_\text{new}[i]\cdot MantMult \rfloor \bigotimes mask }{MantMult}$$ Plus, with
CentProbprobability the random "step in the right direction" operation is performed using the centroid vector, twice. $$m_1=\text{rand}(0\ldots1)\cdot CentSpan$$ $$x_\text{new}[i]=x_\text{new}[i]-m_1(x_\text{new}[i]-x_\text{cent}[i])$$ $$m_2=\text{rand}(0\ldots1)\cdot CentSpan$$ $$x_\text{new}[i]=x_\text{new}[i]-m_2(x_\text{new}[i]-x_\text{cent}[i])$$ With
RandProb2probability an alternative randomization method is used. $$x_\text{new}[i]=x_\text{new}[i]+(-1)^{s}(x_\text{cent}[i]-x_\text{new}[i]), \quad i=1,\ldots,N,\\ \quad s\in\{1,2\}=(\text{rand}(0\ldots1)<0.5 ? 1:2)$$
(Not together with N.2) the "step in the right direction" operation is performed using the random previous solution, current best and worst solutions. This is conceptually similar to Differential Evolution's "mutation" operation. $$x_\text{new}=x_\text{best}-\frac{(x_\text{worst}-x_\text{rand})}{2}$$
With
ScutProbprobability a "short-cut" parameter vector change operation is performed. $$z=x_\text{new}[\text{rand}(1,N)]$$ $$x_\text{new}[i]=z, \quad i=1,\ldots,N$$
After each objective function evaluation, the highest-cost previous solution is replaced using the cost constraint.
You can find this algorithm implemented in the
optimize() function in
biteopt.h on lines 284-395, it does not involve any higher-order math. |
Search
Now showing items 1-10 of 33
The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
(Elsevier, 2018-01)
This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
(Elsevier, 2018-06)
The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ... |
LLVM optimizes power sums, such as
int sum(int count) { int result = 0; for (int j = 0; j < count; ++j) result += j*j; return result; }to code that calculates the result without a loop (godbolt) sum(int): test edi, edi jle .LBB0_1 lea eax, [rdi - 1] lea ecx, [rdi - 2] imul rcx, rax lea eax, [rdi - 3] imul rax, rcx shr rax imul eax, eax, 1431655766 add eax, edi shr rcx lea ecx, [rcx + 2*rcx] lea eax, [rax + rcx] add eax, -1 ret .LBB0_1: xor eax, eax retIt handles more complex cases too (godbolt) – that is, the optimization is not just a simple pattern matcher. This post will show how the optimization is done. Loop analysis – scalar evolution
There are many cases where compilers need to track how values are updated within a loop. For example, the loop vectorizer needs to check that the pointers are moved to the adjacent element in the next iteration, and check that no other pointer indexing may alias the range we are vectorizing.
Both GCC and LLVM does this in the same way in their scalar evolution passes, where each variable at iteration \(i\) (we start enumerating iterations from \(0\)) is represented as a function \(f_0(i)\) defined as a linear recurrence of the form
\[f_j(i) = \begin{cases} \phi_j & \text{if $i = 0$} \\ f_j(i-1) \odot_{j+1} f_{j+1}(i-1) & \text{if $i > 0$} \end{cases}\] where \(\odot\in\{+,*\}\). \begin{cases} 0 & \text{if $i = 0$} \\ f(i-1) + 1 & \text{if $i > 0$} \end{cases}\] \begin{cases} 2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\ f_2(i-1) + 6 & \text{if $i > 0$} \end{cases}\\ f_1(i) & = \begin{cases} k-1 & \text{if $i = 0$} \\ f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$} \end{cases}\\ f(i) = f_0(i) & = \begin{cases} 7 & \text{if $i = 0$} \\ f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$} \end{cases}\end{align}\] One optimization we can see directly from these functions is that the value can be calculated by just three additions within the loop \begin{cases} 2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\ f_2(i-1) + 6 & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{2,+,6\}$}\\ f_1(i) & = \begin{cases} k-1 & \text{if $i = 0$} \\ f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{k-1,+,f_2\}$}\\ f(i) = f_0(i) & = \begin{cases} 7 & \text{if $i = 0$} \\ f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{7,+,f_1\}$}\end{align}\] These can be chained, so \(f(i)\) can be written as a \{\phi_0, +, \phi_1\} + \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 + \psi_0, +, \phi_1 + \psi_1\} \\ \{\phi_0, +, \phi_1\}* \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 * \psi_0, +, \psi_1 * \{\phi_0, +, \phi_1\} + \phi_1 * \{\psi_0, +, \psi_1\} + \phi_1*\psi_1\} \\ \{\phi_0, +, \phi_1,+,0\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0, +, \phi_1\}\end{align}\] So for the loop from the & = \{0 * 0, +, 1 * \{0, +,1\} + 1 * \{0, +, 1\} + 1*1\} \\ & = \{0, +, 1,+,2\}\end{align}\] Similar calculations for
\[f_j(i) =
\begin{cases}
\phi_j & \text{if $i = 0$} \\
f_j(i-1) \odot_{j+1} f_{j+1}(i-1) & \text{if $i > 0$}
\end{cases}\] where \(\odot\in\{+,*\}\).
Example 1The simplest case is a loop such as void foo(int m, int *p) { for (int j = 0; j < m; j++) *p++ = j; }The loop writes \(0\) to
*p++ in the first iteration, \(1\) in the second, etc. So we can express the value written at iteration \(i\) as\[f(i) =
\begin{cases}
0 & \text{if $i = 0$} \\
f(i-1) + 1 & \text{if $i > 0$}
\end{cases}\]
Example 2Polynomials in the iteration variable can also be expressed in this form. void foo(int m, int k, int *p) { for (int j = 0; < m; j++) *p++ = j*j*j - 2*j*j + k*j + 7; }We will see below how to build the functions, but the result for the value stored in this loop is \[\begin{align}f_2(i) & =
\begin{cases}
2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\
f_2(i-1) + 6 & \text{if $i > 0$}
\end{cases}\\
f_1(i) & =
\begin{cases}
k-1 & \text{if $i = 0$} \\
f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases}\\
f(i) = f_0(i) & =
\begin{cases}
7 & \text{if $i = 0$} \\
f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases}\end{align}\] One optimization we can see directly from these functions is that the value can be calculated by just three additions within the loop
void foo(int m, int k, int *p) { int t0 = 7; int t1 = k-1; int t2 = 2; for (int j = 0; j < m; j++) { *p++ = t0; t0 = t0 + t1; t1 = t1 + t2; t2 = t2 + 6; } }which is a useful optimization for architectures were multiplication is expensive. This kind of code is, however, uncommon, so most compilers do not do this optimzation – but they usually do this for simpler cases, such as void foo(int m, int k, int *p) { for (int j = 0; < m; j++) *p++ = k*j + 7; }as constructs of the form
k*j+7 are common in address calculations.
Chains of recurrencesIt is cumbersome to write the recursive functions all the time, so the functions are usually written in the form \(\{\phi_j, \odot_{j+1}, f_{j+1}\}\). For example \[\begin{align}f_2(i) & =
\begin{cases}
2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\
f_2(i-1) + 6 & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{2,+,6\}$}\\
f_1(i) & =
\begin{cases}
k-1 & \text{if $i = 0$} \\
f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{k-1,+,f_2\}$}\\
f(i) = f_0(i) & =
\begin{cases}
7 & \text{if $i = 0$} \\
f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{7,+,f_1\}$}\end{align}\] These can be chained, so \(f(i)\) can be written as a
chain of recurrences(CR) \(\{7,+,\{k-1,+,\{2,+,6\}\}\}\). The inner curly braces are redundant, so the CR is usually written as a tuple \(\{7,+,k-1,+,2,+,6\}\). Building the chains of recurrencesThe chains of recurrences are built by iterating over the code and calculating the CR for the result of each operation (or marking it as unknown if we cannot handle the operation), using simplification rules such as\[\begin{align}c * \{\phi_0, +, \phi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{c * \phi_0, +, c * \phi_1\} \\
\{\phi_0, +, \phi_1\} + \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 + \psi_0, +, \phi_1 + \psi_1\} \\
\{\phi_0, +, \phi_1\}* \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 * \psi_0, +, \psi_1 * \{\phi_0, +, \phi_1\} + \phi_1 * \{\psi_0, +, \psi_1\} + \phi_1*\psi_1\} \\
\{\phi_0, +, \phi_1,+,0\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0, +, \phi_1\}\end{align}\] So for the loop from the
sum function
for (int j = 0; j < count; ++j) result += j*j;we start with
j which we know has the CR \(\{0,+,1\}\) per Example 1. This is then used as
j*j when calculating
result, so we calculate the CR for
j*j using the simplification rules as \[\begin{align}j*j& = \{0,+,1\} * \{0,+,1\} \\
& = \{0 * 0, +, 1 * \{0, +,1\} + 1 * \{0, +, 1\} + 1*1\} \\
& = \{0, +, 1,+,2\}\end{align}\] Similar calculations for
result gives us the CR \(\{0,+,0,+,1,+,2\}\) for the value at the beginning of the iteration, and \(\{0,+,1,+,3,+,2\}\) after adding
j*j.
Doing the optimization
The optimization is done during induction variable simplification, and LLVM has transformed the function to a form more convenient for analysis and optimization
One nice property of the chains of recurrences is that it is easy to calculate the value at a specific iteration – if we have a CR \(\{\phi_0,+,\phi_1,+,\ldots,+,\phi_n\}\), then the value at iteration \(i\) can be calculated as \[\begin{align}f(i) & = \sum_{j=0}^{n}\phi_j{i \choose j} \\ & = \phi_0 + \phi_1i + \phi_2{i(i-1)\over 2!} + \ldots + \phi_n{i(i-1)\cdots(i-n+1)\over n!}\end{align}\] Inserting the values for the CR \(\{0,+,1,+,3,+,2\}\) describing One problem with this optimization is that it is hard for developers to make the compiler generate a loop for this if they know that the majority of values used in reality are small enough for the loop to be the fastest option. GCC does, therefore, not replace the final value of a loop if the expression is expensive int sum(int count) { int result = 0; if (count > 0) { int j = 0; do { result = result + j*j; ++j; } while (j < count); } return result; }or as it looks like in the LLVM IR define i32 @sum(i32) { %2 = icmp sgt i32 %0, 0 br i1 %2, label %3, label %6 ; <label>:3: br label %8 ; <label>:4: %5 = phi i32 [ %12, %8 ] br label %6 ; <label>:6: %7 = phi i32 [ 0, %1 ], [ %5, %4 ] ret i32 %7 ; <label>:8: %9 = phi i32 [ %13, %8 ], [ 0, %3 ] ; {0,+,1} %10 = phi i32 [ %12, %8 ], [ 0, %3 ] ; {0,+,0,+,1,+,2} %11 = mul nsw i32 %9, %9 ; {0,+,1,+,2} %12 = add nuw nsw i32 %11, %10 ; {0,+,1,+,3,+,2} %13 = add nuw nsw i32 %9, 1 ; {1,+,1} %14 = icmp slt i32 %13, %0 br i1 %14, label %8, label %4 }The compiler can see that the function returns
0 if
count <= 0, otherwise it returns the value of
result at loop iteration
count-1.
One nice property of the chains of recurrences is that it is easy to calculate the value at a specific iteration – if we have a CR \(\{\phi_0,+,\phi_1,+,\ldots,+,\phi_n\}\), then the value at iteration \(i\) can be calculated as \[\begin{align}f(i) & = \sum_{j=0}^{n}\phi_j{i \choose j} \\ & = \phi_0 + \phi_1i + \phi_2{i(i-1)\over 2!} + \ldots + \phi_n{i(i-1)\cdots(i-n+1)\over n!}\end{align}\] Inserting the values for the CR \(\{0,+,1,+,3,+,2\}\) describing
result gives us \[f(i) = i + {3i(i-1)\over 2} + {i(i-1)(i-2) \over 3}\] The compiler now only need to insert code that calculates this with \(i =\)
count-1, after the loop
result = count-1 + 3*(count-1)*(count-2)/2 + (count-1)*(count-2)(count-3)/3;but it need to take some care to calculate in the correct precision (as temporary values may not fit in 32-bit integers). Integer division is slow, so it is also doing tricks to replace the divisions with multiplication and shift instructions. The result is the LLVM IR %4 = add i32 %0, -1 %5 = zext i32 %4 to i33 %6 = add i32 %0, -2 %7 = zext i32 %6 to i33 %8 = mul i33 %5, %7 %9 = add i32 %0, -3 %10 = zext i32 %9 to i33 %11 = mul i33 %8, %10 %12 = lshr i33 %11, 1 %13 = trunc i33 %12 to i32 %14 = mul i32 %13, 1431655766 %15 = add i32 %14, %0 %16 = lshr i33 %8, 1 %17 = trunc i33 %16 to i32 %18 = mul i32 %17, 3 %19 = add i32 %15, %18 %20 = add i32 %19, -1Inserting this makes the loop become dead, so it is later removed by dead code elimination, and we eventually end up with the code sum(int): test edi, edi jle .LBB0_1 lea eax, [rdi - 1] lea ecx, [rdi - 2] imul rcx, rax lea eax, [rdi - 3] imul rax, rcx shr rax imul eax, eax, 1431655766 add eax, edi shr rcx lea ecx, [rcx + 2*rcx] lea eax, [rax + rcx] add eax, -1 ret .LBB0_1: xor eax, eax ret PerformanceThis optimization is not always profitable. For example, int sum(int count) { int result = 0; for (int j = 0; j < count; ++j) result += j*j*j*j*j*j; return result; }is calculated by three 32-bit multiplications and one addition within the loop, while the optimized version needs six 64-bit multiplications, five 32-bit multiplications, and a slew of other instructions (godbolt), so the optimized version is slower for small values of
count.I benchmarked on my PC, and
count must be larger than
5 for the optimized version to be faster than the loop. Smaller CPUs with, for example, more expensive 64-bit multiplication, will need a higher
count for the optimization to help. And CPUs not having instructions for 64-bit multiplication (godbolt) will need a
muchhigher
count.
One problem with this optimization is that it is hard for developers to make the compiler generate a loop for this if they know that the majority of values used in reality are small enough for the loop to be the fastest option. GCC does, therefore, not replace the final value of a loop if the expression is expensive
/* Do not emit expensive expressions. The rationale is that when someone writes a code like while (n > 45) n -= 45; he probably knows that n is not large, and does not want it to be turned into n %= 45. */ || expression_expensive_p (def))So GCC not doing this optimization is a feature – not a bug. Further readingsChains of recurrences: Olaf Bachmann, Paul S. Wang, Eugene V. Zima. “Chains of recurrences – a method to expedite the evaluation of closed-form functions” Eugene V. Zima. “On computational properties of chains of recurrences” Robert A. van Engelen. “Symbolic Evaluation of Chains of Recurrences for Loop Optimization” Robert A. van Engelen. “Efficient Symbolic Analysis for Optimizing Compilers” Torbjörn Granlund, Peter L. Montgomery. “Division by Invariant Integers using Multiplication” Updated: The original post incorrectly called the power sums “geometric sums”. |
Here @gung makes reference to the .632+ rule. A quick Google search doesn't yield an easy to understand answer as to what this rule means and for what purpose it is used. Would someone please elucidate the .632+ rule?
I will get to the 0.632 estimator, but it'll be a somewhat long development:
Suppose we want to predict $Y$ with $X$ using the function $f$, where $f$ may depend on some parameters that are estimated using the data $(\mathbf{Y}, \mathbf{X})$, e.g. $f(\mathbf{X}) = \mathbf{X}\mathbf{\beta}$
A naïve estimate of prediction error is $$\overline{err} = \dfrac{1}{N}\sum_{i=1}^N L(y_i,f(x_i))$$ where $L$ is some loss function, e.g. squared error loss. This is often called training error. Efron et al. calls it apparent error rate or resubstitution rate. It's not very good since we use our data $(x_i,y_i)$ to fit $f$. This results in $\overline{err}$ being downward biased. You want to know how well your model $f$ does in predicting new values.
Often we use cross-validation as a simple way to estimate the expected extra-sample prediction error (how well does our model do on data not in our training set?). $$Err = \text{E}\left[ L(Y, f(X))\right]$$
A popular way to do this is to do $K$-fold cross-validation. Split your data into $K$ groups (e.g. 10). For each group $k$, fit your model on the remaining $K-1$ groups and test it on the $k$th group. Our cross-validated extra-sample prediction error is just the average $$Err_{CV} = \dfrac{1}{N}\sum_{i=1}^N L(y_i, f_{-\kappa(i)}(x_i))$$ where $\kappa$ is some index function that indicates the partition to which observation $i$ is allocated and $f_{-\kappa(i)}(x_i)$ is the predicted value of $x_i$ using data not in the $\kappa(i)$th set.
This estimator is approximately unbiased for the true prediction error when $K=N$ and has larger variance and is more computationally expensive for larger $K$. So once again we see the bias–variance trade-off at play.
Instead of cross-validation we could use the bootstrap to estimate the extra-sample prediction error. Bootstrap resampling can be used to estimate the sampling distribution of any statistic. If our training data is $\mathbf{X} = (x_1,\ldots,x_N)$, then we can think of taking $B$ bootstrap samples (with replacement) from this set $\mathbf{Z}_1,\ldots,\mathbf{Z}_B$ where each $\mathbf{Z}_i$ is a set of $N$ samples. Now we can use our bootstrap samples to estimate extra-sample prediction error: $$Err_{boot} = \dfrac{1}{B}\sum_{b=1}^B\dfrac{1}{N}\sum_{i=1}^N L(y_i, f_b(x_i))$$ where $f_b(x_i)$ is the predicted value at $x_i$ from the model fit to the $b$th bootstrap dataset. Unfortunately, this is not a particularly good estimator because bootstrap samples used to produce $f_b(x_i)$ may have contained $x_i$. The leave-one-out bootstrap estimator offers an improvement by mimicking cross-validation and is defined as: $$Err_{boot(1)} = \dfrac{1}{N}\sum_{i=1}^N\dfrac{1}{|C^{-i}|}\sum_{b\in C^{-i}}L(y_i,f_b(x_i))$$ where $C^{-i}$ is the set of indices for the bootstrap samples that do not contain observation $i$, and $|C^{-i}|$ is the number of such samples. $Err_{boot(1)}$ solves the overfitting problem, but is still biased (this one is upward biased). The bias is due to non-distinct observations in the bootstrap samples that result from sampling with replacement. The average number of distinct observations in each sample is about $0.632N$ (see this answer for an explanation of why Why on average does each bootstrap sample contain roughly two thirds of observations?). To solve the bias problem, Efron and Tibshirani proposed the 0.632 estimator: $$ Err_{.632} = 0.368\overline{err} + 0.632Err_{boot(1)}$$ where $$\overline{err} = \dfrac{1}{N}\sum_{i=1}^N L(y_i,f(x_i))$$ is the naïve estimate of prediction error often called training error. The idea is to average a downward biased estimate and an upward biased estimate.
However, if we have a highly overfit prediction function (i.e. $\overline{err}=0$) then even the .632 estimator will be downward biased. The .632+ estimator is designed to be a less-biased compromise between $\overline{err}$ and $Err_{boot(1)}$. $$ Err_{.632+} = (1 - w) \overline{err} + w Err_{boot(1)} $$ with $$w = \dfrac{0.632}{1 - 0.368R} \quad\text{and}\quad R = \dfrac{Err_{boot(1)} - \overline{err}}{\gamma - \overline{err}} $$ where $\gamma$ is the no-information error rate, estimated by evaluating the prediction model on all possible combinations of targets $y_i$ and predictors $x_i$.
$$\gamma = \dfrac{1}{N^2}\sum_{i=1}^N\sum_{j=1}^N L(y_i, f(x_j))$$.
Here $R$ measures the relative overfitting rate. If there is no overfitting (R=0, when the $Err_{boot(1)} = \overline{err}$) this is equal to the .632 estimator.
You will find more information in section 3 of this
1 paper. But to summarize, if you call $S$ a sample of $n$ numbers from $\{1:n\}$ drawn randomly and with replacement, $S$ contains on average approximately $(1-e^{-1})\,n \approx 0.63212056\, n$ unique elements.
The reasoning is as follows. We populate $S=\{s_1,\ldots,s_n\}$ by sampling $i=1,\ldots,n$ times (randomly and with replacement) from $\{1:n\}$. Consider a particular index $m\in\{1:n\}$.
Then:
$$P(s_i=m)=1/n$$
and
$$P(s_i\neq m)=1-1/n$$
and this is true $\forall 1\leq i \leq n$ (intuitively, since we sample with replacement, the probabilities do not depend on $i$)
thus
$$P(m\in S)=1-P(m\notin S)=1-P(\cap_{i=1}^n s_i\neq m)\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=1-\prod_{i=1}^n P(s_i\neq m)=1-(1-1/n)^n\approx 1-e^{-1}$$
You can also carry this little simulation to check empirically the quality of the approximation (which depends on $n$):
n <- 100fx01 <- function(ll,n){ a1 <- sample(1:n, n, replace=TRUE) length(unique(a1))/n}b1 <- c(lapply(1:1000,fx01,n=100), recursive=TRUE)mean(b1)
1. Bradley Efron and Robert Tibshirani (1997). Improvements on Cross-Validation: The .632+ Bootstrap Method.
Journal of the American Statistical Association, Vol. 92, No. 438, pp. 548--560.
In my experience, primarily based on simulations, the 0.632 and 0.632+ bootstrap variants were needed only because of severe problems caused by the use of an improper accuracy scoring rule, namely the proportion "classified" correctly. When you use proper (e.g., deviance-based or Brier score) or semi-proper (e.g., $c$-index = AUROC) scoring rules, the standard Efron-Gong optimism bootstrap works just fine.
Those answers are very useful. I couldn't find a way to demonstrate it with maths so I wrote some Python code which works quite well though:
from numpy import mean from numpy.random import choice N = 3000 variables = range(N) num_loop = 1000 # Proportion of remaining variables p_var = [] for i in range(num_loop): set_var = set(choice(variables, N)) p=len(set_var)/float(N) if i%50==0: print "value for ", i, " iteration ", "p = ",p p_var.append(p) print "Estimator of the proportion of remaining variables, ", mean(p_var) |
Self-Assembly Of Dna Graphs And Postman Tours, 2018 University of North Florida
Self-Assembly Of Dna Graphs And Postman Tours, Katie Bakewell UNF Graduate Theses and Dissertations
DNA graph structures can self-assemble from branched junction molecules to yield solutions to computational problems. Self-assembly of graphs have previously been shown to give polynomial time solutions to hard computational problems such as 3-SAT and k-colorability problems. Jonoska et al. have proposed studying self-assembly of graphs topologically, considering the boundary components of their thickened graphs, which allows for reading the solutions to computational problems through reporter strands. We discuss weighting algorithms and consider applications of self-assembly of graphs and the boundary components of their thickened graphs to problems involving minimal weight Eulerian walks such as the Chinese Postman Problem and ...
Zero Forcing And Maximum Nullity For Hypergraphs, 2018 Iowa State University and American Institute of Mathematics
Zero Forcing And Maximum Nullity For Hypergraphs, Leslie Hogben Mathematics Publications
The concept of zero forcing is extended from graphs to uniform hypergraphs in analogy with the way zero forcing was defined as an upper bound for the maximum nullity of the family of symmetric matrices whose nonzero pattern of entries is described by a given graph: A family of symmetric hypermatrices is associated with a uniform hypergraph and zeros are forced in a null vector. The value of the hypergraph zero forcing number and maximum nullity are determined for various families of uniform hypergraphs and the effects of several graph operations on the hypergraph zero forcing number are determined. The ...
Graphs That Are Cospectral For The Distance Laplacian, 2018 Rice University
Graphs That Are Cospectral For The Distance Laplacian, Boris Brimkov, Ken Duna, Leslie Hogben, Kate Lorenzen, Carolyn Reinhart, Sung-Yell Song, Mark Yarrow Mathematics Publications
The distance matrix D(G) of a graph G is the matrix containing the pairwise distances between vertices, and the distance Laplacian matrix is D
L(G)=T(G)−D(G), where T(G) is the diagonal matrix of row sums of D(G). We establish several general methods for producing DL-cospectral graphs that can be used to construct infinite families. We provide examples showing that various properties are not preserved by D L-cospectrality, including examples of D L-cospectral strongly regular and circulant graphs. We establish that the absolute values of coefficients of the distance Laplacian characteristic polynomial ...
Rigid Linkages And Partial Zero Forcing, 2018 Texas State University - San Marcos
Rigid Linkages And Partial Zero Forcing, Daniela Ferrero, Mary Flagg, H. Tracy Hall, Leslie Hogben, Jephian C.-H. Lin, Seth A. Meyer, Shahla Nasserasr, Bryan Shader Mathematics Publications
Connections between vital linkages and zero forcing are established. Specifically, the notion of a rigid linkage is introduced as a special kind of unique linkage and it is shown that spanning forcing paths of a zero forcing process form a spanning rigid linkage and thus a vital linkage. A related generalization of zero forcing that produces a rigid linkage via a coloring process is developed. One of the motivations for introducing zero forcing is to provide an upper bound on the maximum multiplicity of an eigenvalue among the real symmetric matrices described by a graph. Rigid linkages and a related ...
Spn Graphs And Rank-1 Cp-Completable Graphs, 2018 Iowa State University
Spn Graphs And Rank-1 Cp-Completable Graphs, Leslie Hogben, Naomi Shaked-Monderer Mathematics Publications
A simple graph G is an SPN graph if every copositive matrix having graph G is the sum of a positive semidefinite and nonnegative matrix. SPN graphs were introduced in [Shaked-Monderer, SPN graphs: When copositive = SPN,
Linear Algebra Appl., 509(15):82-113, 2016], where it was conjectured that the complete subdivision graph of K4 is an SPN graph. We disprove this conjecture, which in conjunction with results in the Shaked-Monderer paper show that a subdivision of K4 is a SPN graph if and only if at most one edge is subdivided. We conjecture that a graph is an SPN graph ...
Zero Forcing And Power Domination For Graph Products, 2018 Westminster College - Fulton
Zero Forcing And Power Domination For Graph Products, Katherine F. Benson, Daniela Ferrero, Mary Flagg, Veronika Furst, Leslie Hogben, Violeta Vasilevska, Brian Wissman Mathematics Publications
The power domination number arose from the monitoring of electrical networks, and methods for its determination have the associated application. The zero forcing number arose in the study of maximum nullity among symmetric matrices described by a graph (and also in control of quantum systems and in graph search algorithms). There has been considerable effort devoted to the determination of the power domination number, the zero forcing number, and maximum nullity for specific families of graphs. In this paper we exploit the natural relationship between power domination and zero forcing to obtain results for the power domination number of tensor ...
The Relationship Between K-Forcing And K-Power Domination, 2018 Texas State University
The Relationship Between K-Forcing And K-Power Domination, Daniela Ferrero, Leslie Hogben, Franklin H.J. Kenter, Michael Young Mathematics Publications
Zero forcing and power domination are iterative processes on graphs where an initial set of vertices are observed, and additional vertices become observed based on some rules. In both cases, the goal is to eventually observe the entire graph using the fewest number of initial vertices. The concept of k-power domination was introduced by Chang et al. (2012) as a generalization of power domination and standard graph domination. Independently, k-forcing was defined by Amos et al. (2015) to generalize zero forcing. In this paper, we combine the study of k-forcing and k-power domination, providing a new approach to analyze both ...
Applications Of Analysis To The Determination Of The Minimum Number Of Distinct Eigenvalues Of A Graph, 2018 Iowa State University
Applications Of Analysis To The Determination Of The Minimum Number Of Distinct Eigenvalues Of A Graph, Beth Bjorkman, Leslie Hogben, Scarlitte Ponce, Carolyn Reinhart, Theodore Tranel Mathematics Publications
We establish new bounds on the minimum number of distinct eigenvalues among real symmetric matrices with nonzero off-diagonal pattern described by the edges of a graph and apply these to determine the minimum number of distinct eigenvalues of several families of graphs and small graphs.
On The Strong Chromatic Index Of Sparse Graphs, 2018 Westfield State University
On The Strong Chromatic Index Of Sparse Graphs, Philip Deorsey, Michael Ferrara, Nathan Graber, Stephen G. Hartke, Luke L. Nelsen, Eric Sullivan, Sogol Jahanbekam, Bernard Lidicky, Derrick Stolee, Jennifer White Mathematics Publications
The strong chromatic index of a graph G, denoted χ′s(G), is the least number of colors needed to edge-color G so that edges at distance at most two receive distinct colors. The strong list chromatic index, denoted χ′s,ℓ(G), is the least integer k such that if arbitrary lists of size k are assigned to each edge then G can be edge-colored from those lists where edges at distance at most two receive distinct colors.
We use the discharging method, the Combinatorial Nullstellensatz, and computation to show that if G is a subcubic planar graph with ...
3-Maps And Their Generalizations, 2018 Virginia Commonwealth University
3-Maps And Their Generalizations, Kevin J. Mccall Theses and Dissertations
A 3-map is a 3-region colorable map. They have been studied by Craft and White in their paper 3-maps. This thesis introduces topological graph theory and then investigates 3-maps in detail, including examples, special types of 3-maps, the use of 3-maps to find the genus of special graphs, and a generalization known as n-maps.
On Saturation Numbers Of Ramsey-Minimal Graphs, 2018 University of Central Florida
On Saturation Numbers Of Ramsey-Minimal Graphs, Hunter M. Davenport Honors Undergraduate Theses
Dating back to the 1930's, Ramsey theory still intrigues many who study combinatorics. Roughly put, it makes the profound assertion that complete disorder is impossible. One view of this problem is in edge-colorings of complete graphs. For forbidden graphs
H1,...,Hk and a graph G, we write G "arrows" (H 1,...,Hk) if every k-edge-coloring of Gcontains a monochromatic copy of Hin color i ifor some i=1,2,...,k.If cis a (red, blue)-edge-coloring of G, we say cis a bad coloring if Gcontains no red Kor ... 3
Sparse Trees With A Given Degree Sequence, 2018 Georgia Southern University
Sparse Trees With A Given Degree Sequence, Ao Shen Electronic Theses and Dissertations
In this thesis, we consider the properties of sparse trees and summarized a certain class of trees under some constraint (including with a given degree sequence, with given number of leaves, with given maximum degree, etc.) which have maximum Wiener index and the minimum number of subtrees at the same time. Wiener index is one of the most important topological indices in chemical graph theory. Steiner k Wiener index can be regarded as the generalization of Wiener index, when k = 2, Steiner Wiener index is the same as Wiener index. Steiner k Wiener index of a tree T is the ...
Generalized Matching Preclusion In Bipartite Graphs, 2018 Oakland University
Generalized Matching Preclusion In Bipartite Graphs, Zachary Wheeler, Eddie Cheng, Dana Ferranti, Laszlo Liptak, Karthik Nataraj Theory and Applications of Graphs
The matching preclusion number of a graph with an even number of vertices is the minimum number of edges whose deletion results in a graph that has no perfect matchings. For many interconnection networks, the optimal such sets are precisely sets of edges incident to a single vertex. The conditional matching preclusion number of a graph was introduced to look for obstruction sets beyond these, and it is defined as the minimum number of edges whose deletion results in a graph with neither isolated vertices nor perfect matchings. In this paper we generalize this concept to get a hierarchy of ...
Lights Out! On Cartesian Products, 2017 Iowa State University
Lights Out! On Cartesian Products, Travis Peters, John Goldwasser, Michael Young Electronic Journal of Linear Algebra
The game LIGHTS OUT! is played on a 5 by 5 square grid of buttons; each button may be on or off. Pressing a button changes the on/o state of the light of the button pressed and of all its vertical and horizontal neighbors. Given an initial configuration of buttons that are on, the object of the game is to turn all the lights out. The game can be generalized to arbitrary graphs. In this paper, Cartesian Product graphs (that is, graphs of the form G\box H, where G and H are arbitrary finite, simple graphs) are investigated ...
Upper Bounds On Q-Spectral Radius Of Book-Free And/Or $K_{S,T}$-Free Graphs, 2017 Northwestern Polytechnical University
Upper Bounds On Q-Spectral Radius Of Book-Free And/Or $K_{S,T}$-Free Graphs, Qi Kong, Ligong Wang Electronic Journal of Linear Algebra
In this paper, we prove two results about the signless Laplacian spectral radius $q(G)$ of a graph $G$ of order $n$ with maximum degree $\Delta$. Let $B_{n}=K_{2}+\overline{K_{n}}$ denote a book, i.e., the graph $B_{n}$ consists of $n$ triangles sharing an edge. The results are the following: (1) Let $1< k\leq l< \Delta < n$ and $G$ be a connected \{$B_{k+1},K_{2,l+1}$\}-free graph of order $n$ with maximum degree $\Delta$. Then $$\displaystyle q(G)\leq \frac{1}{4}[3\Delta+k-2l+1+\sqrt{(3\Delta+k-2l+1)^{2}+16l(\Delta+n-1)}$$ with equality if and only if $G$ is a strongly regular graph with parameters ($\Delta$, $k$, $l$). (2) Let $s\geq t\geq 3$, and let $G$ be a connected $K_{s,t}$-free graph of order $n$ $(n\geq s+t)$. Then $$q(G)\leq n+(s-t+1)^{1/t}n^{1-1/t}+(t-1)(n-1)^{1-3/t}+t-3.$$
On The Distance Signless Laplacian Spectral Radius Of Graphs And Digraphs, 2017 Xinjiang University,Urumqi
On The Distance Signless Laplacian Spectral Radius Of Graphs And Digraphs, Dan Li, Guoping Wang, Jixiang Meng Electronic Journal of Linear Algebra
Let \eta(G) denote the distance signless Laplacian spectral radius of a connected graph G. In this paper,bounds for the distance signless Laplacian spectral radius of connected graphs are given, and the extremal graph with the minimal distance signless Laplacian spectral radius among the graphs with given vertex connectivity and minimum degree is determined. Furthermore, the digraph that minimizes the distance signless Laplacian spectral radius with given vertex connectivity is characterized.
Characterizations Of Some Classes Of Graphs That Are Nearly Series-Parallel, 2017 Louisiana State University and Agricultural and Mechanical College
Characterizations Of Some Classes Of Graphs That Are Nearly Series-Parallel, Victoria Fontaine LSU Doctoral Dissertations
A series-parallel graph can be built from a single-edge graph by a sequence of series and parallel extensions. The class of such graphs coincides with the class of graphs that do not have the complete graph K
4 as a minor. This dissertation considers a class M 1 of graphs that are close to being series-parallel. In particular, every member of the class has the property that one can obtain a series-parallel graph by adding a new edge and contracting it out, or by splitting a vertex into two vertices whose neighbor sets partition the neighbor set of the original ...
An Exploration Of The Chromatic Polynomial, 2017 Boise State University
An Exploration Of The Chromatic Polynomial, Amanda Aydelotte Mathematics Undergraduate Theses
In 1912, George Birkhoff was studying the Four Color Problem, and in doing so introduced the concept of the chromatic polynomial. While this did not end up directly contributing to proving that every map could be colored with four colors such that no region shares a border with another region of the same color, the chromatic polynomial has been found to have some very interesting properties. In this paper, it will be our goal to examine some of these properties and use them to determine information about their corresponding graphs.
Optimal Layout For A Component Grid, 2017 California Polytechnic State University, San Luis Obispo
Optimal Layout For A Component Grid, Michael W. Ebert Computer Science and Software Engineering
Several puzzle games include a specific type of optimization problem: given components that produce and consume different resources and a grid of squares, find the optimal way to place the components to maximize output. I developed a method to evaluate potential solutions quickly and automated the solving of the problem using a genetic algorithm.
Classification Of Minimal Separating Sets In Low Genus Surfaces, 2017 Portland State University
Classification Of Minimal Separating Sets In Low Genus Surfaces, J. J. P. Veerman, William Maxwell, Victor Rielly, Austin K. Williams Mathematics and Statistics Faculty Publications and Presentations
Consider a surface
S and let M ⊂ S. If S \ M is not connected, then we say M separates S, and we refer to M as a separating set of S. If M separates S, and no proper subset of M separates S, then we say M is a minimal separating set of S. In this paper we use computational methods of combinatorial topology to classify the minimal separating sets of the orientable surfaces of genus g = 2 and g = 3. The classification for genus 0 and 1 was done in earlier work, using methods of algebraic topology. |
To form a product, you give me $n$ objects, $A_1,\dots,A_n$, and I give you back an object $A_1\times\dots\times A_n$, together with $n$ maps $\pi_i\colon A_1\times\dots\times A_n\to A_i$ (one to each of the $A_i$) satisfying the universal property of the product.
So what happens if $n=0$? Then you give me $0$ objects, and I give you back an object which we call $1$, together with $0$ maps $\pi_i$ (one to each of the $A_i$, of which there aren't any), satisfying the universal property of the product.
What does the universal property say in this case?
For any $X$ given together with $0$ maps $f_i$ (one to each of the $A_i$, of which there aren't any), there is a unique map $!\colon X\to 1$ making all of the triangles commute ($\pi_i\circ ! = f_i$ for all $i$, of which there aren't any).
Removing the vacuous conditions from the definition, we see that the empty product is an object $1$ such that for every object $X$ there is a unique map $!\colon X\to 1$, i.e. $1$ is a terminal object. |
There are tons of examples, but I'll give a few from the context of PDEs.
@daw already gave a very good example. Let me provide a similar one, in the form of the
heat propagator. You may know that the study of the heat equation is fundamental in many fields of PDEs. To that end, let $M$ be a compact, Riemannian manifold without boundary with metric $g$. If $p$ denotes the fundamental solution to the heat equation on $M\times (0,\infty),$ then for $t>0,$ we define the heat propagator as the map $e^{-t\Delta_g}:L^2(M)\rightarrow L^2(M)$ via $$e^{-t\Delta_g} f(x)=\int\limits_M p(x,y,t) f(y)\, \omega_g(y),$$ where $\omega_g$ is the volume form. It's not too hard to show that $e^{-t\Delta_g}$ is a self-adjoint, compact operator on $L^2$.
Here is another example, which is of significant importance in the realm of semi-classical analysis. Note: I'm going to use $\mathcal{S}$ to denote the Schwartz space, and I'm going to use it for both $\mathcal{S}(\mathbb{R}^n)$ and $\mathcal{S}(\mathbb{R}^{2n})$. Symbols will always be in the latter, and everything else in the former.
Let $a\in\mathcal{S},$ and $h>0.$ We define the
Weyl quantization to be the operator $a^w(x,hD)$ on $\mathcal{S}$ so that $$a^w(x,hD)u(x)=\frac{1}{(2\pi h)^n}\int\limits_{\mathbb{R}^{n}}\int\limits_{\mathbb{R}^{n}}e^{\frac{i}{h}(x-y)\cdot\xi} a\left(\frac{x+y}{2},\xi\right)u(y)\, dyd\xi.$$ This is related to applying the conjugation of $a$ by the semi-classical Fourier transform to $u$, but the convex combination of $x$ and $y$ stops this from being quite a semi-classical Fourier multiplier (unless $a$ is independent of $x$). This is called a pseudifferential operator (which can also be defined on manifolds, but I'd rather not make that digression). This operator can be defined as a map $a^w(x,hD):\mathcal{S}'\rightarrow\mathcal{S},$ and it is formally self-adjoint here, provided $a$ is real-valued. In fact, in a proof similar to proving that the Hilbert-Schmidt kernel operator is bounded in $L^2$, you get that the Weyl quantization sends $L^2\rightarrow L^2$, and it's bounded. In a proof similar to that of the Arzela-Ascoli theorem, one can show this is compact, as well.
You also asked about applications of Hilbert-Schmidt operators. First, here's a classical result:
If $T:L^2(X_1,\mu_1)\rightarrow L^2(X_2,\mu_2)$ is HS, then there exists $K\in L^2(X_1\times X_2,\mu_1\times \mu_2)$ such that $$(Tu,v)_{L^2}=\iint K(x_1,x_2)u(x_1)\overline{v(x_2)}\, d\mu_1(x_1)d\mu_2(x_2).$$ Note that $$\|T\|_{\text{HS}}=\|K\|_{L^2}.$$ Conversely, if $K\in L^2(X_1\times X_2,\mu_1\times \mu_2)$, then the above integral defines a Hilbert-Schmidt operator $T$, with the same norm equality.
If a bounded operator on a separable Hilbert space can be written as the product of two HS operators, then we say that this operator is
trace-class. One characterization of such operators is that a bounded operator $C$ is trace-class if and only if $C$ is compact and $(C^*C)^{1/2}$ has summable eigenvalues. The norm on trace-class operators is given by the infimum over all decompositions of the operator (in the way described by the definition of trace-class) of the product of their respective Hilbert-Schmidt norms. This forms the space into a Banach space. A nice thing about trace-class operators is that we can define their determinants in an appropriate sense (although I won't go through this construction). Trace-class operators get their name because we can take their trace, in the sense that if $C$ is trace-class, then we can define its trace as $$\text{Tr}\ C=\sum c_{jj},$$ where $c_{jj}=(Cu_j,u_j)$, and $\{u_j\}$ is the orthonormal basis for the given Hilbert space. Alternatively, if we write $C=AB^*$, then $$\text{Tr}\ C=(A,B)_{\text{HS}}=\sum_{j,k} a_{jk}\overline{b}_{jk},$$ where $a_{jk}=(Au_j,u_k).$ We can talk about the trace of an integral operator as follows: Let $A,B$ be Hilbert-Schmidt on $L^2(X,\mu)$ with integral kernels $K_A,K_B\in L^2(X\times X,\mu\times \mu).$ If $C=AB$, then we can write $$Cu(x)=\iint K_A(x,z)K_B(z,y) u(y)\, d\mu(y)d\mu(z),$$ so we get the trace as $$\text{Tr}\ C=\iint K_A(x,z)K_B(z,y) \, d\mu(y)d\mu(z).$$ The integral kernel of $C$ is given by $$K_C(x,y)=\int K_A (x,z)K_B(z,y)\, d\mu(z).$$ So, we might want the trace to be $$\text{Tr}\ C=\int K_C(x,x)\, d\mu(x).$$ The problem is that the diagonal might have measure zero, so this is a bit of a technical point. Essentially, we can define $K_C(x,x)$ by restriction a.e., and for nice spaces (those commonly looked at), we can get from the first expression of the trace to the second by continuity.
One more thing related to Hilbert-Schmidt operators. Earlier, I mentioned something that is sometimes referred to as the Hilbert-Schmidt kernel theorem. There is a similar result for distributions which is utilized constantly in microlocal analysis and the study of pseduodifferential operators, called the
Schwartz kernel theorem:
Let $A:\mathcal{S}\rightarrow\mathcal{S}'$ be a continuous linear map. Then, there exists $K_A\in\mathcal{S}'(\mathbb{R}^n\times\mathbb{R}^n)$ so that for all $u,v\in\mathcal{S},$ $$\langle Au,v\rangle=\langle K_A, u\otimes v\rangle,$$ where $(u\otimes v)(x,y)=u(x)v(y)\in \mathcal{S}(\mathbb{R}^n\times\mathbb{R}^n).$
We sometimes abuse notation and write this as $Au(x)=\int K_A(x,y) u(y)\, dy,$ so that $$\langle Au,v\rangle=\iint K_A(x,y) v(x)u(y)\, dydx.$$ Seriously, this is used all the time.
References-
Michael Taylor: Partial Differential Equations I
Maciej Zworski: Semiclassical Analysis |
I have found a way to enlarged the Erdos-Szekeres Conjecture. I have computationally confirmed what might be the simplest unknown case for some small values, and was reaching out for how to prove this case in general.
Definitions Given a set of points in the general position on the plane,a subset is a "convex polygon" if each point isa vertex of the subset's convex hull.A cap is a sequence of points with increasing $x$-coordinatesuch that the slope of consecutive pairs is decreasing. A cupis defined similarly but the slopes are increasing.A $n$-polygon (size $n$ convex polygon) can be decomposed into a$k$-cap and an $k'$-cup where $k' = n + 2 - k$.We refer to this as a $(k, k')$-polygon.
For example in the image below, the yellow and red points form a $4$-cap, the yellow and green form $6$-cup, collectively the red, yellow, and green points form both a $(4, 6)$-polygon and a $8$-polygon.
Erdos-Szekeres ConjectureThe tooling in the next section are used to generalize the followingfamous conjecture. If a set contains $2^{n-2}+1$ pointsin the general position on the plane,does it contain an $n$-polygon? Polygon Forcing TheoremsDefine $C(X; k, k')$ to be true if and onlythe set of points $X$ contain a $(k, k')$-polygon. We define a forced polygon theorem is one of the form$$|X| > N \implies \bigcup_i C(X; k_i, k_i')$$We now show some results and problems thatare forced polygon theorems.If the Erdos-Szekeres Conjecture is correct then$$|X| > 2^{n-2} \implies \bigcup_{i = 2}^{n} C(X; i, n+2-i)$$Another example, Erdos and Szekeres show that$$|X| > \binom{k + k' - 4}{k - 2} \implies C(X; 2, k') \cup C(X; k, 2)$$which they use to provide an upper bound to their problem.See "A Combinatorial Problem in Geometry" by Erdos and Szekeres. And in "Ramsey-remainder", by Erdos, Tuza, and Valtr showed that the Erdos-Szekeres conjecture is equivalent to$$|X| > \sum_{i = n - k'}^{k-2} \binom{n - 2}{i}\implies C(X; k, 2) \cup C(X; 2, k') \cup \bigcup_{i = n+2 - (k'-1)}^{k-1} C(X; i, n+2-i)$$And when $k, k' = n$ they are equal. My ProblemI have found a computational technique for creating polygon forcing theorems. Using it I have observed for all $k$ and $k'$ in cases when the program halts:
$$|X| > \frac{(k-1) (k'-1)}{2} \implies C(X; 2, k') \cup C(X; 3, 3) \cup C(X; k, 2)$$
I suspect there is a labeling argument similar to another related problem but I haven't be able to find it and was wondering if I could have help. Thank you for reading this post, as it is quite long.
Edit
Here are all the cases I can compute, in case someone can come up with a comprehensive conjecture. Note that these are upper bounds: If
21 -> [(2, 7), (3, 3), (8, 2)] it is still possible that
20 -> [(2, 7), (3, 3), (8, 2)]. I am curious if there are better bounds for any of these cases.
2 -> [(2, 3), (3, 2)]3 -> [(2, 3), (4, 2)]4 -> [(2, 3), (5, 2)]5 -> [(2, 3), (6, 2)]3 -> [(2, 4), (3, 2)]6 -> [(2, 4), (4, 2)]10 -> [(2, 4), (5, 2)]15 -> [(2, 4), (6, 2)]4 -> [(2, 5), (3, 2)]10 -> [(2, 5), (4, 2)]20 -> [(2, 5), (5, 2)]5 -> [(2, 6), (3, 2)]15 -> [(2, 6), (4, 2)]4 -> [(2, 4), (3, 3), (4, 2)]6 -> [(2, 4), (3, 3), (5, 2)]7 -> [(2, 4), (3, 3), (6, 2)]7 -> [(2, 4), (4, 3), (5, 2)]9 -> [(2, 4), (4, 3), (6, 2)]11 -> [(2, 4), (5, 3), (6, 2)]6 -> [(2, 5), (3, 3), (4, 2)]8 -> [(2, 5), (3, 3), (5, 2)]10 -> [(2, 5), (3, 3), (6, 2)]7 -> [(2, 5), (3, 4), (4, 2)]12 -> [(2, 5), (3, 4), (5, 2)]17 -> [(2, 5), (3, 4), (6, 2)]12 -> [(2, 5), (4, 3), (5, 2)]14 -> [(2, 5), (4, 3), (6, 2)]14 -> [(2, 5), (4, 4), (5, 2)]7 -> [(2, 6), (3, 3), (4, 2)]10 -> [(2, 6), (3, 3), (5, 2)]12 -> [(2, 6), (3, 3), (6, 2)]9 -> [(2, 6), (3, 4), (4, 2)]14 -> [(2, 6), (3, 4), (5, 2)]11 -> [(2, 6), (3, 5), (4, 2)]15 -> [(2, 7), (3, 3), (6, 2)]21 -> [(2, 7), (3, 3), (8, 2)]8 -> [(2, 5), (3, 4), (4, 3), (5, 2)]11 -> [(2, 5), (3, 4), (4, 3), (6, 2)]13 -> [(2, 5), (3, 4), (5, 3), (6, 2)]11 -> [(2, 6), (3, 4), (4, 3), (5, 2)]14 -> [(2, 6), (3, 4), (4, 3), (6, 2)]13 -> [(2, 6), (3, 5), (4, 3), (5, 2)]
Explaining the notation
12 -> [(2, 6), (3, 3), (6, 2)] means$$|X| > 12 \implies C(X;2,6) \cup C(X;3,3) \cup C(X;6,2)$$ |
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1. Measurement of the ZZ production cross section in proton-proton collisions at √s = 8 TeV using the ZZ → ℓ−ℓ+ℓ′−ℓ′+ and ZZ→ℓ−ℓ+νν¯¯¯ decay channels with the ATLAS detector
Journal of High Energy Physics, ISSN 1126-6708, 2017, Volume 2017, Issue 1, pp. 1 - 53
A measurement of the ZZ production cross section in the ℓ−ℓ+ℓ′ −ℓ′ + and ℓ−ℓ+νν¯ channels (ℓ = e, μ) in proton-proton collisions at s=8TeV at the Large Hadron...
Hadron-Hadron scattering (experiments) | Fysik | Physical Sciences | Subatomic Physics | Naturvetenskap | Subatomär fysik | Natural Sciences
Hadron-Hadron scattering (experiments) | Fysik | Physical Sciences | Subatomic Physics | Naturvetenskap | Subatomär fysik | Natural Sciences
Journal Article
2. Search for an additional, heavy Higgs boson in the H→ZZ decay channel at √s = 8 TeV in pp collision data with the ATLAS detector
European Physical Journal C, ISSN 1434-6044, 2016, Volume 76, Issue 1, pp. 1 - 42
Journal Article
3. Combination of searches for WW, WZ, and ZZ resonances in pp collisions at s=8 TeV with the ATLAS detector
Physics Letters B, ISSN 0370-2693, 04/2016, Volume 755, pp. 285 - 305
The ATLAS experiment at the CERN Large Hadron Collider has performed searches for new, heavy bosons decaying to , and final states in multiple decay channels...
PHYSICS, NUCLEAR | ASTRONOMY & ASTROPHYSICS | PHYSICS, PARTICLES & FIELDS | Analysis | Detectors | Fysik | Physical Sciences | Naturvetenskap | Natural Sciences
PHYSICS, NUCLEAR | ASTRONOMY & ASTROPHYSICS | PHYSICS, PARTICLES & FIELDS | Analysis | Detectors | Fysik | Physical Sciences | Naturvetenskap | Natural Sciences
Journal Article
4. Measurement of the Higgs boson mass in the H → ZZ⁎ → 4ℓ and H → γγ channels with s=13 TeV pp collisions using the ATLAS detector
Physics Letters B, ISSN 0370-2693, 09/2018, Volume 784, Issue C, pp. 345 - 366
Journal Article
5. Measurement of ZZ production in pp collisions at \sqrt{s}=7 TeV and limits on anomalous ZZZ and ZZγ couplings with the ATLAS detector
Journal of High Energy Physics, ISSN 1126-6708, 2013, Volume 2013, Issue 3, pp. 1 - 48
A measurement of the ZZ production cross section in proton-proton collisions at $ \sqrt{s}=7 $ TeV using data recorded by the ATLAS experiment at the Large...
Hadron-Hadron Scattering | Quantum Physics | Quantum Field Theories, String Theory | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory
Hadron-Hadron Scattering | Quantum Physics | Quantum Field Theories, String Theory | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory
Journal Article
6. Measurement of the ZZ production cross section in proton-proton collisions at s = 8 $$ \sqrt{s}=8 $$ TeV using the ZZ → ℓ−ℓ+ℓ′−ℓ′+ and Z Z → ℓ − ℓ + ν ν ¯ $$ ZZ\to {\ell}^{-}{\ell}^{+}\nu \overline{\nu} $$ decay channels with the ATLAS detector
Journal of High Energy Physics, ISSN 1029-8479, 1/2017, Volume 2017, Issue 1, pp. 1 - 53
A measurement of the ZZ production cross section in the ℓ−ℓ+ℓ′ −ℓ′ + and ℓ − ℓ + ν ν ¯ $$ {\ell}^{-}{\ell}^{+}\nu \overline{\nu} $$ channels (ℓ = e, μ) in...
Quantum Physics | Quantum Field Theories, String Theory | Hadron-Hadron scattering (experiments) | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory | Nuclear Experiment
Quantum Physics | Quantum Field Theories, String Theory | Hadron-Hadron scattering (experiments) | Classical and Quantum Gravitation, Relativity Theory | Physics | Elementary Particles, Quantum Field Theory | Nuclear Experiment
Journal Article
7. Combined measurement of differential and total cross sections in the H → γγ and the H → ZZ⁎ → 4ℓ decay channels at s=13 TeV with the ATLAS detector
Physics Letters B, ISSN 0370-2693, 11/2018, Volume 786, Issue C, pp. 114 - 133
Journal Article
Physics Letters B, ISSN 0370-2693, 04/2013, Volume 721, Issue 4-5, pp. 190 - 211
Journal Article
9. Constraints on off-shell Higgs boson production and the Higgs boson total width in ZZ → 4ℓ and ZZ → 2ℓ2ν final states with the ATLAS detector
Physics Letters B, ISSN 0370-2693, 11/2018, Volume 786, Issue C, pp. 223 - 244
A measurement of off-shell Higgs boson production in the and decay channels, where stands for either an electron or a muon, is performed using data from...
PHYSICS OF ELEMENTARY PARTICLES AND FIELDS | Fysik | Subatomär fysik | Physical Sciences | Subatomic Physics | Naturvetenskap | Natural Sciences
PHYSICS OF ELEMENTARY PARTICLES AND FIELDS | Fysik | Subatomär fysik | Physical Sciences | Subatomic Physics | Naturvetenskap | Natural Sciences
Journal Article
10. Search for the Standard Model Higgs boson in the decay channel H→ZZ(⁎)→4ℓ with the ATLAS detector
Physics Letters B, ISSN 0370-2693, 11/2011, Volume 705, Issue 5, pp. 435 - 451
A search for the Standard Model Higgs boson in the decay channel , where , is presented. Proton–proton collision data at recorded with the ATLAS detector and...
ATLAS | Leptons | Higgs | LHC | PARTON DISTRIBUTIONS | MASSES | ASTRONOMY & ASTROPHYSICS | PHYSICS, NUCLEAR | TAUOLA | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment | LUMINOSITY | CROSS SECTIONS | STANDARD MODEL | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS | FORECASTING | HIGGS BOSONS | DECAY | SENSITIVITY | PRODUCTION | Fysik | Physical Sciences | Naturvetenskap | Natural Sciences
ATLAS | Leptons | Higgs | LHC | PARTON DISTRIBUTIONS | MASSES | ASTRONOMY & ASTROPHYSICS | PHYSICS, NUCLEAR | TAUOLA | PHYSICS, PARTICLES & FIELDS | Physics - High Energy Physics - Experiment | Physics | High Energy Physics - Experiment | LUMINOSITY | CROSS SECTIONS | STANDARD MODEL | PHYSICS OF ELEMENTARY PARTICLES AND FIELDS | FORECASTING | HIGGS BOSONS | DECAY | SENSITIVITY | PRODUCTION | Fysik | Physical Sciences | Naturvetenskap | Natural Sciences
Journal Article |
This question is a generalization of the question Volume ratio of $\ell_1$ balls and $\ell_1$ surfaces
For any $p\in[1,\infty]$ define $\|x\|_p := (|x_1|^p+\cdots+|x_d|^p)^{1/p}$ for $p\in[1,\infty)$ and $\|x\|_\infty := \max_{1\leq i\leq d}|x_i|$ for $p=\infty$. Denote $B_p^d := \{x\in\mathbb R^d: \|x\|_p\leq 1\}$ as the unit $\ell_p$-ball in $d$ dimension, and let $\partial B_p^d := \{x\in\mathbb R^d: \|x\|_p = 1\}$. The quantity of interest is the following "ratio" $$ \mathfrak d_{p,d} := \frac{\mathrm{vol}_{d-1}(\partial B_p^d)}{\mathrm{vol}_{d}(B_p^d)}. $$
My question is the following:
For any fixed $p\in[1,\infty]$, how does $\mathfrak d_{p,d}$ asymptotically scale with dimension $d$, as $d\to\infty$?
For some special cases, both $\mathrm{vol}_d(B_p^d)$ and $\mathrm{vol}_{d-1}(\partial B_p^d)$ have closed form solutions and the ratio $\mathfrak d_{p,d}$ can be calculated explicitly. Below are three examples:
$p=2$: in this case $\mathrm{vol}_d(B_2^d) = \frac{\pi^{d/2}}{\Gamma(d/2+1)}$, $\mathrm{vol}_{d-1}(\partial B_2^d) = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ and therefore $\mathfrak d_{2,d} = d$;
$p=1$: in this case $\mathrm{vol}_d(B_1^d) = \frac{2^d}{d!}$ and $\mathrm{vol}_{d-1}(\partial B_1^d) = \frac{2^d \sqrt{d}}{(d-1)!}$, and therefore $\mathfrak d_{1,d} = d\sqrt{d}$
$p=\infty$: in this case $\mathrm{vol}_d(B_\infty^d) = 2^d$ and $\mathrm{vol}_{d-1}(\partial B_\infty^d) = 2^d n$. Therefore $\mathfrak d_{\infty,d} = d$.
From the above examples, my (very wild) guess is that $\mathfrak d_{p,d} \asymp d^{1+1/p-1/2}$ for $1\leq p\leq 2$ and $\mathfrak d_{p,d} \asymp d$ for $p\geq 2$. But of course I could be very wrong. The challenge for general $p$ is the apparent difficulty in evaluating the volume of $L_p$ sphere areas (e.g., Surface area of an $\ell_p$ unit ball?), but I'm hoping that the (asymptotic) ratio $\mathfrak d_{p,d}$ for general $p$ is potentially easier to evaluate.
Edit: I realized that, because the volume of a unit $\ell_p$ ball does have closed forms for general $p\in[1,\infty]$, the question can be answered if we know the asymptotic dependency of dimension $d$ of the (unit) $\ell_p$-surface area. This question was explicitly mentioned in the comment under the accepted answer of the following questionSurface area of superellipsoid (dice)Unfortunately, the OP in that question did not pursue this direction. |
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Free keywords: Mathematics, Optimization and Control, math.OC,Computer Science, Distributed, Parallel, and Cluster Computing, cs.DC
Abstract: When solving massive optimization problems in areas such as machine learning,
it is a common practice to seek speedup via massive parallelism. However,
especially in an asynchronous environment, there are limits on the possible
parallelism. Accordingly, we seek tight bounds on the viable parallelism in
asynchronous implementations of coordinate descent.
We focus on asynchronous coordinate descent (ACD) algorithms on convex
functions $F:\mathbb{R}^n \rightarrow \mathbb{R}$ of the form $$F(x) = f(x) ~+~
\sum_{k=1}^n \Psi_k(x_k),$$ where $f:\mathbb{R}^n \rightarrow \mathbb{R}$ is a
smooth convex function, and each $\Psi_k:\mathbb{R} \rightarrow \mathbb{R}$ is
a univariate and possibly non-smooth convex function.
Our approach is to quantify the shortfall in progress compared to the
standard sequential stochastic gradient descent. This leads to a truly simple
yet optimal analysis of the standard stochastic ACD in a partially asynchronous
environment, which already generalizes and improves on the bounds in prior
work. We also give a considerably more involved analysis for general
asynchronous environments in which the only constraint is that each update can
overlap with at most $q$ others, where $q$ is at most the number of processors
times the ratio in the lengths of the longest and shortest updates. The main
technical challenge is to demonstrate linear speedup in the latter environment.
This stems from the subtle interplay of asynchrony and randomization. This
improves Liu and Wright's (SIOPT'15) lower bound on the maximum degree of
parallelism almost quadratically, and we show that our new bound is almost
optimal. |
To extend VorKir's basically correct answer, to derive a priori error estimate for finite element approximations, you need to stack the following three Lego blocks:
The well-posedness of the weak formulation of the PDE. The (global) conformity of the Galerkin approximation. The (local) approximation properties of your finite element space.
(Only the last one is specific to finite elements.) Of course, you need to stack them correctly, so let's look at these in a bit more detail. I'll state them abstractly since you're interested in the blueprint and not the specifics.
Let's assume your problem in weak formulation is to find $u\in V$ such that $$a(u,v) = (f,v) \qquad\text{for all }v\in V \label{weak}\tag{1}$$for some (not necessarily symmetric) bilinear form $a:V\times V\to\mathbb{R}$.
Well-posedness here means that there exist constants $c,C>0$ such that for all $u,v\in V$,$$\begin{aligned} c\|u\|_V^2 &\leq a(u,u), \\ a(u,v)&\leq C\|u\|_V\|v\|_V.\end{aligned}$$The first inequality is called coercivity, the second continuity. The space $V$ in which you can prove Usually $V$ is a Sobolev space such as $H^1(\Omega)$ (which is the standard space for elliptic second order equations).From this, existence, uniqueness and stability of the solution follows using the Lax-Milgram Lemma, which is basically a nonsymmetric variant of the Riesz representation theorem. (If your test function needs to be taken in a different space $W\neq V$ (because you can't show continuity otherwise), the coercivity is replaced by the so-called both is the natural space for the problem and determines the rest of the analysis. inf-sup inequality, and the Lax-Milgram Lemma by the Banach-Nečas-Babuška Theorem.)
Now you pick a finite-dimensional subspace $V_h\subset V$ (this is the key property!) and look for $u_h\in V_h$ such that $$a(u_h,v_h) = (f,v_h) \qquad\text{for all }v_h\in V_h \label{galerkin}\tag{2}.$$Since $v_h\in V_h\subset V$, you can insert $v_h$ in \eqref{weak} as well and subtract the two to obtain the
Galerkin orthogonality$$a(u-u_h,v_h) = 0 \qquad\text{for any }v_h\in V_h.$$ Hence, for arbitrary $v_h\in V_h$, you have $v_h-u_h\in V_h$ and therefore $a(u-u_h,v_h-u_h)=0$. Using coercivity and continuity of the bilinear form, you obtain$$\begin{aligned} c \|{u-u_h}\|_V^2& \leq a(u-u_h,u-u_h)\\ &=a(u-u_h,u-v_h) + a(u-u_h,v_h-u_h)\\ &\leq C \|{u-u_h}\|_V\|{u-v_h}\|_V.\end{aligned}$$Dividing by $\|{u-u_h}\|_V$, rearranging, and taking the infimum over all $v_h\in V_h$ yields$$ \|{u-u_h}\|_V \leq \frac{C}{c} \inf_{v_h\in V_h}\|{u-v_h}\|_V.\label{cea}\tag{3}$$This is Céa's lemma, which says that the error of any (conforming) Galerkin approach is determined by the approximation error of the exact solution in $V_h$. (If the discrete space is not conforming, i.e., $V_h\not\subset V$, you get extra terms you need to estimate and end up with Strang's Lemmas instead.)
The next step is fairly straightforward: By Céa's Lemma, the discretization error is bounded by the best approximation error in $V$, which in turn can be bounded by
any approximation error -- for example, the interpolation error in $V$, which as pointed out above is usually a Sobolev space. The interpolation error of polynomials in Sobolev spaces can be estimated by the , which yields the error estimate$$\|u-\mathcal{I}_h u\|_{H^m} \leq C h^{k-m} \|u\|_{H^k}\label{poly}\tag{4}$$provided $V_h$ contains piecewise polynomials of degree $k-1$ (and of course $u\in H^k$, which is an assumption on the solution of \eqref{weak}). (Basically, this follows from the fact that the interpolation error is bounded on the reference element, combined with a simple scaling using transformation of integrals. Summing over all elements gives a global interpolation estimate.)Note that the choice of $m$ in \eqref{poly} is determined by the condition that $H^m\hookrightarrow V$ so that you can chain \eqref{cea} and \eqref{poly} to obtain the standard a priori error estimate in the $V$ norm (I'm being free here with the constants, which can change from one occurrence to the next):$$\label{apriori}\tag{5} \|{u-u_h}\|_V \leq C \inf_{v_h\in V_h}\|{u-v_h}\|_V \leq C \|u-\mathcal{I}_h u\|_{V} \leq C \|u-\mathcal{I}_h u\|_{H^m} \leq C h^{k-m} \|u\|_{H^k}.$$ Bramble-Hilbert lemma
To get the $L^2$ estimate from this requires the
Nitsche trick. Assume you have a Hilbert space $H$ with $V\hookrightarrow H$ (such as $H=L^2(\Omega)$ for $V=H^m(\Omega)$). Using the fact that you can identify the Hilbert space $H$ with its own dual space and thus write the $H$ norm as an operator norm, you can derive the inequality$$\label{aubin}\tag{6}\|u-u_h\|_{H} = \sup_{\varphi\in H\setminus\{0\}} \frac{(u-u_h,\varphi)_H}{\|\varphi\|_{H}} \leq C \|u-u_h\|_{V} \sup_{\varphi\in H\setminus\{0\}} \frac{\inf_{w_h\in V_h}\|w_\phi-w_h\|_{V}}{\|\varphi\|_{H}},$$where $w_\varphi$ is the solution of the dual (or adjoint) problem$$\label{adjoint}\tag{7}a(v,w_\varphi) = (\varphi,v)_H \qquad\text{for all }v\in V$$(note that the test function $v$ is now the first argument of $a$), which is assumed to be well-posed, too. This is the Aubin-Nitsche Lemma (see my previous answer for a derivation in a specific case).The idea is now to use the same interpolation error estimate for the dual problem as well to get an additional power of $h$. First, you have as before$$ \inf_{w_h\in V_h}\|{w_\varphi-w_h}\|_V \leq \|{w_\varphi - \mathcal{I}_h w_\varphi}_V \| \leq C h\|w_\varphi\|_{{H}^2} \leq Ch \|{\varphi}\|_{H}.$$(In the last step, we have used the well-posedness of the dual problem to estimate the $H^2$ norm of the solution $w_\varphi$ by the $H=L^2$ norm of the right-hand side $\varphi$, which requires some assumptions on the PDE.)Combining this with \eqref{aubin} and \eqref{apriori} gives the estimate$$\|u-u_h\|_{H} \leq C h \|u-u_h\|_{V} \leq C h^{k-m+1} \|u\|_{H^k}.$$
So, to summarize, if you want to create your own exotic finite element spaces and show error estimates, you need to
Find the correct energy space $V$. (For the Poisson equation with homogeneous Dirichlet conditions, $V=H^1_0(\Omega)$.)
Choose the discrete approximation space $V_h$ such that
discrete functions in $V_h$ have the right global continuity properties to be in the $V$ (or you can estimate the additional terms from Strang's Lemma). (For Poisson's equation, having $v_h$ continuously differentiable on each element and continuous across element boundaries to ensure $v_h\in V$.)
Decide on the polynomial degree you want to represent to get the correct power of $h$, and then
choose your degrees of freedom to a) have a unique solution of the interpolation problem in each element and b) ensure the global continuity properties from step 2. (For the Poisson equation and the estimate you gave, this leads to piecewise linear polynomials, and the Lagrange elements correspond to choosing the degrees of freedom at the vertices of the element. Taking $m=1$ and $k=2$ in the interpolation error gives the estimate you stated.)
The Nitsche trick then gives you a free additional power of $h$
if you can show that the dual problem is well-posed and you only measure the error in a weaker norm. (For the Poisson equation which is symmetric, the dual problem is of course another Poisson equation, and that step is a freebie.)
One final remark: In your case, I would recommend Braess's book over Brenner and Scott's; it's just as mathematically rigorous, but it's meant as a textbook instead of a monograph and thus concentrates on the fundamental ideas rather than the more general and technical results:
Dietrich Braess, MR 2322235 Finite elements, ISBN: 978-0-521-70518-9.
(I also can't resist mentioning my lecture notes.) |
Use environments to cluster mathematical expressions
MathJax provides a basic subset of environments for clustering mathematical expressions. For a complete list, see here, but be warned it is a very big page.
Environments are entered with the following syntax, some of them do not need to be explicitly entered with mathmode, but it is no harm doing so.
\begin{<environment>}
...
\end{<environment>}
In such environments a double backslash
\\ or a carriage return
\cr separates rows.
The environments can be used in conjunction with the
mhchem commands.
The most basic one is
gather which is essentially the same as using mathmode without an environment. The only difference is that this environment allows multiple uses of
\tag{...}. If you use
gathered instead, then only one tag for the whole cluster will be created; it does not matter where the tag is placed. This environment comes without alignment possibilities.
Most of the time the typeset can actually benefit from alignment, so this environment is maybe not the best to choose. See here for example uses. Basic alignment for multiple equations
Probably the most useful environment is
align. It can be used to align equations vertically, but also horizontally. Ampersands
& indicate the desired alignment position. Everything left of an odd instance is shoved to the right, and everything to the right is shoved to the left. Even uses of ampersands therefore can be used to separate equations on the same line.
Note that alignment characters also work inside chemical expressions
\ce{...}.
The following example code (
% ... produces a comment and everything following it will be ignored by MathJax, cannot be used within
\ce{...}.
& is the alignment character.)
\begin{align}
\ce{2H2 + O2 &<=> 2H2O} & %no line break
K & % align here
=\frac{a(\ce{H2O})^2}{a(\ce{H2})^2\,a(\ce{O2})} \tag{1}\\ % break line here
\ce{3H2 + N2 &<=> 2NH3} & %no line break
K & %align here
=\frac{a(\ce{NH3})^2}{a(\ce{H2})^3\,a(\ce{O2})} \tag{2} % no \\ required
\end{align}
yields:
\begin{align}
\ce{2H2 + O2 &<=> 2H2O} & %no line break
K & % align here
=\frac{a(\ce{H2O})^2}{a(\ce{H2})^2\,a(\ce{O2})} \tag{1}\\ % break line here
\ce{3H2 + N2 &<=> 2NH3} & %no line break
K & %align here
=\frac{a(\ce{NH3})^2}{a(\ce{H2})^3\,a(\ce{O2})} \tag{2} % no \\ required
\end{align}
You can use this environment to separate equations on the same line. The code
\begin{align}
c^2 &= a^2 + b^2 & % separate here
&\implies&
c &= \pm\sqrt{a^2 + b^2}
\end{align}
yields:
\begin{align}
c^2 &= a^2 + b^2 & % separate here
&\implies&
c &= \pm\sqrt{a^2 + b^2}
\end{align}
Equations too long for one line
Occasionally you have an equations that are too long for one line. Her the environment
multline comes in handy. In this the first line will be aligned to start at the left margin, while the last line will be aligned to end at the right margin. Lines between those will be centered. A
\tag{...} will be produced for the whole set, the placement does not matter. The following code
\begin{multline}\tag{3}
\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^{n} = \\ % aligned left
f(a) +\frac{f'(a)}{1!}(x-a) +\frac{f''(a)}{2!}(x-a)^2\\ % centred
+\frac{f'''(a)}{3!}(x-a)^3 +\frac{f''''(a)}{4!}(x-a)^4 +\cdots. % right
\end{multline}
yields
\begin{multline}\tag{3}
\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^{n} = \\ % aligned left
f(a) +\frac{f'(a)}{1!}(x-a) +\frac{f''(a)}{2!}(x-a)^2\\ % centred
+\frac{f'''(a)}{3!}(x-a)^3 +\frac{f''''(a)}{4!}(x-a)^4 +\cdots. % right
\end{multline}
A different possibility is tweaking the
align environment, like it is shown at the end of this answer. Alternatively you can use the
split environment, which works almost the same; however, it does not allow a tag. The following code
\begin{split}
\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^{n} &=
f(a) +\frac{f'(a)}{1!}(x-a) +\frac{f''(a)}{2!}(x-a)^2\\
&\quad +\frac{f'''(a)}{3!}(x-a)^3 +\frac{f''''(a)}{4!}(x-a)^4 +\cdots.
\end{split}
yields:
\begin{split}
\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x-a)^{n} &=
f(a) +\frac{f'(a)}{1!}(x-a) +\frac{f''(a)}{2!}(x-a)^2\\
&\quad +\frac{f'''(a)}{3!}(x-a)^3 +\frac{f''''(a)}{4!}(x-a)^4 +\cdots.
\end{split}
This environment can be encased in an
align environment if necessary. It might be beneficial if you want to sort terms. Entries before the first alignment character will be oriented to the right, but in every other instance it will be oriented to the left. In this environment multiple tags are allowed, hence the following code
\begin{align}
f(x):&= (1+x)\mathrm{e}^x \tag{4}\\ % first line
& = \mathrm{e}^x + x\mathrm{e}^x \tag{4a}\\
& = \sum_{n=0}^\infty \frac{x^n}{n!} %
+ \sum_{n=0}^\infty \frac{x^{n+1}}{n!} \tag{4b}\\
& = \left(%
\begin{split} % the following counts as one line
& 1 &+ x &+ \frac{x^2}{2!} &+ \frac{x^3}{3!} &+ \cdots\\
+~& x &+ x^2 &+ \frac{x^3}{2!} &+ \frac{x^4}{3!} &+ \cdots\\
\end{split}\right)\tag{4c}\\
& = \sum_{n=0}^\infty \frac{n+1}{n!}x^n \tag{4d}\\ % last line
\end{align}
yields:
\begin{align}
f(x):&= (1+x)\mathrm{e}^x \tag{4}\\ % first line
& = \mathrm{e}^x + x\mathrm{e}^x \tag{4a}\\
& = \sum_{n=0}^\infty \frac{x^n}{n!} %
+ \sum_{n=0}^\infty \frac{x^{n+1}}{n!} \tag{4b}\\
& = \left(%
\begin{split} % the following counts as one line
& 1 &+ x &+ \frac{x^2}{2!} &+ \frac{x^3}{3!} &+ \cdots\\
+~& x &+ x^2 &+ \frac{x^3}{2!} &+ \frac{x^4}{3!} &+ \cdots\\
\end{split}\right)\tag{4c}\\
& = \sum_{n=0}^\infty \frac{n+1}{n!}x^n \tag{4d}\\ % last line
\end{align}
Adding labels and referring to them
In a maths environment you can add labels
\label to tags
\tag, which then can be referred
\eqref or
\ref to later.
For example the following code
The equilibrium constant is defined by the expression
$$
K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.
\tag{5}\label{equ-constant}
$$
[...] As seen in equation \ref{equ-constant} [...]
[...] insert into \eqref{equ-constant} [...]
produces the following output:
The equilibrium constant is defined by the expression
$$
K_x=\prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}}.
\tag{5}\label{equ-constant}
$$
[...] As seen in equation \ref{equ-constant} [...]
[...] insert into \eqref{equ-constant} [...]
These references can be inserted directly without
$ signs, but they can also be used in conjunction with them.
Labels can be applied wherever tags can also be applied. They will also insert a link that brings you back to where the label was defined.
Since MathJax is rendered per page and not per post, the labels will affect other posts, too. If you get an error, try a different label. Conclusion
Especially aligning equations can be very beneficial to the readability of a post.
There are quite a few environments that might be useful to achieve this. In most cases, however, you don't need such extensive maths and the
align environment is quite sufficient. |
This question is about group/phase velocities and also De Brogilie wavelength.
What I would like to know is how to derive ratio $\lambda_e/\lambda_p$ ($\lambda_e$ and $\lambda_p$ are De Broglie wavelength for electron and proton) if we know that electron and proton have same velocities?
I know that when we say "velocity" $v$, this velocity is the same as "group velocity" $v_g$. So for start I can calculate ratio for group velocities:
\begin{align} \boxed{v_e = v_p} \longrightarrow \frac{v_{ge}}{v_{ge}} = \frac{v_e}{v_p} = 1 \end{align}
and similarly I can do for phase velocities $v_p$:
\begin{align} \frac{v_{pe}}{v_{pp}} = \dfrac{\tfrac{c^2}{v_{ge}}}{\tfrac{c^2}{v_{gp}}} = \frac{v_{gp}}{v_{ge}} = 1 \end{align}
But when i try to calculate the ratio for wavelengths I got stuck:
\begin{align} \frac{\lambda_{e}}{\lambda_{p}} = \frac{\tfrac{h}{p_e}}{\tfrac{h}{p_p}} = \frac{p_p}{p_e} = ~\longleftarrow \substack{\text{I got stuck here where i don't know}\\\text{how to use relation $\boxed{v_p = v_e}$}} \end{align}
Can anyone give me a hint? I was trying to use relation $p = \frac{\sqrt{{E_k}^2 + 2E_0E_k}}{c}$ but I got lost... |
Regarding the edit: I think natural density should still be easier for Frobenius. Let $G$ be $\mathrm{Gal}(F/\mathbb{Q})$ and let $c$ be a class function $G \to \mathbb{C}$.
Cebatarov with natural density is the statement$$ \sum_{p \leq N} c(\mathrm{Frob}(p)) \sim \left( \sum_{g \in G} c(g) \right) \cdot \pi(N) $$Frobenius is the special case of a class function where $c(g)=c(h)$ whenever $g$ and $h$ generate the same cyclic subgroup.
In both cases, the set of such class functions is a vector space, so it is enough to prove this for a spanning set of class functions. In the Cebatarov case, the spanning set is inductions of one dimensional characters of subgroups $H$ of $G$. In the Frobenius case, it is enough to induct only trivial characters.
Let $H$ be a subgroup of $G$, let $L$ be the fixed field and let $\chi: H \to \mathbb{C}^{\ast}$ be a character. Let $K \subset H$ be the kernel of $\chi$ and let $M$ be the fixed field of $K$, so $M/L$ is an abelian extension.
Then I believe (this is the part I am not sure of) that Cebatarov for $\mathrm{Ind}_H^G \chi$ should reduce to showing that $L(M/L, \chi)(s)$ has no zeroes or poles on $\mathrm{Re}(s)=1$, except for a simple pole at $s=1$ for $\chi$ trivial. (For Dirichlet density, we only need to show this at $s=1$.) Frobenius amounts proving this only in the case that $\chi$ is trivial. In this case, we are trying to show that $\zeta_L(s)$ has no zeroes or poles on $\mathrm{Re}(s)=1$, except a simple pole at $s=1$.
So the key question is, are the easiest proofs that $\zeta_L(s)$ has no zeroes or poles on the critical line easier than the corresponding proofs for $L(M/L, \chi)(s)$? I think the answer is yes. The proofs that I am aware involve two main parts
(1) Show that the function in question has no poles (other than the stated one at $s=1$) on $\mathrm{Re}(s)>0$.
(2) Write down some clever product of various $L$-functions which is manifestly non-vanishing, and conclude that none of the factors can vanish. This argument requires (1) to have been already done, to rule out the possibility that a zero and pole cancel.
Now, I don't know whether (2) gets significantly easier when $\chi$ is trivial. But (1) definitely does! The fact that $\zeta_H(s)$ extends to $\{ \mathrm{Re}(s)>0,\ s \neq 1 \}$ follows from the same sort of computations that lead to the class number formula for number fields. For $L$ functions, the way I know how to do it requires first proving that the character $\chi$ on $\mathcal{O}_L$ is periodic, so that we can group together sums over a complete period and guarantee that they converge. This is equivalent to Artin reciprocity, which as far as I know is as hard as all of class field theory.
Chebatarov's original proof (see the last section of Lenstra and Stevenhagen) avoids this. To my limited understanding of this, Cebatarov first proves the result when $M$ is a cyclotomic extension of $L$, where Artin reciprocity is easy, and then some how cleverly reduces to this case. But, even so, it seems to me that it must be easier when we work with just trivial characters. |
2 0
Hello,
Consider I have a linear time-invariant (LTI) system, with ##x(t)##, ##y(t)##, and ##h(t)##, as input, output, and impulse response functions, respectively. I have two choices to write the convolution integral to get ##y(t)##: $$ 1)\ \ \ y(t) = \int_{0}^{t} h(t-t')x(t')dt' $$ and $$ 2)\ \ \ y(t) = \int_{-\infty}^{t} h(t-t')x(t')dt' .$$ What are the differences between these two, and are initial conditions important factors in the decision of choosing one of these? Is it related to causality of the system?
Also, suppose I am just given the frequency response of the system, $$ Y(\omega)=H(\omega)X(\omega) . $$ Using Fourier transform and convolution integral theory, I want to change from frequency domain to time domain. Which one of the convolution integrals above should I pick? Because I need to work in the freuqency domain, I am trying not to use Laplace transform.
Thank you for sharing your ideas.
Consider I have a linear time-invariant (LTI) system, with ##x(t)##, ##y(t)##, and ##h(t)##, as input, output, and impulse response functions, respectively.
I have two choices to write the convolution integral to get ##y(t)##:
$$ 1)\ \ \ y(t) = \int_{0}^{t} h(t-t')x(t')dt' $$
and
$$ 2)\ \ \ y(t) = \int_{-\infty}^{t} h(t-t')x(t')dt' .$$
What are the differences between these two, and are initial conditions important factors in the decision of choosing one of these? Is it related to causality of the system?
Also, suppose I am just given the frequency response of the system,
$$ Y(\omega)=H(\omega)X(\omega) . $$
Using Fourier transform and convolution integral theory, I want to change from frequency domain to time domain. Which one of the convolution integrals above should I pick?
Because I need to work in the freuqency domain, I am trying not to use Laplace transform.
Thank you for sharing your ideas. |
Given a real finite-dimensional vector space $V$ with a symmetric bilinear form $b$, we define the
$Cl(V,b)$ as the quotient of the tensor algebra $\bigotimes V$ by the two sided ideal generated by $v\otimes v + b(v,v) \mathbb{1}$ for all $v \in V$. $Cl(V,b)$ is a $\mathbb{Z}_2$-graded unital real associative algebra. Clifford algebra
Every such $(V,b)$ is isomorphic to $\mathbb{R}^{r+s+t}$ with bilinear form defined by polarisation from the quadratic form $$ q(x) = \sum_{i=1}^{r + s + t} \varepsilon_i x_i^2, $$ where $$ \varepsilon_i = \begin{cases} 0 & i = 1,\dots,r\\ 1 & i = r+1,\dots,r+s\\ -1 & i = r+s+1, \dots, r+s+t .\end{cases} $$ Let $Cl(r,s,t)$ denote the Clifford algebra of $\mathbb{R}^{r+s+t}$ and the above bilinear form.
As $\mathbb{Z}_2$-graded unital real associative algebras, $$ Cl(r,s,t) \cong \Lambda \mathbb{R}^r \hat\otimes Cl(s,t), $$ where $\hat\otimes$ is the $\mathbb{Z}_2$-graded tensor product and where $Cl(s,t):= Cl(0,s,t)$ are the standard Clifford algebras associated to non-degenerate bilinear forms.
The representations of $Cl(s,t)$ are well-known: there are either one or two simple modules (up to isomorphism) depending on $s,t$ and every finite-dimensional module is a direct sum of simples.
I am interested in the representations of $Cl(r,s,t)$ for $r=1$, but more generally for $r>0$.
For $(s,t) = (1,0)$ and $(0,1)$, it is easy to work this out "by hand". The resulting category of representations is no longer semisimple, but it is not hard to show that any finite-dimensional module is a direct sum of indecomposable (but not simple) modules.
But before attempting to study the case of general $(s,t)$, I wonder whether there is some technology out there which can be brought to bear on this problem.
More concretely, I have a couple of
Questions
Would a knowledge of the indecomposable modules of $\Lambda \mathbb{R}$ and $Cl(s,t)$ be sufficient to determine the indecomposable modules of their $\mathbb{Z}_2$-graded tensor product? If so, how?
Is there a classification of indecomposable finite-dimensional modules of the exterior algebra $\Lambda \mathbb{R}^r$ for $r>1$? If so, where? |
March 18th, 2017, 11:46 AM
# 1
Senior Member
Joined: Jan 2017
From: Toronto
Posts: 209
Thanks: 3
Calculating optimal isosceles trapezoid
Trough is to be formed by bending up two sides of a long metal rectangle so that the
cross-section of the trough is an isosceles trapezoid. If the width of the
metal sheet is 2 meters, how should it be bent to maximize the volume of the trough?
Answer: The sides and bottom should all be 2/3 meter, and the sides should be bent up at angle pi/3.
Any tips would be much appreciated.
March 18th, 2017, 12:58 PM
# 2
Math Team
Joined: Jul 2011
From: Texas
Posts: 3,031
Thanks: 1620
$A = \dfrac{h}{2}(b_1+b_2)$
$h = x\sin{\theta}$
$b_1 = 2(1-x)$
$b_2 = 2(1-x) + 2x\cos{\theta}$
I started by setting $\dfrac{\partial A}{\partial x} = 0$, solved for $x$ in terms of $\theta$ ...
$x=\dfrac{1}{2-\cos{\theta}}$.
Set $\dfrac{\partial A}{\partial \theta} = 0$, substituted for $x$, found $\cos{\theta} = \dfrac{1}{2} \implies x = \dfrac{2}{3} \text{ and } \theta = \dfrac{\pi}{3}$
Give it a go ...
March 18th, 2017, 01:23 PM
# 4
Senior Member
Joined: Jan 2017
From: Toronto
Posts: 209
Thanks: 3
Simple and beautiful. I need more practices.
Quote:
March 19th, 2017, 09:38 AM
# 5
Senior Member
Joined: Jan 2017
From: Toronto
Posts: 209
Thanks: 3
Here is my work..
P = 2 - 2x
H = x * cos(θ)
Q = 2x * sin(θ) + P = 2 * xsin(θ) + 2 - 2x
A = 1/2 (P + Q) * H
A = 1/2 * (4 - 4x + 2x * sin(θ)) * x * cos(θ)
A = (2 - 2x + x * sin(θ)) * x * cos(θ)
∂A/∂x = 2 * cos(θ) - 4x * cos(θ) + 2x sin(θ) cos(θ) = 0
∂A/∂θ = -2x * sin(θ) + 2x^2 * sin(θ) + 2x * (cos(θ)^2 - sin(θ)^2) = 0
∂A/∂x = cos(θ) - 2x * cos(θ) + x sin(θ) cos(θ) = 0
∂A/∂θ = -x * sin(θ) + x^2 * sin(θ) + x * cos(θ)^2 - x * sin(θ)^2 = 0
2 unknowns, but with 2 complicated equations...
Any tips to help me moving forward would be much appreciated...
March 19th, 2017, 10:57 AM
# 6
Math Team
Joined: Jul 2011
From: Texas
Posts: 3,031
Thanks: 1620
$A = \dfrac{x\sin{t}}{2}\bigg[4(1-x)+2x\cos{t}\bigg]$
$A = \sin{t}\bigg[2x-2x^2+x^2\cos{t}\bigg]$
$\dfrac{dA}{dx} = \sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg]$
note $\sin{t} \ne 0$ ...
$\sin{t}\bigg[2 - 4x + 2x\cos{t}\bigg] = 0 \implies 2 - 4x + 2x\cos{t} = 0 \implies x = \dfrac{1}{2-\cos{t}}$
$A = \dfrac{x}{2}\bigg[4(1-x)\sin{t} + x \cdot 2\sin{t}\cos{t}\bigg]$
$A = \dfrac{x}{2}\bigg[4(1-x)\sin{t} + x\sin(2t)\bigg]$
$\dfrac{dA}{dt} = \dfrac{x}{2}\bigg[4(1-x)\cos{t} + 2x\cos(2t)\bigg]$
note $x \ne 0$ ...
$4(1-x)\cos{t} + 2x\cos(2t) = 0$
$4\cos{t} - 4x\cos{t} + 2x(2\cos^2{t}-1) = 0$
$2\cos{t} - 2x\cos{t} + x(2\cos^2{t}-1) = 0$
substitute $\dfrac{1}{2-\cos{t}}$ for $x$ ...
$\dfrac{(2\cos{t})(2-\cos{t})}{2-\cos{t}} - \dfrac{2\cos{t}}{2-\cos{t}} + \dfrac{2\cos^2{t}-1}{2-\cos{t}} = 0$
$4\cos{t}-2\cos^2{t} - 2\cos{t} + 2\cos^2{t}-1 = 0$
$2\cos{t} - 1 = 0 \implies \cos{t} = \dfrac{1}{2} \implies t = \dfrac{\pi}{3}$
$x = \dfrac{1}{2-\cos\left(\frac{\pi}{3}\right)} = \dfrac{2}{3}$
March 19th, 2017, 01:10 PM
# 7
Senior Member
Joined: Jan 2017
From: Toronto
Posts: 209
Thanks: 3
You sir are a miracle worker! I finally got the answer.
Quote:
December 4th, 2017, 09:52 PM
# 8
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Joined: Dec 2017
From: China
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You can find other explanations in this post:
https://machinemfg.com/press-brake-ultimate-guide/
Last edited by skipjack; December 5th, 2017 at 03:16 AM.
Tags calculating, isosceles, optimal, trapezoid
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You start from wrong assumptions, which explains your doubts. The line
the induced drag is due to the tip vortices
is as true as saying that wet streets cause rain. Also, the opinion that the
tip vortices strength will be as same as the bound vortex
is wrong. Unfortunately, many authors don't understand the topic themselves and copy what others have written before without thinking the issue through. Ideally, you would forget about all what you have heard about vortices and lifting lines, but since you ask I will try to explain potential flow theory a little.
In potential flow theory, lift is caused by vortices which are caused by the movement of a wing through air. These vortices run along a closed line: Within the wing they form the bound vortex, then they leave the wing backwards as trailing vortices and are connected at the point where the movement started by the starting vortex.
Now comes the important part which most authors conveniently leave out: There is no single vortex; instead, potential flow assumes an infinite number of infinitesimally small vortices which form out of nowhere when lift is increased or speed is reduced. Consequently, no single vortex leaves the wing at the tips, instead, a sheet of vortices leaves the wing at the trailing edge. The change in strength of the bound vortices over span is equivalent to the strength of the vortices leaving the wing, so the vortices fade out towards the tips.
My advice is: If you do not want to operate or to write a potential flow code, do yourself a favor and forget all that. It is much better to interpret lift as the consequence of a pressure field around a wing which accelerates the air flowing around this wing downwards. Induced drag is simply the component of the resulting pressure forces parallel to the direction of movement, while the perpendicular component is lift. Please make sure to follow at least the last link; it gives a very good explanation what induced drag really is.
Tip vortices are the consequence of air filling the void above the downward moving air behind the wing. They are not originating from the wingtips, but the consequence of the vortex sheet rolling up (if you want to stay in that picture). Note that the distance between the cores of the vortices is much smaller than the wingspan. For an elliptic wing of span $b$, it is actually only $\frac{\pi}{4}\cdot b$
A higher wingspan allows to capture more air for lift creation, so less downward acceleration is needed. Lower downwash speed also causes a less powerful trailing vortex. Note that the mass of air affected by the wing grows with the square of the wingspan! |
I carefully read all circa 70 pages of FEniCS tutorial and I still do not understand how to solve electrostatic problems when I have materials with different dielectric constant. The self contained system of equations in electrostatic $$ \mathrm{div}\boldsymbol{\mathcal{D}}=4\pi\rho,\\ \boldsymbol{\mathcal{D}}=\varepsilon\mathbf{E},\\ \mathbf{E}=-\nabla\varphi. $$ works only in regions where the functions are differentiable and smooth. At the interface between two regions with different materials these equations have to be augmented with the following boundary conditions: $$ \varphi_1=\varphi_2,\\ \varepsilon_1\frac{\partial\varphi_1}{\partial n}=\varepsilon_2\frac{\partial\varphi_2}{\partial n}. $$ How can I specify the second boundary condition? The example in the tutorial considers only a simple case of $$ \frac{\partial\varphi}{\partial n}=0 $$ between two regions.
Vatiational formulation of Poisson problem with $\varepsilon \in L^\infty$
Both interfacial conditions are incorporated in following variational problem. Given
$\Omega$ Lipschitz domain $\partial\Omega = \Gamma_\mathrm{D} \cup \Gamma_\mathrm{N}$ $V=\{v\in H^1(\Omega); v|_{\Gamma_\mathrm{D}}=0 \} $ $\varepsilon\in L^\infty(\Omega)$ $g \in L^2(\Gamma_\mathrm{N}) $
find $\varphi\in V$ such that
$$ \int_\Omega \varepsilon \nabla \varphi \cdot \nabla \psi \;\mathrm{d}x = \int_\Omega \rho \psi \;\mathrm{d}x + \int_{\Gamma_\mathrm{N}} g \psi \;\mathrm{d}S \quad \forall \psi\in V$$
Here $g$ represents Neumann condition on $\varepsilon\frac{\partial \varphi}{\partial \mathbf{n}} = g$ on $\Gamma_N$. Both your interfacial conditions are handled automatically by this formulation. You check it by dividing $\Omega$ into subdomains where $\varepsilon$ takes two different values and testing by something close to characteristic functions of these subdomains. Moreover $\varepsilon$ can be much more arbitrary - essential boundedness suffices. Note that I threw away $4\pi$ which corresponds to SI units. If you would like to derive this formulation from physicist point of view, just take Gateaux derivative of field energy minus charge energy in appropriate (accounting to Dirichlet condition) space.
For implementation in FEniCS just pick conforming FE space, i.e. arbitrary order Lagrange/CG elements. Problem is implemented in
subdomains-poisson demo assuming discontinuity in $\epsilon$ matches mesh facets.
Discontinuity in $\varepsilon$ non-matching with mesh
If you need discontinuity non-matching mesh, you could try
DOLFIN-PUM library which implemets XFEM/PUM method. It is not maintained now and Garth said that there are plans for better implementation directly into FEniCS. Library is able to integrate exactly this type of discontinuity in $\epsilon$. But there is additional problem that XFEM/PUM method introduces dicontinuity also to your solution and you need to enforce somehow continuity. It can be solved with Lagrange multiplier, penalty, etc. but I can't rememeber if I succedeed doing this.
You can also solve problem with non-matching mesh using standard formulation and increase quadrature degree or interpolate $\varepsilon$ to some higher-order space prior to computation. I also use $C^k$ approximation to Heaviside function with $k=0,1,2$ to define material discontinuities. You can implement this using
conditional. None of these approaches can be integrate $\varepsilon$ exactly but it could enhance solution somehow.
Since the problem is treated in
subdomains-poisson for the case, where the mesh is aligned with the interface, I assume in the following that in your case the jump can occur anywhere.
I think the first step here is not to think about implementation FEniCS, but rather to find an adequate variational form for this problem, where the interface conditions are properly incorporated. The problem looks like it can be treated with a variant of the following method:
Hansbo, A. and Hansbo, P. An unfitted finite element method, based on Nitsche’s method, for elliptic interface problems (Comp. Meth. in Appl. Mech. and Eng., Volume 191, Issues 47–48, 22 November 2002, Pages 5537–5552)
Disclaimer: There may be others and maybe also better alternatives for your specific case, but I am mostly familiar with Nitsche-type techniques.
Once you have an idea about how to discretize your problem, you can think about implementation in FEniCS. As far as I know, it is not possible to implement this method straightforwardly in vanilla FEniCS. dolfin-olm may be used to implement this method. I have not used it so far, I don't know if it works with the current version of FEniCS, and I don't know if they merged (parts of) it into dolfin in the meantime.
Again, note that in standard finite element methods the interface condition can be taken into account by aligning your mesh with the interface. If that is an option, you should go for that. |
This question arises from HNN Embedding Theorem for Amenable Groups?
Recall that a group $G$ is called
SQ-universal if every countable group is isomorphic to a subgroup of a quotient of $G$. The first non-trivial example of an SQ-universal group was provided by Higman, Neumann and Neumann in 1949. They proved that the free group of rank $2$ is SQ-universal, which is equivalent to the statement that every countable group embeds into a $2$-generated one. Presently many other examples of $SQ$-universal groups are known (e.g., hyperbolic and relatively hyperbolic groups).
It is straightforward to see that any SQ-universal group contains a non-abelian free subgroup and hence is non-amenable. However the following problem seems open.
Problem 1. Does there exist a finitely generated amenable group $A$ such that every countable amenable group embeds into a quotient of $A$?
I believe, the answer is "no". One way to disprove it would be to use the Folner functions, defined by Vershik in 70's. Recall that for a finitely generated amenable group $A$, $Fol_A\colon \mathbb N\to \mathbb N$ is defined by $Fol_A(n)$ = the size of a smallest finite subset $S \subseteq A$ satisfying $|\partial S|/|S|\le 1/n$. The asymptotic growth of $Fol_A(n)$ is independent of the choice of a finite generating set of $A$ up to a natural equivalence.
It is not hard to show that, when we pass to subgroups and quotient groups, this function does not decrease in the sense of the natural relation
$$ f\preceq g \; {\rm iff}\; \exists\, C>0\; {\rm such\; that}\; f(n) \le Cg(Cn)\; \forall\, n. $$ Thus to answer Problem 1 negatively it would be sufficient to prove the following. Conjecture 2. For any function $f\colon \mathbb N\to \mathbb N$, there exists a finitely generated amenable group $A$ such that $f\preceq Fol_A$.
Erschler [On isoperimetric profiles of finitely generated groups,
Geom. Dedicata 100 (2003), 157–171] showed the existence of amenable groups with $Fol$ growing faster than any iterated exponential function. She also announced the proof of Conjecture 2 there, but I did not find it in her later papers.
Final remark: Problem 1 also makes sense if we replace "finitely generated" with "countable". |
I read in reliable sites that GR and classical physics calculate the angle of deflection in the same manner. The formula is almost identical: $$\theta = \frac{4GM}{c^2*r} \rightarrow \frac{4GM}{c*r} = v *\frac{1}{c}$$GR modifies the formula adding a factor $ \theta = \frac{4GM}{c^2r}\times \frac{1+\gamma}{2}$ where gamma has value
1 for energy and 0 for mass, so that the classical laws apply to orbits $ \theta = \frac{2GM}{c^2r}. $Now if I did not misinterpret the procedure, they find the g-pull at point of impact R $g = \frac{GM}{R^2} $ and then integrating over time they get the normal velocity the ray acquires from gravity $v = \frac{[2*]2GM}{R\times c}$In the case of the Sun this value is [2*]127400 cm/s. Then, considering v and c the legs of a right triangle, they consider v/c (254800/3*10^10) the tangent of the angle theta (the angle of deflection in radians) and since theta is a small angle, they identify the value of the tangent with the angle itself. Also John Rennie did this here:(Deflection of light by the Sun)
the equation I gave is an approximate equation that works when θ is small. At these small angles the difference between θ and tanθ is negligible.
If my report is correct (please forgive any blunders), I have a couple of questions:
If you consider the result of the integral (254800 cm/s) a vector of velocity v, why don't you follow the regular addition of vectors?
Why should the angle between two vectors be intended in radians and not in degrees. What is the rationale?
If you think that $v/c$ is really the tangent of the angle of deflection, why use the subterfuge of
approximationand not find the exact angle using the appropriate function of $\mathrm{arctan}$, $\mathrm{tan}^{-1}(v/c)$?
@NaturalPhilosopy, that "
doesn't make sense" I wrote in the question. Of course you are only reporting the literature, but I hope you, or someone here, realize that that is not rational, scientific. Why sould you assume anything? why should you deviate from the standard optimal math procedure? what is the logic, the rationale? you know the tangent, want to find the angle : arctan. And what if the angle is not so small? suppose v = c, then tan theta = 1, what do you do? Lastly, if you calculate the effective angle theta with tan= 254800/3*10^10, you discover that the difference is not that negligible.
@Jerry Schirner, sure, radians has a meaning but so do many other things in math. The fact that it is meaningful doesn't mean that it is relevant, appropriate here. That is not enough to justify, logically and physically that the ratio between two normal vectors corresponds to as many radians. Has anybody ever proved that? Can you do that? |
Defining parameters
Level: \( N \) = \( 9450 = 2 \cdot 3^{3} \cdot 5^{2} \cdot 7 \) Weight: \( k \) = \( 2 \) Character orbit: \([\chi]\) = 9450.dn (of order \(18\) and degree \(6\)) Character conductor: \(\operatorname{cond}(\chi)\) = \( 945 \) Character field: \(\Q(\zeta_{18})\) Sturm bound: \(4320\) Dimensions
The following table gives the dimensions of various subspaces of \(M_{2}(9450, [\chi])\).
Total New Old Modular forms 13104 2592 10512 Cusp forms 12816 2592 10224 Eisenstein series 288 0 288 Decomposition of \(S_{2}^{\mathrm{new}}(9450, [\chi])\) into irreducible Hecke orbits
The newforms in this space have not yet been added to the LMFDB. |
Bungee Drop
Bungee Drop Engineering & Build Event Forum Threads 2015 2014 There are no tests available for this event Images Images (needs submissions) There are no question marathons for this event This event was not held recently in Division B Division C Champion Mentor High School Bungee Drop is a Division C event last held in the 2015 season. The object is to drop a mass attached to an elastic cord from a given height and to get the mass as close to the ground without letting it touch the ground. Contents 1 Objective 2 General Advice 2.1 Practicing 2.2 A Mathematical Approach 2.3 Materials 2.4 Competition Tips 3 Scoring 4 Resources Objective
The object is to drop a mass attached to an elastic cord from a given height and to get the mass as close to the ground without letting it touch the ground. The heights for the two drops can be anywhere from two to five meters (Regionals/States) or ten meters (Nationals), with the same height for both drops (different from 2014 rules). After all devices are impounded impound, both the drop height and object mass will be announced. These weights will be anywhere from 50 to 300 grams and will be placed in a 500-591 mL plastic bottle.
General Advice
Read the event description carefully! The cord can be made from any materials, but the bottom meter must pass the 'elasticity' test. With the 2015 rules for elasticity tests, only the bottom meter of the device needs to be stretch, so the remainder of the cord can be unelastic string. While all sorts of materials can be used to make the cord elastic, ranging from metal springs to Slinkys to elastic or rubber bands, some will be more reliable and will work better.
Practicing
The most important task to be done is calibration. Constructing a device may not be as difficult for Bungee Drop as it is for other events, but practice is key. Since elastic cords are not necessarily consistent in elasticity throughout, it is important to practice with multiple drop heights.
Creating a more reliable testing apparatus will allow you to have better data points to rely on at the competition. A typical bungee testing-device utilizes a soda bottle with mass contained in it and a clamp to hold the cord in place. It is recommended to also have a controlled method of releasing the bottle so that trials are systematic. For an example of a testing apparatus, see the image to the right.
Calibration this year promises to be more complicated than it has been in past years. If, say, you start calibration with a height of 3.00 meters, you can't only find one mark to make on our bungee for that height - you must make a mark for that height with 25 grams, for that height with 50 grams, etc. Getting the same degree of precision as was possible in past years will now, theoretically, take more the time (assuming you calibrate for, say 15 heights, and 15 masses). Of course, the real solution will to be to find a pattern - a formula. With a formula, you can simply input a mass and a height to know how long your cord should be.
A Mathematical Approach
There are many different ways one can tackle the Bungee Drop problem. However, if you want to compete nationally, there really is one approach that trumps all. Math. Math can never break down. 1+1 will never suddenly equal 3 the day before the competition. Math lets you determine the intervals in between. Math is love, math is life.
Partially Elastic Bungee Equations Please note: I strongly discourage the usage of partially elastic bungee cords, and instead deeply advocate fully elastic cords, mainly because partially elastic cords deviate from Hooke's law very easily due to higher strain at higher drop heights, while fully elastic bungees retain low strain. However, this section will remain as a reference section for the fully elastic bungees. Furthermore, the fully elastic equation was derived based off the partially elastic equation, so many principles need to be understood here before tackling the fully elastic bungee equations. Finally, if you really want to use the partially elastic equation, here you go.
This formula sheet (courtesy of Joseph Liba (forum name: joiemoie) of Acton-Boxborough Regional High), uses basic physics equations to create the ideal bungee equation.
Requirements: Your bungee cord has an elastic portion and a non-elastic portion. Graphing Calculator
Note: If x₂ is less than the total length of the elastic portion, this k value will be slightly larger than your equation produces. This will only happen in the case of high masses and low drop heights. However, the following calculations indicate that the probability of such an event occurring is rather small and negligible:
Even if this case were to occur in a tournament, it would only bring you slightly further away from tiering. Maybe if you were planning on going about, say 1cm from the ground, instead you'd be going probably 5 cm from the ground.
Calculating the "K" Constant
Use the first equation, [math]k=\frac{2ma(x_1+x_2)}{x_2^2}[/math] to calculate the "k" constant. To make things faster, be sure use an equation that allows you simply input x₁, x₂, and D, where D is not the drop height, but rather distance traveled. A simple way to calculate D is just to get the drop height and subtract how far off the ground it was. I would suggest running as few trials as you can to determine "k".
Also you can just use a force probe, F = kx, and calculate k that way.
K Isn't Perfect. Fix or Doom
As the Wikipedia page notes, "In fact, many materials will noticeably deviate from Hooke's law well before those elastic limits are reached." This is absolutely true. Your bungee's k value won't be constant forever. For example, at roughly 3 meters, our bungee's K value started to go down. This is when you need a NEW EQUATION for k! WOW.
K is really a function of x2, as demonstrated by the following image: [1]
Note: K is really just the slope of this graph. Notice how as x2 goes up, at a certain point, k starts to go down. Truly showing k isn't perfect. HOW TO SOLVE: Make a graph of k vs x2. Find an equation. I'm sorry, but there's no magical equation denoting k vs x2. That's the truly experimental part. It may be linear. It may wiggle. Who knows. Now how do you do this? Just use the old equations and calculate a bunch of k values for different x2 values.
The Main Equation
Use the second equation [math]x_1=D-\frac{\sqrt{2Dkma}}{k}[/math]. "x₁" is going to be how long you measure the bungee from the bottom of the ring top of the key ring upwards. Note that D should be the distance from the bottom of the bottle to the floor. The length of the bottle has absolutely zero effect.
Now, what do you do if your k value varied, as denoted by the previous section? Easy! Graphing calculator. Let [math]k=f(x_2)[/math] where f(x2) is some function that you determined earlier. Let [math]x_2=\frac{\sqrt{2Df(x_2)ma}}{f(x_2)}[/math] Use the solve function on your graphing calculator to solve for x2. Let [math]x_1=D-x_2[/math] Done (Oh, and if you're planning on using the error propagation equations, calculate k and store that somewhere. You'll need it
Error Propagation Equations
This is only if you are going extremely uber-hardcore. This will calculate your predicted error in your equation based on your drop height and mass, and also your error in measurements. This uses some calculus concepts, so just use this equation and shove it into your calculator. Note that this time, the Δx₁ is going to be the error in measuring x₁. Basically means that if you have shaky hands, and think you can only be precise with your measuring tape to half a centimeter, then you would input this as your Δx₁. The Δk would indicate your error in k, which can simply be found by calculating the standard deviation of your k values.
Note, this is assuming that the testing rig is perfectly measured, so there is no error in D or m. If you have some awful event directing going down, that's quite unfortunate (personal experience).
[math]\Delta D=|(1+(m^2a^2+2kmax_1)^{-1/2}ma)|\Delta x_1+|k^{-2}(kmax_1(m^2a^2+2kmax_1)^{-1/2}-(m^2a^2+2kmax_1)^{1/2}-ma)|\Delta k[/math]
If you are distrustful of your event administration, here's the equation that also takes into account the error in mass and drop height. When I say error in mass, this means like their scale looks pretty shoddy, or is only accurate to say the thousands so you have a +/- .0005 error. Error in drop height means like, say they're just whipping out a ruler and eyeballing the height, and you're pretty sure there's an error of 1cm in the drop height. (personal story. At the MIT invitationals, they tried to use a laser that reflected back to calculate the drop height, and the real drop height was lower than what they said by about .5 meters, so basically almost everyone tiered, and they realized the screwup and asked everyone to retest. However, when they remeasured, they had to do it on the spot and whipped out their lil old ruler, giving an error of about 1cm. Always be prepared for event administration error ;) "
Anyways here's the full on equation for error in D based on error in mass, k, x1, and drop height.
[math]\Delta D=|((2a^2m+2ax_1k)(2(a^2m^2+2ax_1m)^{1/2})^{-1}+a)k^{-1}|\Delta m+|(1+(m^2a^2+2kmax_1)^{-1/2}ma)|\Delta x+|k^{-2}(kmax_1(m^2a^2+2kmax_1)^{-1/2}-(m^2a^2+2kmax_1)^{1/2}-ma)|\Delta k+Error In Drop Height[/math]
Accounting for Stretch over Time
Notice as you test your bungee more and more, it'll tend to get less elastic and become longer. This will slightly throw off your "k" value every time you test. How do you compensate for this? Use measuring tape for the non elastic portion of your rope! When your bungee stretches out over time, and becomes less elastic, it will become longer. However, the measuring tape will shift along with the stretch, simulating a constant "k" value.
In summary, use flexible measuring tape attached to the elastic portion of your bungee.
Entirely Elastic Bungee Equations
Please refer to the previous section, except instead of k, we will be using lamda "λ". λ is Young's Modulus of elasticity. λ is a constant. The reason we use λ instead of k compared to the previous section is because in the previous section, we assume that the length of the elastic portion is constant. However, with a bungee that is entirely elastic, the length of the elastic portion is always changing. And the higher up the bungee you go, the more elastic the part under it will be. I mean, think about it logically, too. If you try to stretch a tiny piece of rubber, it's really hard, but if you try to stretch a long piece, it's much easier. λ takes into account the length and the k value at that particular length.
Once again, x1 = length of bungee you're measuring out, x2 is the amount it stretches, D is the distance from floor to bottom of bottle, m is mass, and a is acceleration due to gravity (9.8). Also please remember to use metric. All units are in meters, kilograms, seconds.
All we gotta do is experimentally calculate the λ. [math]\lambda=\frac{2max_1(x_1+x_2)}{x_2^2}[/math]
Keep in mind, λ may not be perfect either, at which point you'll have to find λ as a function of f(x2) as I mentioned in an earlier section.
Cool. You have λ. Now here's the equation that you'll be using at competitions. [math]x_1=D+\frac{Dam-D\sqrt{a^2m^2+2am\lambda}}{\lambda}[/math]
Also, adjust the equation if λ is a function of x2.
Error Equation: [math]\Delta D=|(a^2mx_1\lambda^{-1}(a^2m^2+2am\lambda)^{-1/2}+ax_1(a^2m^2+2am\lambda)^{-1/2}+ax_1\lambda^{-1})|\Delta m+|((a^2m^2+2am\lambda)^{1/2}\lambda^{-1}+am\lambda^{-1}+1)|\Delta x_1+|(amx_1\lambda^{-1}(a^2m^2+am\lambda)^{-1/2}-x_1((a^2m^2+2am\lambda)^{1/2}+am)\lambda^{-2})|\Delta k+Error In Drop Height[/math]
Disclaimer: I haven't tested a fully elastic equation yet. So whoever wants to step up and try it out, feel free to be the lab rat, and if it works, remove this disclaimer ;)
Full Elastic vs. Partially Elastic
So now the question is: should you use a fully elastic or partially elastic bungee? What are the pros and cons of both?
Full Elastic: + Slower drop so that the camera is more likely to catch it if it gets close to the ground (one time at a regionals tournament the guy was recording the drops with his iphone that had a 30fps camera and sometimes the bungee dropped so fast that even though it went really close to the ground, the frame only caught the times when it was about 6cm from the ground) + The strain is guaranteed to be roughly constant throughout all drop heights and hooke's law will not break. + Bungee stretches out slower over time, allowing you to do more trials - Can be hard to measure out the x1 if using rubber bands - I haven't tested it yet. - Very good chance that hooke's law will break given high distances. Partially elastic: + Deriving the equation was easier lol + You can use a measuring tape as the non-elastic portion to account for bungee that stretches out over time. -Drops faster so the camera might not catch it
Materials
The 2015 rules allow students to use any materials as long as they pass the 'elasticity' test. While this does give a lot of freedom in deciding how to make your device, obviously some work better than others.
Springs Springs can be fairly reliable, possibly more reliable than elastic cords. However, they face the issue of tangling during a drop if not positioned correctly. If you decide to use a spring, see which methods of dropping prevent it from tangling. Also, make sure the spring constant is not too large or small (a slinky may not be a good choice because its spring constant is small enough for it to stretch under its own weight. Rubber/Latex/etc. The simplest cord you can make from rubber/latex/etc. is by tying rubber bands together to form a long enough cord, given that it passes the elasticity test. However, depending on how you tie it, the cord may be inconsistently stretch in different parts. This issue can be alleviated if you buy a long enough cord to meet your needs, but even that may be slightly inconsistent (though better). Competition Tips At the competition, always be more cautious for new drop heights. Adjust the cord so that it is slightly shorter than you would expect. If the two drop heights are the same, you can adjust the cord for the second drop depending on how close the first drop was. If you use graphs to determine a formula, consider adding error bars to your graph so you can predict how careful you should be. Some event supervisors will not disclose how close each of your drops was, so consider having your partner stand near the surface to observe how close each drop is. The rules permit students to verify the mass of the bottle on their own. Since scales will vary in accuracy, consider using your own scale to measure the mass so that it fits with your data. Scoring
The goal, as before, is to get the device to drop as close as possible to the floor below, without touching. The team with the lowest score for the sum of the two drops from the ground, wins. However, if one drop hits the surface, a team will be ranked below (Tier 2) all teams that have no touches, and if a team has two touches, it will be ranked below (Tier 3) all teams that had one or no touches.
Tiebreakers are broken in the following order: 1. Single best overall drop; 2. Cord with the greatest elasticity. |
I am wondering if one can relate the KL divergence to the probability of error in a Bayesian binary hypothesis testing setting. That is, we have to decide between hypotheses $A$ and $B$ given observationx, and we have prior probabilities of either being true $p(A)$ and $p(B)$ such that $p(A)+p(B)=1$. Under either $A$ or $B$, observation induces (different) probability distributions $p_A(x)$ and $p_B(x)$.
The setting is identical to a previous question, and this question may be considered its extension. The following restatement of eq. (3) from the previous question lower-bounds the probability of error of the hypothesis test:
$$P_e = \frac{1-||p_A(x)p(A) - p_B(x)p(B)||_1}2\tag{1}$$
where $||\cdot||_1$ denotes the $L_1$ norm. When the prior probabilities are equal, that is $P(A)=P(B)=\frac{1}{2}$, (1) reduces to the expression involving the total variation distance between probability distributions induces by the two hypothesis:
$$P_e = \frac{1-\frac{1}{2}||p_A(x) - p_B(x)||_1}2\tag{2}$$
Pinsker's inequality states that
$$||p_A(x) - p_B(x)||_1\leq\sqrt{2D(p_A(x)||p_B(x))}\tag{3}$$ where $D(p_A(x)||p_B(x))=\sum_x p_A(x)\log\frac{p_A(x)}{p_B(x)}$.
Of course (3) can be used to lower-bound (2) as follows:
$$P_e \geq \frac{1-\frac{1}{\sqrt{2}}\sqrt{D(p_A(x)||p_B(x))}}2\tag{4}$$
However, I am wondering if a similar KL-divergence based lower bound can be derived for (1). Does weighting by the priors work in this case, i.e. is (5) true:
$$||p_A(x)p(A) - p_B(x)p(B)||_1\leq\sqrt{2D(p_A(x)p(A)||p_B(x)p(B))}\tag{5}$$
I don't think it is true. So, is there a KL-divergence based lower-bound for (1)? |
Most of your question is unclear to me, but I the answer to what I think is the core of your question is
yes: For any hermitian operator $A$ and any well-behaved-enough function $f:\mathbb R\to\mathbb R$, it is possible to construct a new operator $f(A)$ which acts in essentially all important respects as the action of $f$ on $A$.
There are different ways to construct $f(A)$, and they depend on exactly what $f$ is, what "well-behaved-enough" means, and how nice $A$ is. In general, the keyword to search for is function of an operator.
For example, if $f$ is analytic in a large enough region - one that includes all the spectrum of $A$, for example, then you can define it as$$f(A)=\sum_{n=0}^\infty \frac{f^{(n)}(a_0)}{n!}(A-a_0)^n.$$
Note, in particular, that this includes functions of the form $A+A^2$, which are perfectly allowed. If $A$ is a linear operator then so is $A^2$, and adding two linear operators is bread and butter in linear algebra. There are a few caveats - for example, if the domain of $A$ is smaller than the Hilbert space then the domain of $A^2$ will typically be smaller, so the sum only makes sense in that restricted domain - but this is only the standard measure of care one needs to take when infinite-dimensional spaces are involved.
Alternatively, if $A$ has an eigenvector expansion as $A=\sum_k a_k|a_k\rangle\langle a_k|$, then you can define$$f(A)=\sum_k f(a_k)|a_k\rangle\langle a_k|.$$If everything behaves well, then both definitions will match.
Finally, note that one should keep an eye on the dimensional analysis of the whole thing. If $A=x$ has dimensions of position, then it does not make sense to add $x+x^2$, any more than it does to do this in classical mechanics. This is particularly the case with, say, exponentials of operators, like the displacement operator$$e^{ix_0\hat p}=\sum_{n=0}^\infty i^n\frac{x_0^n}{n!}\hat p^n$$which only makes sense in units where $\hbar=1$. (Otherwise, you need to add in the $\hbar$ explicitly, as $\exp(ix_0\hat p/\hbar$.) |
Maass forms of weight $0$ with trivial character
A
Maass form on a subgroup \(\Gamma\) of \(\GL_{2}(\R)\)is a smooth, square-integrable, automorphic eigenfunction of the Laplace-Beltrami operator $\Delta$. In other words, $$f\in C^\infty(\mathcal{H}),\quad f\in L^2(\Gamma\backslash{\mathcal H}),\quad f(\gamma z)=f(z)\ \forall\gamma\in\Gamma,\quad (\Delta+\lambda)f(z)=0 \textrm{ for some } \lambda \in \C.$$ Maass forms of weight $k$ with trivial character
A Maass form $f$ of weight $k$ and multiplier system $v$ on a group $\Gamma$ is a smooth function $f:\mathcal{H} \rightarrow\C$ with the following properties:
it transforms according to a unitary weight \( k\) slash-action: $ f|[\gamma,k]( z) = v(\gamma) f(z)$ for all $\gamma \in \Gamma$ where \( f|[\gamma,k]( z) = \exp(-ik\mathrm{Arg}(c z+d)) f(\gamma z)\) for $\gamma=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$
it is an eigenfunction of the corresponding weight $k$ Laplacian \( \Delta_k = y^2\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) - iky\frac{\partial}{\partial x} \)
If the level is equal to 1, the only possible multiplier systems are given by the Dedekind eta function. A compatible multiplier system for weight $k\in \R$ is given by
\[ v(A) = v_{\eta}^{2k}:=\eta(Az)^{2k}/\eta(z)^{2k} \]
or one of its 6 conjugates. One can show that $v(A)$ is well-defined and independent of the point $z$ in the upper half-plane.
Remark:
One can also consider Maass forms transforming with the usual "holomorphic" weight $k$ slash-action. In this case the corresponding weight $k$ Laplacian will look slightly different: \( \Delta_k = y^2\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) - iky\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right). \) This convention is usually used in the context of Harmonic weak Maass forms.
Authors: Knowl status: Review status: reviewed Last edited by Nathan Ryan on 2019-05-01 11:07:32 Referred to by: lfunction.underlying_object mf.maass.mwf.dimension mf.maass.mwf.eigenvalue mf.maass.mwf.fourierexpansion mf.maass.mwf.level mf.maass.mwf.precision mf.maass.mwf.spectralparameter mf.maass.mwf.symmetry mf.maass.mwf.weight rcs.cande.lfunction rcs.rigor.lfunction.maass rcs.source.lfunction.maass lmfdb/lfunctions/templates/MaassformGL2.html (line 5) lmfdb/modular_forms/maass_forms/maass_waveforms/views/templates/mwf_browse_all_eigenvalues.html (line 156) lmfdb/modular_forms/maass_forms/maass_waveforms/views/templates/mwf_navigate.html (line 26) History:(expand/hide all) 2019-05-01 11:07:32 by Nathan Ryan (Reviewed) 2019-04-29 23:35:41 by Nathan Ryan 2018-12-19 06:34:33 by Alex J. Best Differences(show/hide) |
№ 9
All Issues On One Convolution Equation in the Theory of Filtration of Random Processes Abstract
We study the problems of analytic theory and the numerical-analytic solution of the integral convolution equation of the second kind $$ \begin{array}{cc}\hfill {\varepsilon}^2f(x)+{\displaystyle \underset{0}{\overset{r}{\int }}K\left(x-t\right)f(t)dt=g(x),}\hfill & \hfill x\in \left[0,r\right)\hfill \end{array}, $$ where $$ \begin{array}{cccc}\hfill \varepsilon >0,\hfill & \hfill r\le \infty, \hfill & \hfill K\in {L}_1\left(-\infty, \infty \right),\hfill & \hfill K(x)={\displaystyle \underset{a}{\overset{b}{\int }}{e}^{-\left|x\right|s}d\sigma (s)\ge 0.}\hfill \end{array} $$ The factorization approach is used and developed. The key role in this approach is played by the V. Ambartsumyan nonlinear equation.
English version (Springer): Ukrainian Mathematical Journal 66 (2014), no. 8, pp 1220-1235. Citation Example: Barsegyan A. G., Engibaryan N. B. On One Convolution Equation in the Theory of Filtration of Random Processes // Ukr. Mat. Zh. - 2014. - 66, № 8. - pp. 1092–1105. Full text |
I have a short question about the cap product in cohomology.
Let $X$ be a topological space, $R$ a commutative ring with unit $1_R$. We define the cap product on singular chain- and cochain modules $$\cap: C^p(X;R)\otimes _R C_{p+q}(X;R)\to C_q(X;R)$$ by $$\alpha\cap \sigma:=(-1)^{pq}\alpha (\sigma _{|[q,..,p+q]})\sigma _{|[0,..,q]}$$for $\alpha\in C^p(X;R)$ and $\sigma\in C_{p+q}(X;R)$.
This maps induces a map in (co-)homology $$\cap: H^p(X;R)\otimes _R H_{p+q}(X;R)\to H_q(X;R),$$ $$[\alpha]\cap [\sigma]:=[\alpha\cap\sigma ]=[(-1)^{pq}\alpha (\sigma _{|[q,..,p+q]})\sigma _{|[0,..,q]}].$$
Now let $A\subseteq X$, in lecture the professor states that $\cap$ induces a map $$H^p(X,A;R)\otimes _R H_{p+q}(X,A;R)\to H_q(X;R).$$ Here my question is, why he doesn't say that $\cap$ induces a map $$H^p(X,A;R)\otimes _R H_{p+q}(X,A;R)\to H_q(X,A;R)?$$Would it be wrong and if yes, why is it wrong to say that $\cap$ maps to $H_q(X,A;R)$ in this case?
Best |
As I have heard, tangent vector to a smooth manifold $M$ in $p \in M$ is the operator $D_{\xi}$:$f \to D_{\xi}f$, where $f$ is a smooth function $f: M \to R$, with the following properties:
$D_{\xi}(f+g)=D_{\xi}f+D_{\xi}g $
$D_{\xi}(fh)=(D_{\xi}f)h(p)+f(p)(D_{\xi}h) $
So, the basis of the tangent space is:
${\frac{\partial }{\partial x^{1}}}, \frac{\partial }{\partial x^{1}}, ..., \frac{\partial }{\partial x^{n}}$, where ${x^{1},...,x^{n}}$ are coordinates of the local map $(U_{p}, \phi_{p})$.
and any vector in tangent space has the following form:
$\xi^{1}\frac{\partial }{\partial x^{1}}+...+\xi^{n}\frac{\partial }{\partial x^{n}} \in TM_{p}$, where the set $\xi^{1},...,\xi^{n}$ is called the coordinates of tangent vector.
How does this definition relate to intuitive graphical interpretation? |
The force of gravity is constantly being applied to an orbiting object. And therefore the object is constantly accelerating. Why doesn't gravity eventually "win" over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas? I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it?
Newtonian mechanics explains that they
do fall toward the object they're orbiting, they just keep missing.
Quick and dirty derivation for a circular orbit.
Let the primary have mass $M$ and the satellite mass $m$ such that $m \ll M$ (it can also be done for other cases, but this saves on mathiness).
Assume we start with an initial circular orbit on radius $r$, velocity $v = \sqrt{G\frac{M}{r}}$. The acceleration of the satellite due to gravity is $a = G\frac{M}{r^2}$ which means we can also write $v = \sqrt{\frac{a}{r}}$. The period of the orbit is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r}{a}}$.
Chose a coordinate system in which the initial position is $r\hat{i} + 0\hat{j}$ and the initial velocity points in the $+\hat{j}$ direction. Chose a short time $t \ll T$ and lets see how far from the primary the satellite ends up after that time.
If we have chosen $t$ short enough, we can approximate gravity as having uniform strength through the time period (and we shall show later that that
is justified).
The new position is $\left(r - \frac{1}{2}at^2\right)\hat{i} + vt\hat{j}$ which lies at a distance $$ r_2 = \sqrt{r^2 - r a t^2 + \frac{1}{4}a^2 t^4 + v^2 t^2} $$ pulling our at factor of $r$ we get $$ r_2 = r \sqrt{1 - \frac{a}{r} t^2 + \frac{1}{4}\frac{a^2}{r^2} t^4 + \frac{v^2}{r^2} t^2} $$ and converting all the $\frac{a}{r}$ and $\frac{v}{r}$ terms into expressions of the period we get $$ r_2 = r \sqrt{1 - \left(2\pi\frac{t}{T}\right)^2 + \frac{1}{4}\left(2\pi\frac{t}{T}\right)^4 + \left(2\pi\frac{t}{T}\right)^2}$$ Finally, we drop the $(t/T)^4$ term as negligible and note that the $(t/T)^2$ terms cancel so the result is $$r_2 = r$$ or the radius never changed (which justified the constant magnitude for acceleration, and a small enough $t$ justifies both the constant direction and the dropping of the fourth degree term).
The force of gravity has little to do with friction. As dmckee says, what is happening is that the body falls, but precisely because it has enough momentum, it falls around the object towards which it gravitates instead of into it. Of course, this is not always the case, collisions do happen. Also systems of astronomical bodies are complicated and the combined effect of the action of several different bodies on one can destabilize trajectories that in a simple 2-body case would be stable ellipses. The result could be collision or escape of the body.
In the 2-body case however, the crucial aspect of gravity which guarantees the stability of the system is the fact that gravitation is a
centripetal force. It always acts towards the center of the other gravitating mass. One can show that this feature implies the conservation of angular momentum, which means that if the 2-body system had some angular momentum to begin with, it will keep the same angular momentum indefinitely.
(Extra note, even in the 2-body case, there can be collisions and escape to infinity, the first if there is not enough angular momentum (for instance one body having velocity directed towards the other body, like an apple falling from a tree), the other if there is too much angular momentum, resulting in parabolic or hyperbolic trajectories.)
I always think of it this way: gravity and the centrifugal force generated by the orbiting object are exactly in balance. If you tie a rope to an object and spin it around you, the centrifugal force will pull on the rope. You pull back with the same force to keep the object orbiting around you. That's exactly what gravity does to orbiting objects.
You can also see that the speed of the object forces it into a particular orbit. If the object slows down, it will fall and either reach a new lower orbit, or crash into the object they're rotating around. |
So not quite sure what you're missing, but here's how you go about doing this sort of thing.
So first, I am assuming this is 1D due to your description. Second, I'm assuming you know the relationship between points in the physical domain, $x$, and the computational domain, $\xi$, something along the lines of $x=x(\xi)$. Given you have this relationship, you can do the following for some function $\phi(\cdot)$:
$$\begin{align}\frac{\partial \phi}{\partial x} &= \frac{\partial \phi}{\partial \xi} \frac{\partial \xi}{\partial x}\\\frac{\partial \phi}{\partial x} &= \frac{\partial \phi}{\partial \xi} \left(\frac{\partial x}{\partial \xi}\right)^{-1}\\\end{align}$$
With this representation, since you know $x(\xi)$, you can produce the second term exactly in the multiplication. So now you just need to approximate the first term, $\frac{\partial \phi}{\partial \xi}$. This term can be approximated using normal finite difference schemes. Using a central difference, for example, you could get the following:
$$\frac{\partial \phi}{\partial x} = \left(\frac{\phi_{i+1}-\phi_{i-1}}{2\Delta \xi}\right) \left(\frac{\partial x}{\partial \xi}\right)^{-1}$$
In this case, $\phi_{i}$ is associated to both $x_i$ and $\xi_i$, where $\phi_i = \phi(x_i)$, and where, in addition, $\phi_i = \phi(x(\xi_i))$. This should help you understand how to compute the necessary quantity you're after. |
I've been trying to convince myself that "coherent sheaf" is a natural definition. One way I might be satisfied is the following: for modules over a Noetherian ring $A$, coherent and finitely presented modules agree. For quasicoherent sheaves over a locally Noetherian scheme $X$, coherent and locally finitely presented sheaves agree. In general, coherent sheaves are locally finitely presented, and hence they pull back along morphisms $f : X \to Y$ where $X$ is locally Noetherian.
Is this already enough information to tell me what coherent sheaves must be?
More precisely, if $Y$ is a scheme, let $N_Y$ be the category whose objects are pairs $(X, f)$ of a locally Noetherian scheme and a morphism $f : X \to Y$ and whose morphisms are commuting triangles. The
category of coherent sheaves $\text{Coh}(N_Y)$ on $N_Y$ is the category whose objects are assignments, to each $(X, f) \in N_Y$, of a coherent sheaf $F_X \in \text{Coh}(X)$ and assignments, to each morphism $g : (X_1, f_1) \to (X_2, f_2)$ in $N_Y$, of an isomorphism
$$F_{X_1} \cong g^{\ast} F_{X_2}$$
satisfying the obvious compatibility condition, with the obvious notion of morphism. Since coherent sheaves are locally finitely presented, they pull back to locally finitely presented sheaves in a way satisfying the obvious compatibility conditions, so there is a natural functor
$$\text{Coh}(Y) \to \text{Coh}(N_Y)$$
given by taking pullbacks along all morphisms $f : X \to Y$ where $X$ is locally Noetherian. The more precise version of my question is:
Is this functor an equivalence?
I think there is a slicker way to ask this using descent and another slicker way to ask this using Kan extensions, but I'll refrain from both to be on the safe side. If the above is true, I'd also be interested in knowing to what extent I can restrict "locally Noetherian" to a smaller subcategory. Does it suffice to use Noetherian schemes? Affine Noetherian schemes? Over an affine Noetherian base $\text{Spec } S$, does it suffice to use $\text{Spec } R$ where $R$ is finitely generated over $S$? |
Answer
5 seconds
Work Step by Step
We use the equation for period to find the length of half of a period: $\frac{1}{2}T = \pi \sqrt{\frac{L}{g}} = \pi \sqrt{\frac{25}{9.81}} \approx 5 s$
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LaTeX is a great tool for printable professional-looking documents, but can be also used to generate PDF files with excellent navigation tools. This article describes how to create hyperlinks in your document, and how to set up LaTeX documents to be viewed with a PDF-reader.
Contents
Let's start with a minimal working example, by simply importing the
hyperref package all cross-referenced elements become hyperlinked.
\documentclass{book} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{hyperref} \begin{document} \frontmatter \tableofcontents ... \end{document}
The lines in the table of contents become links to the corresponding pages in the document by simply adding in the preamble of the document the line
\usepackage{hyperref} One must be careful when importing hyperref. Usually, it has to be the last package to be imported, but there might be some exceptions to this rule.
The default formatting for links can be changed so the information in your documents is more clearly presented. Below you can see an example:
\documentclass{book} \usepackage[utf8]{inputenc} \usepackage[english]{babel} \usepackage{hyperref} \hypersetup{ colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan, } \urlstyle{same} \begin{document} \tableofcontents \chapter{First Chapter} This will be an empty chapter and I will put some text here \begin{equation} \label{eq:1} \sum_{i=0}^{\infty} a_i x^i \end{equation} The equation \ref{eq:1} shows a sum that is divergent. This formula will later be used in the page \pageref{second}. For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} or open the next file \href{run:./file.txt}{File.txt} It's also possible to link directly any word or \hyperlink{thesentence}{any sentence} in your document. \end{document}
This is a complete example, it will be fully explained in the rest of the article. Below is a description of the commands related to the colour and styling of the links.
\hypersetup{ ... }
\colorlinks=true
\linkcolor=blue
\filecolor=magenta
\urlcolor=cyan
\urlstyle{same}
Links to a web address or email can added to a LaTeX file using the
\url command to display the actual link or
\href to use a hidden link and show a word/sentence instead.
For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com}
There are two commands in the example that generate a link in the final document:
\href{http://www.sharelatex.com}{Something Linky}
\url{http://www.sharelatex.com}
The commands
\href and
\url presented in the previous section can be used to open local files
For further references see \href{http://www.sharelatex.com}{Something Linky} or go to the next url: \url{http://www.sharelatex.com} or open the next file \href{run:./file.txt}{File.txt}
The command
\href{run:./file.txt}{File.txt} prints the text "File.txt" that links to a local file called "file.txt" located in the current working directory. Notice the text "run:" before the path to the file.
The file path follows the conventions of UNIX systems, using . to refer the current directory and .. for the previous directory.
The command
\url{} can also be used, with the same syntax described for the path, but it's reported to have some problems.
It was mentioned before that all cross-referenced elements become links once
hyperref is imported, thus we can use
\label anywhere in the document and refer later those labels to create links. This is not the only manner to insert hyperlinks manually.
It's also possible to link directly any word or \hyperlink{thesentence}{any sentence} in you document. If you read this text, you will get no information. Really? Is there no information? For instance \hypertarget{thesentence}{this sentence}.
There are two commands to create user-defined links.
\hypertarget{thesentence}{this sentence}
\hyperlink{thesentence}{any sentence}
Links in a document are created having in mind a document that will be read in PDF format. The PDF file can be further personalized to add additional information and change the way the PDF viewer displays it. Below an example:
\hypersetup{ colorlinks=true, linkcolor=blue, filecolor=magenta, urlcolor=cyan, pdftitle={Sharelatex Example}, bookmarks=true, pdfpagemode=FullScreen, }
Using the command
\hypersetup, described in the section styles and colours, accepts extra parameters to set up the final PDF file.
pdftitle={Sharelatex Example}
bookmarks=true
pdfpagemode=FullScreen
See the reference guide for a full list of options that can be passed to
\hypersetup.
Linking style options
Option Default value Description
hyperindex
true Makes the page numbers of index entries into hyperlinks
linktocpage
false Makes the page numbers instead of the text to be link in the Table of contents.
breaklinks
false Allows links to be broken into multiple lines.
colorlinks
false Colours the text for links and anchors, these colours will appear in the printed version
linkcolor
red Colour for normal internal links
anchorcolor
black Colour for anchor (target) text
citecolor
green Colour for bibliographical citations
filecolor
cyan Colour for links that open local files
urlcolor
magenta Colour for linked URLs
frenchlinks
false Use small caps instead of colours for links PDF-specific options
Option Default value Description
bookmarks
true Acrobat bookmarks are written, similar to the table of contents.
bookmarksopen
false Bookmarks are shown with all sub-trees expanded.
citebordercolor
0 1 0 Colour of the box around citations in RGB format.
filebordercolor
0 .5 .5 Colour of the box around links to files in RGB format.
linkbordercolor
1 0 0 Colour of the box around normal links in RGB format.
menubordercolor
1 0 0 Colour of the box around menu links in RGB format.
urlbordercolor
0 1 1 Colour of the box around links to URLs in RGB format.
pdfpagemode
empty Determines how the file is opened. Possibilities are UseThumbs (Thumbnails), UseOutlines (Bookmarks) and FullScreen.
pdftitle
Sets the document title.
pdfauthor
Sets the document Author.
pdfstartpage
1 Determines on which page the PDF file is opened.
For more information see |
(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) |
Is there ever a practical difference between the notions induction and strong induction?
Edit: More to the point, does anything change if we take strong induction rather than induction in the Peano axioms?
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The terms "weak induction" and "strong induction" are not commonly used in the study of logic. The terms are commonly used only in books aimed at teaching students how to write proofs.
Here are their prototypical symbolic forms:
weak induction: $(\Phi(0) \land (\forall n) [ \Phi(n) \to \Phi(n+1)]) \to (\forall n) \Phi(n)$
strong induction: $(\Psi(0) \land (\forall n) [ (\forall m \leq n) \Psi(m) \to (\forall m \leq n+1) \Psi(m)]) \to (\forall n) \Psi(n)$
The thing to notice is that "strong" induction is almost exactly weak induction with $\Phi(n)$ taken to be $(\forall m \leq n)\Psi(n)$. In particular, strong induction is not actually stronger, it's just a special case of weak induction modulo some trivialities like replacing $\Psi(0)$ with $(\forall m \leq 0 )\Psi(m)$. Of course you can write variations of the symbolic forms, but the same point applies to all of them: "strong" induction is essentially just weak induction whose induction hypothesis has a bounded universal quantifier.
So the question is not why we still have "weak" induction - it's why we still have "strong" induction when this is not actually any stronger.
My opinion is that the reason this distinction remains is that it serves a pedagogical purpose. The first proofs by induction that we teach are usually things like $\forall n \left [ \sum_{i=0}^n i = n(n+1)/2 \right ]$. The proofs of these naturally suggest "weak" induction, which students learn as a pattern to mimic.
Later, we teach more difficult proofs where that pattern no longer works. To give a name to the difference, we call the new pattern "strong induction" so that we can distinguish between the methods when presenting a proof in lecture. Then we can tell a student "try using strong induction", which is more helpful than just "try using induction".
In terms of logical strength in formal arithmetic, as you can see above, the two forms are equivalent over some weak base theory as long as you are looking at induction for a class of formulas that is closed under bounded universal number quantification. In particular, all the syntactic classes of the analytical and arithmetical hierarchies have that property, so weak induction for $\Sigma^0_k$ formulas is the same as strong induction for $\Sigma^0_k$ formulas, weak induction for $\Pi^1_k$ formulas is the same as strong induction for $\Pi^1_k$ formulas, and so on. This equivalence will hold under any reasonable formalization of "strong" induction - I chose mine above to make the issue particularly obvious.
Addendum I was asked in a comment why $$(1)\colon (\forall t)[(\forall m < t)\Phi(m) \to \Phi(t)]$$implies$$(2)\colon (\forall n)[(\forall m \leq n)\Phi(m) \to (\forall m \leq n+1) \Phi(m)].$$I'm going to give a relatively formal proof to show how it goes. The proof is not by induction, instead it just uses universal generalization to prove the universally quantified statements.
For the proof, I will assume (1) and prove (2). Working towards that goal, I fix a value of $n$ and assume: $$ (3)\colon (\forall m \leq n)\Phi(m). $$
I first want to prove $(\forall m < n+1)\Phi(m)$, which is an abbreviation for $(\forall m)[m < n+1 \to \Phi(m)]$. Pick an $m$. If $m < n+1$, then $m \leq n$, so I know $\Phi(m)$ by assumption (3). So, by universal generalization, I obtain $(\forall m < n+1)\Phi(m)$.
Next, note that a substitution instance of (1) gives $(\forall m < n+1)\Phi(m) \to \Phi(n+1)$. I have proved $(\forall m < n+1)\Phi(m)$ so I can assert $\Phi(n+1)$.
So now I have assumed $(\forall m \leq n)\Phi(m)$ and I have also proved $\Phi(n+1)$. Another proof by cases establishes $(\forall m \leq n+1)\Phi(m)$.
By examining the proof, you can see which axioms I need in my weak base theory. I need at least the following two axioms:
$(m \leq t) \leftrightarrow (m < t) \lor (m=t)$
$(m < t+1) \to (m \leq t)$
I believe those are the only two axioms I used in the proof.
One way of looking at the question is to say that all induction is over some well-founded set, and strong induction is useful when the order type of that well-founded set is not the same as that of the natural numbers. Rather than elaborate on this general remark, let me simply give two examples of slightly more sophisticated induction that would appear in many undergraduate courses.
The first is the statement that every positive integer has a prime factorization. Here if we are trying to factorize n it is convenient to assume that all smaller numbers can be factorized, since if n is not prime then we will not use the factorization of n-1 but rather the factorizations of two factors of n about which we know nothing. In a sense we could say that the well-founded set that underlies this proof is N with the relation "is a proper factor of". (The simplest proof that this is well founded is to use the fact that if a is a proper factor of b then a is less than b, though once we know the fundamental theorem of arithmetic then this seems slightly unnatural. But we don't want to assume the fundamental theorem of arithmetic in order to prove the easy part of the fundamental theorem of arithmetic.)
The second is the statement that every tree with n vertices has n-1 edges. Here the proof is that every tree must have at least one vertex of degree 1 (or you could keep following a path until you hit a point you've hit before -- to make this completely rigorous will itself involve induction but let's forget that one) and that if you remove that vertex and its edge then you must still have a tree, which by induction has n-2 edges. Here we could either do straightforward induction on the number of vertices or we could do induction on the more complicated well founded set of all graphs under strict containment. The advantage of the first is that it is straightforward induction, but the disadvantage is that we have to create a slightly artificial inductive hypothesis -- that all trees with n vertices have n-1 edges. Graph theory beginners often trip up here and try to prove results like this by arguing that if you take a tree and add a new vertex to it, joining it to one of the existing vertices, then by induction the old tree has n-2 edges so the new one must have n-1 edges. In other words, they assume the result for n-1 and prove it for n, but unfortunately the result they assume for n-1 is not the right one and strictly speaking all they prove is that for each n there is a tree with n vertices and n-1 edges.
The second example isn't exactly strong induction, but I think it fits into the general discussion and helps to give some idea of what forms of induction are appropriate when.
Induction and strong induction, or what I would prefer to call the least number principle, are not equivalent formula by formula relative to, say, the standard algebraic axioms $\text{PA}^-$ underlying PA expressing that the structure is a discretely ordered semiring whose least element is $0$. However, induction for the class of all $\Sigma_n$ formulas is equivalent to the least number principle for the class of all $\Sigma_n$ formulas for each $n$.
Let me give a few details.
For a formula $\phi(x)$ define $$I(\phi) := (\phi(0) \& (\forall x)(\phi(x) \to \phi(x+1))) \to (\forall x)\phi(x)$$
while $$L(\phi) := (\exists x) \neg \phi(x) \to (\exists y)(\neg \phi(y) \& (\forall z < y) \phi(z))$$
It is easy to see that $\text{PA}^- \vdash L(\phi) \to I(\phi)$ while $\text{PA}^- \vdash I(\tilde{\phi}) \to B(\phi)$ where $\tilde{\phi}(x) := (\forall y \leq x) \phi(y)$.
If $\phi$ is of class $\Sigma_n$ (expressible as a formula with $n$ alternations of unbounded quantifiers starting with $\exists$ and then a string of bounded quantifiers followed by a quantifier free formula) then while $\tilde{\phi}$ not explicitly $\Sigma_n$, it is equivalent to a $\Sigma_n$ formula. Thus, relative to $\text{PA}^-$, we have a level by level equivalence of these principles.
It is fairly easy to see by considering structures that are not well-ordered that no formula by formula equivalence can be expected.
This is in response to Andrej Bauer's comment on Ricky Derner post, as the answer does not seem to fit in the comment section.
Suppose we take out the Induction Schema and replace it with: $$(SI)\quad \forall k\Bigl(\bigl(\forall n(n\lt k\rightarrow \phi(n))\bigr)\longrightarrow \phi(k)\Bigr)\Rightarrow \forall m(\phi(m)).$$
Now, consider the theory that has the first four Peano Axioms and (SI), instead of the usual Induction schema. The statement "Every natural number is either $0$ or a successor" is a theorem under the usual Peano Axioms, but is not a theorem under this modified axiomatic system. To see this, take $\omega+\omega$ as a model. It satisfies the first four axioms using the usual ordinal successor function, and it satisfies the "strong induction" schema (SI) as well. However, $\omega$ itself, as an element of $\omega+\omega$, is neither a successor nor $0$. (Think of having two copies of the natural numbers, a "red" copy going first and a "blue" copy going second; then the "blue $0$" is neither $0$ nor a successor; you can apply regular induction to the proposition "$n$ is red", but the set you obtain is not all of your set of numbers).
So the two theories are not equivalent. (SI) is a theorem in Peano Arithmetic, but regular induction is not a theorem in the theory that has the first four Peano Axioms and (SI).
However, if you take the first four Peano Axioms, you add "Every number is either $0$ or a successor" as a "fourth-and-a-half" axiom, and then you take (SI) instead of the usual Induction Schema, then you can prove the usual Induction Schema as a theorem in this system; so the two systems (usual Peano Axioms, and first four Axioms plus the "fourth-and-a-half" axiom plus (SI)) are equivalent. If you want (SI) to be equivalent to regular induction, you need a bit more than just the first four Peano Axioms.
If we take strong induction rather than induction in the Peano axioms, we would also need to add an axiom stating that every natural is either zero or a successor. |
In the book "Quantum Field theory and the Standard Model" by Matthew Schwartz, page 23-24, the position space wavefunction is defined as
$$\psi(x)=\langle 0|\phi(x)|\psi\rangle, \tag{2.82+2.83}$$
where $|\psi\rangle$ is any state in the Fock space. Then he uses the equations (i) $\partial_t^2\phi_0=(\nabla^2-m^2)\phi_0$ (i.e., the Klein-Gordon equation for the free massive scalar field $\phi_0(\textbf{x},t)$) and (ii) $[H,\phi_0]=-i\partial_t\phi_0$ to derive Eq. 2.85, in following 3 steps:
$$i\langle 0|\partial_t\phi_0(\textbf{x},t)|\psi\rangle=\langle 0|\int \frac{d^3\textbf{p}}{(2\pi)^3}\frac{\sqrt{\textbf{p}^2+m^2}}{\sqrt{2\omega_\textbf{p}}}(a_p e^{-ip\cdot x}-a^\dagger_p e^{ip\cdot x})|\psi\rangle\\ =\langle 0|\sqrt{m^2-\nabla^2}\phi_0(\textbf{x},t)|\psi\rangle.\tag{2.85}$$
This equation is used to successfully derive the Schrodinger equation in quantum mechanics for the state $\psi(\textbf{x},t)\equiv \langle 0|\phi_0(\textbf{x},t)|\psi\rangle$.
The first equality follows from differentiation of $\phi_0(x)$ w.r.t $t$. How does the second equality follow from the first?
How are the inputs (i) and (ii) are utilized to derive the second equality from the first?
Is it true that the operator $$i\partial_t\sim\sqrt{m^2-\nabla^2}~?$$ If yes, what is the use of doing the middle step? |
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Now showing items 21-30 of 51
D-meson nuclear modification factor and elliptic flow measurements in Pb–Pb collisions at $\sqrt {s_{NN}}$ = 5.02TeV with ALICE at the LHC
(Elsevier, 2017-11)
ALICE measured the nuclear modification factor ($R_{AA}$) and elliptic flow ($\nu_{2}$) of D mesons ($D^{0}$, $D^{+}$, $D^{⁎+}$ and $D^{s+}$) in semi-central Pb–Pb collisions at $\sqrt{s_{NN}} =5.02$ TeV. The increased ...
Charmonium production in Pb–Pb and p–Pb collisions at forward rapidity measured with ALICE
(Elsevier, 2017-11)
The ALICE collaboration has measured the inclusive charmonium production at forward rapidity in Pb–Pb and p–Pb collisions at sNN=5.02TeV and sNN=8.16TeV , respectively. In Pb–Pb collisions, the J/ ψ and ψ (2S) nuclear ...
Net-baryon fluctuations measured with ALICE at the CERN LHC
(Elsevier, 2017-11)
First experimental results are presented on event-by-event net-proton fluctuation measurements in Pb- Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV, recorded by the ALICE detector at the CERN LHC. The ALICE detector is well ...
Investigations of anisotropic collectivity using multi-particle correlations in pp, p-Pb and Pb-Pb collisions
(Elsevier, 2017-11)
Two- and multi-particle azimuthal correlations have proven to be an excellent tool to probe the properties of the strongly interacting matter created in heavy-ion collisions. Recently, the results obtained for multi-particle ...
$\phi$ meson production in Pb-Pb collisions at $\sqrt s_{NN}$=5.02 TeV with ALICE at the LHC
(Elsevier, 2017-11)
Strangeness production is a key tool to understand the properties of the medium formed in heavy-ion collisions: an enhanced production of strange particles was early proposed as one of the signatures of the Quark-Gluon ...
Jet-hadron correlations relative to the event plane at the LHC with ALICE
(Elsevier, 2017-11)
In ultra relativistic heavy-ion collisions at the Large Hadron Collider (LHC), conditions are met to produce a hot, dense and strongly interacting medium known as the Quark Gluon Plasma (QGP). Quarks and gluons from incoming ...
Measurements of the dielectron continuum in pp, p-Pb and Pb-Pb collisions with ALICE at the LHC
(Elsevier, 2017-11)
Dielectrons produced in ultra-relativistic heavy-ion collisions provide a unique probe of the whole system evolution as they are unperturbed by final-state interactions. The dielectron continuum is extremely rich in physics ...
Exploring jet substructures with jet shapes in ALICE
(Elsevier, 2017-11)
The characterization of the jet substructure can give insight into the microscopic nature of the modification induced on high-momentum partons by the Quark-Gluon Plasma that is formed in ultra-relativistic heavy-ion ...
Measurements of the nuclear modification factor and elliptic flow of leptons from heavy-flavour hadron decays in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 and 5.02 TeV with ALICE
(Elsevier, 2017-11)
We present the ALICE results on the nuclear modification factor and elliptic flow of electrons and muons from open heavy-flavour hadron decays at mid-rapidity and forward rapidity in Pb--Pb collisions at $\sqrt{s_{\rm NN}}$ ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ... |
While preparing some lecture notes, I had a basic point of confusion come up that I haven't been able to settle.
The $BP$-Adams spectral sequence (or $p$-local Adams-Novikov spectral sequence) for the sphere begins with $E_2$-page $$E_2^{*, *} = \operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, BP_*)$$ and converges to $\pi_* \mathbb{S} \otimes_{\mathbb{Z}} \mathbb{Z}_{(p)}$.
There are a variety of cool periodicities visible in this $E_2$-page, which we can organize via the following secondary spectral sequence. There is an ascending chain of $(BP_* BP)$-invariant ideals for $BP_*$ given by $I_r = (p, v_1, \ldots, v_{r-1})$, connected to one another by the short exact sequences $$0 \to BP_* / I_r^\infty \to v_r^{-1} BP_* / I_r^\infty \to BP_* / I_{r+1}^\infty \to 0.$$ The quotient $BP_* / I_r^\infty$ is thought of as the closed substack of the moduli of formal groups detected by the ideal sheaf corresponding to $I_r$ together with its formal neighborhood inside the parent stack. Applying $\operatorname{Ext}$ and stringing the resulting long exact sequences together, one arrives at the (trigraded) chromatic spectral sequence (CSS): $$E_1^{r, *, *} = \operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, v_r^{-1} BP_* / I_r^\infty) \Rightarrow \operatorname{Ext}^{*, *}_{BP_* BP}(BP_*, BP_*).$$ Much of the fun in chromatic homotopy theory after this point comes from identifying the groups in this $E_1$-page as other sorts of things, like certain group cohomologies.
Shifting gears somewhat, Bousfield localization at the Johnson-Wilson $E(r)$-theories and the Morava $K(r)$-theories is meant to perform the same organization at the level of homotopy types. The spectra $E(\infty)$ and $E(0)$ correspond to $BP$ and to $H\mathbb{Q}$ respectively, so the sequence of localization functors $L_{E(r)}$ are meant to interpolate between rational homotopy theory and the sort of homotopy theory visible to the $p$-local Adams-Novikov spectral sequence.
There are two ways to study these functors as $r$ increases. First, there there is a natural map $L_{E(r)} X \to L_{E(r-1)} X$. Its homotopy fiber detects the difference between these two spectra, denoted $M_r X$ and called the $r$th monochromatic layer of $X$. Second, there is a pullback square, dubbed chromatic fracture: $$\begin{array}{ccc} L_{E(r)} X & \to & L_{K(r)} X \\ \downarrow & & \downarrow \\ L_{E(r-1)} X & \to & L_{E(r-1)} L_{K(r)} X. \end{array}$$ In both of these situations, you can hope to inductively study the filtering spectra $L_{E(r)} X$ by studying the "filtration layers", which are either $M_r X$ or $L_{K(r)} X$ depending upon your approach.
My question is:How exactly do these two approaches connect to the chromatic spectral sequence?
I suspect that the CSS for $L_{E(R)} X$ looks like the CSS for $X$, after quotienting out the information in $r$-degrees $r > R$. I also suspect that the CSS for one of the two of $M_R X$ and $L_{K(R)} X$ looks like that for $L_{E(R)} X$, after additionally quotienting out the information in $r$-degrees $r < R$. However, I can't seem to make the pieces line up. For instance, Prop. 7.4 of Hopkins, Mahowald, and Sadofsky's
Constructions of elements in Picard groups suggests that this description holds for $L_{K(R)} \mathbb{S}$, as that statement matches their Adams-Novikov spectral sequence converging to $\pi_* L_{K(R)} \mathbb{S}$ --- just as one would expect from a collapsing chromatic spectral sequence. On the other hand, the bottom corner of the fracture square is of the form $L_{E(R-1)} L_{K(R)} X$, and this description seems to say that its CSS is empty, which doesn't sound right.
I'd appreciate someone setting me straight about this. Thanks! |
I work on modelling high intensity discharge xenon-filled lamps. The model governing the discharge is quite complex and sadly includes fluid dynamics. After some time, I managed to implement a finite-difference particle-tracking lagrangian fluid dynamics for axisymmetric 1D discharge:
$$ \frac{\partial}{\partial t} \left(\frac{1}{\rho}\right) = \frac{\partial(rv)}{\partial s} \\ \frac{\partial v}{\partial t} = - r \frac{\partial p}{\partial s} \\ \frac{\partial r}{\partial t} = v \\ \frac{\partial}{\partial t} \left( \mathcal{E} + \frac{v^2}{2}\right) = - \frac{\partial}{\partial s}(prv) - \frac{\partial(rW)}{\partial s} + \frac{f}{\rho} \\ W = -\lambda \frac{\partial T}{\partial r} \\ p = p(\rho, T) \\ \frac{\partial\mathcal{E}}{\partial T} = c_v(\rho, T) $$
Computational cells are organised so that lagrangian particle speeds and eulerian coordinates are defined on cell-interfaces and energy equation parameters (temperature, density, pressure, thermal capacity, thermal conductivity) are defined on cell centers (image below). Domain boundaries coincide with cell interfaces. .
Right-side boundary conditions are dirichlet for speed $$v(r)|_{r=R} = 0$$, keeping the scheme homogenous. For energy balance equation it gets a bit trickier as there is no boundary node to specify dirichlet condition at, so I intruduce fictious node in the depth of the wall material and interpolate boundary condition at cell interface:
$$ \frac{\hat{T}_{last} + \hat{T}_{last + 1}}{2} = T_{boundary} \\->\\ \hat{T}_{last + 1} = 2T_{boundary} - \hat{T}_{last} $$
For computational efficiency reason I solve the equations separately. There's iterational process on energy equation; first I solve energy equation with implicit O(t^2)O(h^2) scheme, then I advance gas dynamics part on interpolated temperature profile with explicit scheme with time step several orders smaller to refine coefficients for energy equation. This is dictated by the extremely high cost to recalculate energy source function as there's radiative transfer over complex spectra involved.
Model works great for pulses with current of about 500A and duration greater than 50e-6 seconds, however once I go into kA ranges waves rebounding from the lamp wall tend to spuriously oscillate and destroy the model upon converging in the center ().
I tried some less radical methods of dealing with these oscillations like introducing artificial viscosity, however I found that my implementation of them greatly reduced stability due to loss of conservation.
With great dread I look forward to rewriting everything from scratch in terms of TVD scheme or with nested grids or with something equally time-consuming, and my question is: what are some less intrusive approaches to eliminating spurious oscillations that I overlooked?
Edit 1: added more information about energy balance equation, schema, and boundary conditions used. |
The Annals of Probability Ann. Probab. Volume 23, Number 4 (1995), 1523-1556. The "True" Self-Avoiding Walk with Bond Repulsion on $\mathbb{Z}$: Limit Theorems Abstract
The "true" self-avoiding walk with bond repulsion is a nearest neighbor random walk on $\mathbb{Z}$, for which the probability of jumping along a bond of the lattice is proportional to $\exp(-g \cdot$ number of previous jumps along that bond). First we prove a limit theorem for the distribution of the local time process of this walk. Using this result, later we prove a local limit theorem, as $A \rightarrow \infty$, for the distribution of $A^{-2/3}X_{\theta_{s/A}}$, where $\theta_{s/A}$ is a random time distributed geometrically with mean $e^{-s/A}(1 - e^{-s/A})^{-1} = A/s + O(1)$. As a by-product we also obtain an apparently new identity related to Brownian excursions and Bessel bridges.
Article information Source Ann. Probab., Volume 23, Number 4 (1995), 1523-1556. Dates First available in Project Euclid: 19 April 2007 Permanent link to this document https://projecteuclid.org/euclid.aop/1176987793 Digital Object Identifier doi:10.1214/aop/1176987793 Mathematical Reviews number (MathSciNet) MR1379158 Zentralblatt MATH identifier 0852.60083 JSTOR links.jstor.org Subjects Primary: 60F05: Central limit and other weak theorems Secondary: 60J15 60J55: Local time and additive functionals 82C41: Dynamics of random walks, random surfaces, lattice animals, etc. [See also 60G50] Citation
Toth, Balint. The "True" Self-Avoiding Walk with Bond Repulsion on $\mathbb{Z}$: Limit Theorems. Ann. Probab. 23 (1995), no. 4, 1523--1556. doi:10.1214/aop/1176987793. https://projecteuclid.org/euclid.aop/1176987793 |
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Forward-backward multiplicity correlations in pp collisions at √s = 0.9, 2.76 and 7 TeV
(Springer, 2015-05-20)
The strength of forward-backward (FB) multiplicity correlations is measured by the ALICE detector in proton-proton (pp) collisions at s√ = 0.9, 2.76 and 7 TeV. The measurement is performed in the central pseudorapidity ...
Rapidity and transverse-momentum dependence of the inclusive J/$\mathbf{\psi}$ nuclear modification factor in p-Pb collisions at $\mathbf{\sqrt{\textit{s}_{NN}}}=5.02$ TeV
(Springer, 2015-06)
We have studied the transverse-momentum ($p_{\rm T}$) dependence of the inclusive J/$\psi$ production in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV, in three center-of-mass rapidity ($y_{\rm cms}$) regions, down to ...
Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at $\sqrt{s}$ = 7 TeV
(Springer, 2015-09)
Prompt D meson and non-prompt J/$\psi$ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of $\sqrt{s}=7$ TeV. The results ...
Coherent $\rho^0$ photoproduction in ultra-peripheral Pb-Pb collisions at $\mathbf{\sqrt{\textit{s}_{\rm NN}}} = 2.76$ TeV
(Springer, 2015-09)
We report the first measurement at the LHC of coherent photoproduction of $\rho^0$ mesons in ultra-peripheral Pb-Pb collisions. The invariant mass and transverse momentum distributions for $\rho^0$ production are studied ...
Inclusive, prompt and non-prompt J/ψ production at mid-rapidity in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2015-07-10)
The transverse momentum (p T) dependence of the nuclear modification factor R AA and the centrality dependence of the average transverse momentum 〈p T〉 for inclusive J/ψ have been measured with ALICE for Pb-Pb collisions ... |
In Hajime Urakawa's monograph
The Spectral Geometry of the Laplacian on page 41, we make an assumption that I can't quite justify on my own. The following is our setup:
Let $(M^n,g)$ be a closed connected Riemannian manifold, with $\Delta_g$ the Laplacian (a negative operator), and $0=\lambda_0<\lambda_1\le\lambda_2\le\cdots$ the eigenvalues of $-\Delta_g$ with corresponding eigenfunctions $u_0=\hbox{const},u_1,u_2,u_3,\ldots$. Then for $N\gg0$, $$\iota:M\ni x\mapsto(u_1(x),\ldots,u_N(x))\in\mathbb R^N$$ defines an embedding.
But why is this true? I know that $(u_i)_{i\ge0}$ forms an orthonormal basis for $L^2(M,\mu_g)$, but this doesn't seem to be quite enough to prove that we can have an embedding. Presumably we need to show that we can also approximate functions in $H^1(M,g)$? |
Path integral via discretization
So let me start with what seems to be the point of view of physicists (corrections are highly appreciated since this is what I understood!). Let a quantum system with coordinates $q_a$ and momenta $p_b$ be given satisfying commutation relations $$[q_a,p_b]=i\delta_{ab}.$$
Further suppose the system has a Hamiltonian $H$ which is time-independent. Usually in that setting the path integral is introduced as a means to compute the transition amplitude
$$\langle q',t'|q,t\rangle=\langle q'|e^{-iH(t'-t)}|q\rangle=\int\mathfrak{D}x(t) \exp\left\{iS[x(t)]\right\}$$
This is usually defined by a discretization procedure allied to a Wick rotation to Euclidean time $\tau = it$ to deal with convergence. The right discretization seems to be derived by slicing the time interval, evaluating $\langle q',t'|q,t\rangle$ to first order in $t'-t$, and imposing some ordering convention. So, for example with a Lagrangian $L = T - V$ the above integral would be
defined as something of the form
$$\int\mathfrak{D}x(t) \exp\left\{iS[x(t)]\right\}=\lim_{N\to \infty} C_N\int \prod_{k=1}^N dx_k \exp \left\{i\sum_{k=1}^N \frac{m}{2}\frac{(x_k-x_{k-1})^2}{\epsilon_N^2}-V(x_k)\right\}\tag{1}$$
So: a path integral in Physics is
defined by the continuum limit of these aforementioned discretizations. Cylinder set measures
Now there's the mathematicians point of view on which one studies integration over locally convex vector spaces which are infinite dimensional. In that case, if $E$ is such a space we perform two definitions:
Definition: Let $E$ be a locally convex vector space. A cylinder setis defined to be a subset $C\subset E$ of the form $$C=\{x\in E : (\ell_1(x),\dots,\ell_n(x))\in C_0\}$$ where $C_0\subset \mathbb{R}^n$ is a Borel subset and $\ell_k\in E^\ast$ are continuous linear functionals. Equivalently, it is a preimage $C = P^{-1}(C_0)$ of a Borel set under a continuous linear map $P : E\to \mathbb{R}^n$.
Definition:Let $E$ be a locally convex vector space. A cylinder set measure $\nu$ is a nonnegative additive set function defined on the $\sigma$-algebra generated by cylinder sets of $E$ such that for any continuous linear $P : E\to \mathbb{R}^n$ the set function $$\nu\circ P^{-1} : B\mapsto \nu(P^{-1}(B))$$ is countably additive. Comparison
If we now compare there are a few points to mention:
It seems that cylinder sets capture discretizations. If $x$ is a continuous path, $(\ell_1(x),\dots, \ell_n(x))$ is an $n$-point discretization. To be even more precise in the case of paths we could take $\ell_k(x) = x(t_k)$ for some $t_1,\dots, t_n$ in the interval. In the same way we could take $\ell_k(x) = a_k$ some Fourier coefficient of $x(t)$. I've seem both things done in Physics.
It seems cylinder set measures are in fact a way to define "a measure per discretization". So for each discretization we give a measure - integrate over $n$ points, integrate over $n$ Fourier coefficients, so forth.
Still, the connection doesn't feel complete for me. The issue is that to define a cylinder set measure we must define $\nu$ on
the whole algebra generated by cylinder sets.
The Physicist approach seems to do this exactly for a specific collection of cylinder sets. Either for the ones with $\ell_k(x) = x(t_k)$ or for the ones with $\ell(x_k)=a_k$ a Fourier coefficient.
But there are infinitely many other choices of the $\ell_k$ which give rise to many more cylinder sets. And one would need still to define $\nu$ on the $\sigma$-algebra itself.
The question: is there really a relation between cylinder set measures and the Physicist discretization of a path integral? If so, how the relation can be made more precise? If not, why not, considering the similarities?
For this discussion, please let us consider the Euclidean path integral. So the issue here
is not the imaginary exponent. |
Hey guys! I built the voltage multiplier with alternating square wave from a 555 timer as a source (which is measured 4.5V by my multimeter) but the voltage multiplier doesn't seem to work. I tried first making a voltage doubler and it showed 9V (which is correct I suppose) but when I try a quadrupler for example and the voltage starts from like 6V and starts to go down around 0.1V per second.
Oh! I found a mistake in my wiring and fixed it. Now it seems to show 12V and instantly starts to go down by 0.1V per sec.
But you really should ask the people in Electrical Engineering. I just had a quick peek, and there was a recent conversation about voltage multipliers. I assume there are people there who've made high voltage stuff, like rail guns, which need a lot of current, so a low current circuit like yours should be simple for them.
So what did the guys in the EE chat say...
The voltage multiplier should be ok on a capacitive load. It will drop the voltage on a resistive load, as mentioned in various Electrical Engineering links on the topic. I assume you have thoroughly explored the links I have been posting for you...
A multimeter is basically an ammeter. To measure voltage, it puts a stable resistor into the circuit and measures the current running through it.
Hi all! There is theorem that links the imaginary and the real part in a time dependent analytic function. I forgot its name. Its named after some dutch(?) scientist and is used in solid state physics, who can help?
The Kramers–Kronig relations are bidirectional mathematical relations, connecting the real and imaginary parts of any complex function that is analytic in the upper half-plane. These relations are often used to calculate the real part from the imaginary part (or vice versa) of response functions in physical systems, because for stable systems, causality implies the analyticity condition, and conversely, analyticity implies causality of the corresponding stable physical system. The relation is named in honor of Ralph Kronig and Hans Kramers. In mathematics these relations are known under the names...
I have a weird question: The output on an astable multivibrator will be shown on a multimeter as half the input voltage (for example we have 9V-0V-9V-0V...and the multimeter averages it out and displays 4.5V). But then if I put that output to a voltage doubler, the voltage should be 18V, not 9V right? Since the voltage doubler will output in DC.
I've tried hooking up a transformer (9V to 230V, 0.5A) to an astable multivibrator (which operates at 671Hz) but something starts to smell burnt and the components of the astable multivibrator get hot. How do I fix this? I check it after that and the astable multivibrator works.
I searched the whole god damn internet, asked every god damn forum and I can't find a single schematic that converts 9V DC to 1500V DC without using giant transformers and power stage devices that weight 1 billion tons....
something so "simple" turns out to be hard as duck
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it?
If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@AaronStevens Yeah, I had a good laugh to myself when he responded back with "Yeah, maybe they considered it and it was just too complicated". I can't even be mad at people like that. They are clearly fairly new to physics and don't quite grasp yet that most "novel" ideas have been thought of to death by someone; likely 100+ years ago if it's classical physics
I have recently come up with a design of a conceptual electromagntic field propulsion system which should not violate any conservation laws, particularly the Law of Conservation of Momentum and the Law of Conservation of Energy. In fact, this system should work in conjunction with these two laws ...
I rememeber that Gordon Freeman's thesis was "Observation of Einstein-Podolsky-Rosen Entanglement on Supraquantum Structures by Induction Through Nonlinear Transuranic Crystal of Extremely Long Wavelength (ELW) Pulse from Mode-Locked Source Array "
In peskin book of QFT the sum over zero point energy modes is an infinite c-number, fortunately, it's experimental evidence doesn't appear, since experimentalists measure the difference in energy from the ground state. According to my understanding the zro pt energy is the same as the ground state, isn't it? If so, always it is possible to substract a finite number (higher exited state for e.g.) from this zero point enrgy (which is infinte), it follows that, experimentally we always obtain infinte spectrum.
@ACuriousMind What confuses me is the interpretation of Peskin to this infinite c-number and the experimental fact
He said, the second term is the sum over zero point energy modes which is infnite as you mentioned. He added," fortunately, this energy cannot be detected experm., since the experiments measure only the difference between from the ground state of H".
@ACuriousMind Thank you, I understood your explanations clearly. However, regarding what Peskin mentioned in his book, there is a contradiction between what he said about the infinity of the zero point energy/ground state energy, and the fact that this energy is not detectable experimentally because the measurable quantity is the difference in energy between the ground state (which is infinite and this is the confusion) and a higher level.
It's just the first encounter with something that needs to be renormalized. Renormalizable theories are not "incomplete", even though you can take the Wilsonian standpoint that renormalized QFTs are effective theories cut off at a scale.
according to the author, the energy differenc is always infinite according to two fact. the first is, the ground state energy is infnite, secondly, the energy differenc is defined by substituting a higher level energy from the ground state one.
@enumaris That is an unfairly pithy way of putting it. There are finite, rigorous frameworks for renormalized perturbation theories following the work of Epstein and Glaser (buzzword: Causal perturbation theory). Just like in many other areas, the physicist's math sweeps a lot of subtlety under the rug, but that is far from unique to QFT or renormalization
The classical electrostatics formula $H = \int \frac{\mathbf{E}^2}{8 \pi} dV = \frac{1}{2} \sum_a e_a \phi(\mathbf{r}_a)$ with $\phi_a = \sum_b \frac{e_b}{R_{ab}}$ allows for $R_{aa} = 0$ terms i.e. dividing by zero to get infinities also, the problem stems from the fact that $R_{aa}$ can be zero due to using point particles, overall it's an infinite constant added to the particle that we throw away just as in QFT
@bolbteppa I understand the idea that we need to drop such terms to be in consistency with experiments. But i cannot understand why the experiment didn't predict such infinities that arose in the theory?
These $e_a/R_{aa}$ terms in the big sum are called self-energy terms, and are infinite, which means a relativistic electron would also have to have infinite mass if taken seriously, and relativity forbids the notion of a rigid body so we have to model them as point particles and can't avoid these $R_{aa} = 0$ values. |
Assume you have a smooth quasi-projective scheme $X$ (you can actually assume $X$ is projective over an affine scheme of finite type) defined over $\mathbb Z$ (or if you prefer, a discrete valuation ring), and let $F$ be a locally free tilting sheaf on $X$, that is, a locally free sheaf such that $\mathrm{Ext}^i(F,F)=0$ for all $i>0$. Assume that this sheaf is a generator of the derived category of coherent sheaves after base change to $\mathbb{Q}$ (or basically equivalently, to $\mathbb{F}_p$ for all sufficient large primes); does it follow that it is a generator over $\mathbb{Z}$ or modulo all primes?
Let $X = \lim X_i$ be a scheme written as a cofiltered limit of qcqs schemes with affine transition morphisms. Then the category of perfect objects in $D(O_X)$ is the colimit of the categories of perfect objects in $D(O_{X_i})$. Moreover, if $E$ is a perfect object and generator for $D_{QCoh}(O_X)$, then for some $i$ there exists an $E_i$ over $X_i$ which is perfect, a generator, and pulls back to $E$ (we will see an argument in a special case for this below --- but the argument in general is the same).
More specifically, suppose that we have a ring $A = \text{colim}_{i \in I} A_i$, we have a smallest element $0 \in I$ we have a projective morpihsm $X_0 \to \text{Spec}(A_0)$ projective and $X_i$, resp. $X \to \text{Spec}(A)$ is the base change of $X_0$ via the ring maps $A_0 \to A_i$, resp. $A_0 \to A$. Then we know there is a generator of the form $F_0 = O_{X_0} \oplus O_{X_0}(1) \oplus \ldots \oplus O_{X_0}(n)$ for a suitable $n \geq 0$ which
also is a generator after restriction to the fibres of $X_0 \to \text{Spec}(A_0)$. Moreover the same thing holds for the pullbacks $F_i$ and $F$ of $F_0$ to $X_i$ and $X$.
Going back to our $E$, we can construct $F$ out of $E$ using a finite number of cones and direct summands. Hence we find that for sufficiently large $i$ we can construct $F_i$ out of $E_i$ by a finite number of cones and direct summands. Since taking cones and direct summands commute with restriction to fibres we see that, for some $i$, the restriction of $E_i$ to the fibres of $X_i \to \text{Spec}(A_i)$ generates.
Apply with $A = \mathbf{Q}$ and $A_i = \mathbf{Z}[1/i]$ and $0 = 1$ to get what you want in case $X$ is projective over $\mathbf{Q}$. I think the same argument works if $X$ is just quasi-projective over $\mathbf{Q}$. For general varieties you first have to show a priori that there is
some generator which generates on all the fibres. Presumably you can do this with some argument using Noetherian induction, but I didn't try. |
With a graduate student, I'm going through the paper (Proc. London Math. Soc. (3) 47 (1983), no. 2, 193–224.)
Here's the background and notation.
We have a quadratic character $\chi$ modulo $q$, with Siegel zero $\beta_0$. $\eta=((1-\beta_0)\log q)^{-1}$, so $3\le \eta\ll q$ is known. Let $L=\log q$. We take $$ q^{250}\le x\le q^{500}. $$
Lemma 3 (stated on p. 198) says $$ \sum_{p<x,\chi(p)=1}\frac{\log(p)}{p}\ll L\log(\eta)^{-1/2} $$ Heath-Brown says "This is not necessarily the best bound of its type, but suffices for our purposes." I think the proof he gives is flawed. I know results like this are found elsewhere. My question is
Can this proof be fixed? If not, can you provide a reference for a more robust proof?
(Side note: In a recent blog post on this same paper, Tao proves his Lemma 5 which is similar, summing instead $1/p$, up to a $o(1)$ error. "For more precise estimates on the $o(1)$ error, see the paper of Heath-Brown (particularly Lemma 3).")
The proof starts in Section 4 on p. 206. The overall structure is to write $$ \frac{L^\prime}{L}(s,\chi)-\frac{L^\prime}{L}(s^\prime,\chi) $$ as a both sum over zeros and as a sum over primes. Here $s=1+L^{-1}$ and $s^\prime=1+aL^{-1}$, where $a$ is to be chosen later. The zeros side comes down to estimating $$ aL^{-1}\sum_{\rho\ne\beta_0}|\rho-1|^{-2}. $$ The zeros $\rho$ at height $\ge 1$ give no trouble. Heath-Brown quotes Prachar to estimate the number of zeros in the disk $|s-1|\le r$ as $$ \ll 1+r\log q $$ for $r\le 2$. To count zeros below height $1$ it would make sense to take $r=\sqrt{2}$ here.
With $r_0$ the closest non-Siegel zero to $1$, he quotes the Deuring-Heilbronn phenomenon to say $r_0\gg L^{-1}\log\eta$.
One would expect then the bound to involve the $aL^{-1}$ term, the number of zeros at height below $1$, and the worst case for the term being summed, namely $r_0^{-2}\ll L^2\log(\eta)^{-2}$, i.e. a bound of $$ aL(1+\sqrt{2}L)\log(\eta)^{-2}\ll aL^2\log(\eta)^{-2}. $$ Heath-Brown uses instead the count of zeros inside the circle of radius $r_0$ (which makes no sense, by definition of $r_0$) and gets a better estimate $\ll aL/\log\eta$. The correct (I think) estimate does not suffice for the error bound the lemma claims. |
I'm reading a technical report on Simulated Annealing: On the Convergence Time of Simulated Annealing, by Sanguthevar Rajasekaran. You may find it following this link.
Given $G=(V, E)$ is the graph whose vertices are possible solutions to the given problem, and $E$ is the set of edges between the neighbours, the goal is to prove convergence of SA. For that purpose, the author assumed that $G$ is strongly connected and denoted the diameter of $G$, its degree and maximal energy difference between vertices by $D$, $d$ and $\Delta$, respectively.
There is a lemma on the 6th page:
If $X$ is any state in $V$, then the expected number of steps before a global optimal state is visited starting from $X$ is $\le (\frac 1 d \times e^{-\Delta / T})^{-D}$.
The probability to visit a neighbour $S_j$ after $S_i$ is:
$$\frac 1 d \times e^{\frac{-\Delta E}{T}}$$ where $\frac 1 d$ is the probability of selecting $S_j$ of all $d$ neighbours of $S_i$, and $e^{\frac{-\Delta E}{T}}$ is the probability to actually move to it. $T$ is held constant in the model.
In the proof of the lemma, the author determines the probability to visit a global optimum $g$ from any state $X$ as $\ge [\frac 1 d \times e^{-\frac \Delta T}]^D$ and finishes the proof:
This implies that the expected number of steps before $g$ is visited is $\le [d e^{\frac \Delta T}]^D$.
Would anyone care to explain how do we get the estimate for the number of steps? I'm stuck with it, but am really, really intrigued to figure this out. :) |
2.8. Density Estimation¶
Density estimation walks the line between unsupervised learning, featureengineering, and data modeling. Some of the most popular and usefuldensity estimation techniques are mixture models such asGaussian Mixtures (
sklearn.mixture.GaussianMixture), andneighbor-based approaches such as the kernel density estimate(
sklearn.neighbors.KernelDensity).Gaussian Mixtures are discussed more fully in the context ofclustering, because the technique is also useful asan unsupervised clustering scheme.
Density estimation is a very simple concept, and most people are already familiar with one common density estimation technique: the histogram.
2.8.1. Density Estimation: Histograms¶
A histogram is a simple visualization of data where bins are defined, and the number of data points within each bin is tallied. An example of a histogram can be seen in the upper-left panel of the following figure:
A major problem with histograms, however, is that the choice of binning can have a disproportionate effect on the resulting visualization. Consider the upper-right panel of the above figure. It shows a histogram over the same data, with the bins shifted right. The results of the two visualizations look entirely different, and might lead to different interpretations of the data.
Intuitively, one can also think of a histogram as a stack of blocks, one block per point. By stacking the blocks in the appropriate grid space, we recover the histogram. But what if, instead of stacking the blocks on a regular grid, we center each block on the point it represents, and sum the total height at each location? This idea leads to the lower-left visualization. It is perhaps not as clean as a histogram, but the fact that the data drive the block locations mean that it is a much better representation of the underlying data.
This visualization is an example of a
kernel density estimation, in this casewith a top-hat kernel (i.e. a square block at each point). We can recover asmoother distribution by using a smoother kernel. The bottom-right plot showsa Gaussian kernel density estimate, in which each point contributes a Gaussiancurve to the total. The result is a smooth density estimate which is derivedfrom the data, and functions as a powerful non-parametric model of thedistribution of points. 2.8.2. Kernel Density Estimation¶
Kernel density estimation in scikit-learn is implemented in the
sklearn.neighbors.KernelDensity estimator, which uses theBall Tree or KD Tree for efficient queries (see Nearest Neighbors fora discussion of these). Though the above exampleuses a 1D data set for simplicity, kernel density estimation can beperformed in any number of dimensions, though in practice the curse ofdimensionality causes its performance to degrade in high dimensions.
In the following figure, 100 points are drawn from a bimodal distribution, and the kernel density estimates are shown for three choices of kernels:
It’s clear how the kernel shape affects the smoothness of the resulting distribution. The scikit-learn kernel density estimator can be used as follows:
>>> from sklearn.neighbors.kde import KernelDensity>>> import numpy as np>>> X = np.array([[-1, -1], [-2, -1], [-3, -2], [1, 1], [2, 1], [3, 2]])>>> kde = KernelDensity(kernel='gaussian', bandwidth=0.2).fit(X)>>> kde.score_samples(X)array([-0.41075698, -0.41075698, -0.41076071, -0.41075698, -0.41075698, -0.41076071])
Here we have used
kernel='gaussian', as seen above.Mathematically, a kernel is a positive function \(K(x;h)\)which is controlled by the bandwidth parameter \(h\).Given this kernel form, the density estimate at a point \(y\) withina group of points \(x_i; i=1\cdots N\) is given by:
The bandwidth here acts as a smoothing parameter, controlling the tradeoff between bias and variance in the result. A large bandwidth leads to a very smooth (i.e. high-bias) density distribution. A small bandwidth leads to an unsmooth (i.e. high-variance) density distribution.
sklearn.neighbors.KernelDensity implements several common kernelforms, which are shown in the following figure:
The form of these kernels is as follows:
Gaussian kernel (
kernel = 'gaussian')
\(K(x; h) \propto \exp(- \frac{x^2}{2h^2} )\)
Tophat kernel (
kernel = 'tophat')
\(K(x; h) \propto 1\) if \(x < h\)
Epanechnikov kernel (
kernel = 'epanechnikov')
\(K(x; h) \propto 1 - \frac{x^2}{h^2}\)
Exponential kernel (
kernel = 'exponential')
\(K(x; h) \propto \exp(-x/h)\)
Linear kernel (
kernel = 'linear')
\(K(x; h) \propto 1 - x/h\) if \(x < h\)
Cosine kernel (
kernel = 'cosine')
\(K(x; h) \propto \cos(\frac{\pi x}{2h})\) if \(x < h\)
The kernel density estimator can be used with any of the valid distancemetrics (see
sklearn.neighbors.DistanceMetric for a list of available metrics), thoughthe results are properly normalized only for the Euclidean metric. Oneparticularly useful metric is theHaversine distancewhich measures the angular distance between points on a sphere. Hereis an example of using a kernel density estimate for a visualizationof geospatial data, in this case the distribution of observations of twodifferent species on the South American continent:
One other useful application of kernel density estimation is to learn a non-parametric generative model of a dataset in order to efficiently draw new samples from this generative model. Here is an example of using this process to create a new set of hand-written digits, using a Gaussian kernel learned on a PCA projection of the data:
The “new” data consists of linear combinations of the input data, with weights probabilistically drawn given the KDE model.
Examples:
Simple 1D Kernel Density Estimation: computation of simple kernel density estimates in one dimension. Kernel Density Estimation: an example of using Kernel Density estimation to learn a generative model of the hand-written digits data, and drawing new samples from this model. Kernel Density Estimate of Species Distributions: an example of Kernel Density estimation using the Haversine distance metric to visualize geospatial data |
257 23 Homework Statement There is an infinite charged plate in yz plane with surface charge density ##\sigma = 8*10^8 C/m^2## and negatively charged particle at coordinate (4,0,0) Find magnitude of efield at coordinate (4,4,0) Homework Equations E= E1+E2
So I figured to get e-field at point (4,4,0), I need to find the resultant e-field from the negatively charged particle and the plate
##E_{resultant}=E_{particle}+E_{plate}##
##E_{particle}=\frac{kq}{d^2}=\frac{(9*10^9)(-2*10^-6)}{4^2}=-1125N/C##
Now for the plate is where I'm confused.
If this was a wire, it would have been okay for me since I only need to deal with one dimension.
Since what they requested was a plate in yz plane, does this means that my:
##\sigma=dy*dy*x?## where ##dy## is the 'slice' I take and x is the width of the plate? Is that accurate?
If it is true, then to find the e-field created by that slice at the point,
##dE=\frac{kdq}{R^2}##
##dE=\frac{k\sigma *x*dy}{a^2+y^2}##
I know that the vertical components of the resultant e-field will cancel out because there are same amount of segments on top and below the point.
So need to find ##dE_{x}##, which = ##dEcos\theta##, where ##\theta## is shown:
So ##dE_{x} = dEcos\theta = (\frac{k\sigma *x*dy}{a^2+y^2}) (\frac{a}{\sqrt{y^2+a^2}})##,
Now the problem is I can't integrate this to find my resultant e-field because I do not know what the value of x is. If this was a wire in a plane it will have been solvable for me, but now I'm kind of stuck.
Any clues/help? Thanks :) |
In the restated problem both $X$ and $Y$ have rank at most $k \ll n$, and the same is true of Gram matrices $A$ and $B$. Also $C=(X+Y)(X+Y)^T$ will have rank at most $k$.
The goal of forming $C$ from $A$ and $B$, even approximately, seems unattainable. To briefly enlarge upon Jack Poulson's example, let $X$ and $Y$ be any two matrices such that:
$$ XX^T = YY^T = \begin{pmatrix} I & 0 \\ 0 & 0 \end{pmatrix} $$
where the given nonzero block $I$ is $k \times k$. Thus $X,Y$ might be of this form:
$$ X = \begin{pmatrix} P \\ 0 \end{pmatrix} $$$$ Y = \begin{pmatrix} Q \\ 0 \end{pmatrix} $$
where $P,Q$ are $k \times k$ orthogonal matrices.
Taking $Q = I$ for simplicity, we see that:
$$ C = (X+Y)(X+Y)^T = 2 \begin{pmatrix} I+P & 0 \\ 0 & 0 \end{pmatrix} $$
So Jack's point is that the problem is not well-defined without further constraints. Knowing $A = XX^T$ and $B = YY^T$ is insufficient to approximate $C$, as $P$ might be any orthogonal matrix of the right size.
Let me suggest that computing $C$ from $A$ and $B$ might be the wrong goal. After all, storing $A$ rather than $X$ (resp. $B$ than $Y$) takes more space, since $X$ is only $n \times k$, but $A$ would generally be a full $n \times n$.
Ordinarily a programmer would be pleased to store the
factors $X$ (resp. $Y$) and to do the fairly easy update $X+Y$ rather than compute and store $C$. Saves time, saves space.
Of course one might fear that further operations done
implicitly with $C$ using $X+Y$ are less efficient. This is not so, at least with a wide range of studied numeric applications. Consider for example forming a matrix-vector product $Cv$. With explicitly formed $C$ this takes $n^2$ multiplications (and roughly equal number of adds). Doing the implicit product by associativity $(X+Y)(X+Y)^Tv$ only takes $2kn$ multiplies (resp. adds). |
Background
I am trying to analyse fourier characteristics of a derivative. For example if I have a first order derivative approximated as following: $$\frac{\partial \Psi(x)}{\partial x} = \frac{\Psi_{i+1}-\Psi_{i-1}}{\Delta x}$$
to get its fourier characteristics, substitute $\Psi(x)= e^{j\, kx}$ in LHS, and its discrete counter part $\Psi(x=i\Delta x) = \Psi_{i}= e^{j \,ki\Delta x}$ RHS. Here $j=\sqrt{-1}$ in and k represents the wavenumber. It gives
$$F \left[\frac{\partial \Psi(x)}{\partial x}\right] = \frac{e^{jk(i+1)\Delta x}-e^{jk(i-1)\Delta x}}{\Delta x}$$ Comparing both sides gives $$k= \sin(j \, k\Delta x)$$
Problem statement
I like to carry out similar analysis for
implicit derivatives of lele (1992), given as following$$f_i' + \alpha (f_{i+1}' + f_{i-1}') + \beta (f_{i+2}' + f_{i-2}') \nonumber \\ =a\frac{f_{i+1}- f_{i-1}}{(2h)^2} + b\frac{f_{i+2}- f_{i-2}}{(4h)^2} + c\frac{f_{i+3}- f_{i-3}}{(6h)^2} $$The coefficients are determined after Taylor series expansion of both sides and then comparing coefficients.
My problem contains a derivative term of type $\frac{\partial}{\partial x} \left[a(x) \frac{\partial}{\partial x} \right]\psi $. It can be simplified if I assume constant media for Fourier analysis:
$$\frac{\partial}{\partial x} \left[a(x) \frac{\partial}{\partial x} \right]\psi = a(x) \frac{\partial}{\partial x} \frac{\partial\psi}{\partial x} \psi= a(x) \frac{\partial^2\psi}{\partial x^2} $$
In simulation I am using first derivative two times for computing the second derivatives as following:
step computes : $\zeta = a(x) \frac{\partial}{\partial x} $ step compute : $\frac{\partial \zeta}{\partial x}$
For
explicit derivatives (example given in background part) the equality holds i.e. $\partial_x.\partial_x ==\partial_x^2$ is valid in numerical form.
So my question is
whether $\partial_x.\partial_x$ is equal to $ \partial_x^2$ for implicit derivative. If it doesn't then how can I carry out its analysis. How to handle the cross derivative terms e.g. $\partial_x.\partial_y$.
Any help or suggestion is appreciated! |
Logistic operator
Logistic operator (or
LogisticOperator) is quadratic function of specific kind determined with the single parameter \(s\), which is usulally assumed to be real, \(s\!>\!1\): \(\!\!\!\!\!\!\!\!\!\!\!(1) ~ ~ ~ ~ \mathrm{LogisticOperator}_s(z) = s\,z \,(1\!-\!z)\)
(Parameter \(s\) is often denoted also with letters \(u\) or \(q\)). In such a way, \(\mathrm{LogisticOperator}_s\) is entire function.
Contents Logistic operator as transfer function
The explicit plots of the Logistic operator for \(s\!=\!3\), \(s\!=\!4\) and \(s\!=\!5\) are showm in pictures at right with black curves.
They are built up with regular iteration at the fixed point \(L\!=\!0\).
These superfunciton and Abel function are used to plot the non-integer iterations of the LogisticOperator; these iterates are shown with colored curves in figure above.
Superfunction of the LogisticOperator
In TORI, the LogisticOperator is used as an example of transfer function that allows construction of superfunction by regular iteration at the fixed point \(L\!=\!0\). The specific superfunction called LogisticSequence\(_s\) (without spacebar) refers to the solution \(F\) of the transfer equation
\(\!\!\!\!\!\!\!\!\!\!\!(2) ~ ~ ~ ~ F(z\!+\!1)=T(F(z))\) \(\!\!\!\!\!\!\!\!\!\!\!(3) ~ ~ ~ ~ F(z)=s^z+O(s^{2z})\).
Often, the argument \(z\) is supposed to be a real number. In principle, it could be taken also from any set such that operations of multiplication (denoted with spacebar) and substraction (denoted with symbol "-" are somehow defined, but the properties of superfunctions in this case still need the additional research. In TORI, at least up to year 2012, the LogisticOperator is used for complex variables (and not yet for quaternions).
In principle, other superfunctions of the LogisticOperator can be constructed in the similar way. That called LogisticSequance seems to be the simplest one. According to the Axiom TORI number 6, this superfunction is considered as "principal", and all others superfunctions of the LogisticOperator are treated as alternatives.
Abel function of the Logistic operator
The inverse finction of the LogisticSequence is called ArcLogisticSequence. It is plotted in figure at right. Many superfuncitons for the given transfer function exist; and many Abel functions exist too. That called ArcLogisticSequance seems to be the simplest one. According to the Axiom TORI number 6, this Abel function is considered as "principal", and all other as alternative.
Function \(G=\mathrm{ArcLogisticSequence}_s\) satisfies the Abel equation
\(\!\!\!\!\!\!\!\!\!\!\!(4) ~ ~ ~ ~ G(T(z))=G(z)+1\)
for the transfer function \(T=\mathrm{LogisticOperator}_s\).
Inverse function
The inverse function of the Logistic operator can be expressed as follows
[1]: \(\!\!\!\!\!\!\!\!\!\!\!(5) ~ ~ ~ ~\displaystyle \mathrm{LogisticOperator}_s^{\, -1}(z)= \frac{1}{2} - \sqrt{\frac{1}{4}-\frac{z}{s}}\) Iteration of the Logistic operator
For integer \(c\), the \(c\)th iteration of function \(T=\mathrm{LogisticOperator}_s\) can be evaluated by the straightforward application of \(T\) or \(T^{-1}=\mathrm{LogisticOperator}_s^{\, -1}\), repeated \(c\) or \(-c\) times:
\(T^{-2}(z)=T^{-1}(T^{-1}(z))\) \(T^{-1}(z)=\mathrm{LogisticOperator}_s^{\, -1}(z)\) \(T^0(z)=z\) \(T^1(z)=T(z)=\mathrm{LogisticOperator}_s(z)\) \(T^2(z)=T(T(z))\)
and so on. For non-integer values, the representation through the LogisticSequence is suggested;
\(\!\!\!\!\!\!\!\!\!\!\!(6) ~ ~ ~ ~ T^c(z)=\mathrm{LogisticSequence}_s\Big( c + \mathrm{ArcLogisticSequence}_s(z)\Big)\)
following the general rule of construction of iterations of a transfer function with specified Superfunction and Abel function. Properties of the superfunction LogisticSequence and the Abel function ArcLogisticSequence are known; the efficient algorithms for the evaluation are supplied. With these functions, the explicit plot of some iterations of the logistic operators are shown in the figure.
While the LogisticOperator is not monotonous function, the relation
\(\!\!\!\!\!\!\!\!\!\!\!(6) ~ ~ ~ ~ \mathrm{LogisticOperator}_s^{\,c}\Big( \mathrm{LogisticOperator}_s^{\,d}(x) \Big)=\mathrm{LogisticOperator}_s^{\,c+d}(x)\)
is valid only for moderate values of \(x\), that should not exceed the branchpoint of function \(\mathrm{ArcLogisticSequence}_s\). For various s, these brancpoints are shown with black circles in figure at right.
References http://mizugadro.mydns.jp/PAPERS/2010logistie.pdf D.Kouznetsov. Holomorphic extension of the logistic sequence. Moscow University Physics Bulletin, 2010, No.2, p.91-98. (Russian version: p.24-31) |
Guitar Speaker Power Handling
The power rating of a guitar speaker is an indication of how much power it can handle without being damaged thermally or mechanically. It is not an indication of how loud the speaker will sound in comparison to other speakers. Let us examine the basics of how speakers work, how the power rating is determined and look at things from the perspective of the guitar amplifier so that we can choose guitar speakers that will last a lifetime.
How an Electrodynamic Loudspeaker Works
Guitar speakers are a type of loudspeaker known as electrodynamic or "moving coil" loudspeakers. The magnetic circuit (composed of front plate, back plate, pole piece and magnet) and voice-coil make up the motor of a guitar speaker. An alternating electrical current flowing through the voice-coil generates an alternating magnetic field perpendicular to flow of current through the coil. The magnetic circuit creates a strongly focused magnetic field in the air gap between the front plate and the pole piece on which the voice-coil is centered. The voice-coil is pushed and pulled through the air gap based on the interaction between these two magnetic fields. Since the speaker cone is connected to the voice-coil, it now has a mechanical force with which to push air particles and make sound.
Thermal Damage
Speakers are transducers. They convert electrical energy provided by an amplifier into acoustical energy. They are actually very inefficient transducers because most of the electrical energy is converted into heat instead of sound. The reference efficiency (ratio of acoustic power out to electrical power in) for most guitar speakers is around 2% to 6%, which means that 98% to 94% of the electrical energy is dissipated in the form of heat.
From the aspect of power dissipation, a guitar speaker can be modeled as a resistor. Most guitar amp enthusiasts are familiar with the equation for electrical power and how it can be used to determine the power dissipated across a resistor. Resistors can be thought of as transducers that intentionally convert electrical energy to heat in order to create a voltage drop. Resistors have a power rating that indicates how much power they can dissipate before being damaged and this rating is analogous to the speaker power rating.
Equation for Power$$P = \frac{V^2}{R}$$
where ~P~ = power, ~V~ = voltage, ~R~ = resistance
For example:$$P = \frac{V^2}{R} = \frac{{20}^2}{8} = \frac{400}{8} = 50\text{ W}$$
The voice-coil is the electrical interface of the speaker and is given a "nominal impedance" specification (e.g. 4, 8 or 16 ohm) which can be used to approximate power dissipation when connected to an amplifier. (The actual impedance of the voice-coil varies with frequency). Just as a resistor will eventually burn up if its power rating is not high enough, the speaker's voice-coil will burn up if it is overpowered by the amplifier. One of the most common symptoms of an overpowered speaker is a burned voice-coil, which usually measures as an open circuit on an ohm meter. No sound can be produced by a speaker with an open voice-coil. An overheated voice-coil former may also become warped and begin to rub against the pole piece causing the speaker to buzz loudly.
Mechanical Damage
Loudspeakers can be damaged mechanically by over-excursion of the voice-coil and cone. This is more common with old speakers that have worn suspensions and adhesives, but may also occur at extreme low frequencies outside of the speaker's useable frequency range. When over-excursion occurs, the voice-coil can become misaligned or bottom out. The cone and suspension (surround and spider) can also become stretched or torn.
How the Speaker Power Rating is Determined
Many speaker manufacturers rate their speakers based on industry standards similar to IEC 60268-5 or AES2-1984. These standards specify a pink noise test signal with a crest factor of two (i.e. 6 dB) which is meant to simulate the transient character of music having an average value, as well as frequent instantaneous voltage spikes that swing up to twice the average value. Pink noise is a particular type of random noise with equal energy per octave and actually sounds like a space shuttle launch. The test signal is applied to the test speaker for a few hours, allowing for a reasonable way to test the speaker's real world thermal and mechanical capabilities. After testing, the speaker must be in working order, without permanent alteration of its technical features. The power rating is calculated using the RMS value of the applied voltage and the minimum value of electrical impedance within the working range of the speaker.
Guitar Amplifier Power Output Ratings
The power output rating of a guitar amp is mostly a ballpark figure for what it can put out. Amp specifications commonly list power output in a form similar to the following:
Power Output: 50W into 8ohm at 5% THD
This type of power output rating is obtained by using a sine wave from a signal generator (usually 1 kHz) as the input signal. The 5% THD (total harmonic distortion) figure means that the sine wave was able to generate 50W of power output with relatively low distortion (near the threshold of clipping or overdrive). THD measurements were one of the first conventions used to objectively compare the fidelity of audio amplifiers.
Guitar amps are unconventional audio amplifiers. While most audio amplifiers are designed to keep distortion as low as possible, guitar amplification has evolved to where overdrive distortion is usually a requirement. For example, the Marshall® JCM800 2203 is a 100W tube amp that has a highly regarded overdrive sound. The owner's manual lists the power output as follows:
Typical power at clipping, measured at 1kHz, average distortion 4% 115 watts RMS into 4, 8, 16 ohms. Typical output power at 10% distortion 170 watts into 4 ohms.
This example shows that for many guitar amplifiers, the power output rating (100W in this case) is not a maximum power output rating, but more of a ballpark clean power output specification.
RMS and Overdrive Distortion
RMS (root mean square) is a kind of average value that can be used to compare the power dissipation from different signals on equal terms. For example, a 20 VDC power supply dissipates the same amount of heat across an 8 ohm resistor as a sine wave with an RMS value of 20 VAC.
General Equation for the RMS value of a periodic function$$V_{RMS} = \sqrt{\frac{1}{T} \int_{0}^{T} [V(t)]^2 dt}$$
where the
Square portion of the equation is
and the
Mean portion of the equation is
and the
Root portion of the equation is
Guitar amp output ratings are usually based on a sine wave at low distortion, but if the volume is turned up further or a gain boosting effect is used, the sine wave becomes more overdriven and can approach the shape of a square wave. The RMS value of a square wave is equal to its amplitude, while the RMS value of a sine wave is equal to its amplitude divided by the square root of two.
Plugging the RMS values into the equation for power shows that a square wave dissipates twice as much power across the same load as a sine wave with the same amplitude.
Power calculation for a sine wave$$P = \frac{(\frac{V_m}{\sqrt{2}})^2}{R}$$$$P = \frac{\frac{{V_m}^2}{2}}{R}$$$$P = \frac{1}{2} \times \frac{{V_m}^2}{R}$$ Power calculation for a square wave$$P = \frac{{V_m}^2}{R}$$
This simplified overdrive distortion model illustrates how the 100 watt Marshall® amp which puts out 115 watts at 4% THD could put out an additional 50 watts at 10% THD.
Tube vs. Solid State Outputs
Many tube guitar amps use output transformers with secondary taps connected to an impedance switch allowing for the same power output when connected to 4, 8 or 16 ohm load impedances. Solid state amps do not use output transformers and do not have the same power output when connected to different load impedances.
For tube outputs, it is important to match the load impedance to the amp\'s output impedance. For solid state outputs, it is important to use a load that is greater than or equal to the rated minimum load impedance and to know the amp's power output at that load. For example, the Fender M-80 is a solid state amp rated for 69 W(RMS) at 5% THD into 8 ohms and 94 W(RMS) at 5% THD into 4 ohms (the minimum load impedance).
With solid-state amps, overdrive distortion generated by the power-amp is not generally considered musically pleasing, so most people will not exceed the amp's low THD power rating. Tube power amps, on the other hand, are often played well beyond their low THD rating.
Amps with Multiple Speakers
When an amp uses multiple speakers the output power is divided between them. The nominal impedance of each speaker should be the same value so that power is distributed equally and so that the output impedance of the amplifier can be matched.
Formula for calculating the equivalent overall impedance of speakers wired in parallel
~Z_{\text{total}}~ = Equivalent Overall Impedance
~Z_1~ = Impedance of speaker 1, etc.$$ Z_{\text{total}} = \frac{1}{\frac{1}{Z_1} + \frac{1}{Z_2} + \frac{1}{Z_3} + \ldots + \frac{1}{Z_n}}$$
Example: Two Speakers in Parallel$$Z_{\text{total}} = \frac{1}{\frac{1}{Z_1} + \frac{1}{Z_2}} = \frac{1}{\frac{1}{16Ω} + \frac{1}{16Ω}} = \frac{1}{\frac{1}{8}} = 8Ω$$ Formula for calculating the equivalent overall impedance of speakers wired in series
~Z_{\text{total}}~ = Equivalent Overall Impedance
~Z_1~ = Impedance of speaker 1, etc.$$ Z_{\text{total}} = Z_1 + Z_2 + Z_3 + \ldots + Z_n$$
Example: Two Speakers in Series$$Z_{\text{total}} = Z_1 + Z_2 = 4Ω + 4Ω = 8Ω$$ Choosing Guitar Speakers to Last a Lifetime
There is no standard method used by all amp manufacturers when selecting an appropriate speaker power rating. If you want to choose a speaker to last a lifetime, you will want to choose a speaker that can handle the maximum amount of preamp and power amp overdrive distortion that can possibly be put into it and safely avoid exceeding the speaker's thermal limits. In the case of single speaker setups, this means choosing a speaker rated for at least twice the rated output power of the amp. For multiple speakers, choose twice the rated power that would be distributed to it.
You might decide to go with a lower power rating because you know that you will never be cranked at full power and love the sound of a lower power rated speaker. In the same way you may choose a speaker with a much higher power rating because of the way it sounds.
A Real World Example: Speakers for a Fender® '65 Twin Reverb® Reissue
1) Determine the rated output power of the amp.
Amplifiers have two power ratings: power consumption and power output. The power consumption is always much higher than the power output. In this case the output power is 85 watts RMS and the power consumption is 260 watts.
2) Determine the output impedance for that output power rating.
In this case it is 4 ohm.
3) Determine the number of speakers.
In this case there are two 12" (8 ohm) speakers wired in parallel for an overall impedance of 4 ohms.
For this amp, speaker choices to last a lifetime should be rated for at least 85 watts each. There are a lot of speaker choices rated for 100 watts and this rating would be very safe. Actually, the stock speaker for this amp is the Jensen C12K and it is rated for 100 watts. |
When a charged particle goes through liquid argon (LAr), it takes about 23.6 eV energy to ionize an electron in average. For minimal ionizing particle (heavy ionizing particle), the mean energy loss is about 2.1 (25) MeV/cm leading to about 100 (10) nm average separation between two adjacent ions.
The
mean rate of energy loss by moderately relativistic charged heavy particles is well-described by the Bethe formula. Considering the actual energy transfers might be restricted to $T \le T_{cut} \le T_{max}$, this formula can be expressed in a unified way,
\[
-< \frac{dE}{dx} > = Kz^2 \frac{Z}{A}\frac{1}{\beta^2}[\frac{1}{2}ln\frac{2m_ec^2\beta^2\gamma^2T_{cut}}{I^2}-\frac{\beta^2}{2}(1+\frac{T_{cut}}{T_{max}})-\frac{\delta(\beta\gamma)}{2}].
\]
Details of the parameters can be seen in PDG.
Note that the mean $dE/dx$ has no dependency on the thickness of the absorbers. Calculations regarding the LAr are shown in the central plot on the right side (from Milind Diwan).
The flucatuations in enegy loss can be adequately described by the highly-skewed Landau or Landau-Vavilov distribution (see the upper pannel of the right plot), though the Landau distribution fails to describe energy loss in thin absorbers.
The
most probable energy loss can be calculated with the equation,
\[
\Delta_p = {\xi} [ln\frac{2mc^2\beta^2\gamma^2}{I} + ln\frac{{\xi}}{I} +j -\beta^2 - \delta(\beta\gamma)],
\]
where $\xi = (K/2)< Z/A >(x/\beta^2)$ MeV for a detector with a thickness $x$ in $g~cm^{-2}$. Details can be seen in Ref. [1], especailly the density effect $\delta$ which was not included in Landau's or Vavilov's work. Note that the most probable energy loss is generally smaller than the mean value and do have dependency on the thickness of the absorbers as indicated by $\xi$. MPV vs. muon momentum (or thickness) are shown in the central (or lower) plot on the right side.
Useful numbers for liquid argon: LAr density 1.38/1.40 g/cm$^3$ at 89/87 K. The minimum ionizing muon (about 260 MeV kinetic energy) has the mean energy loss of 2.1 MeV/cm and 5 GeV $\mu$ in 1 (0.03) cm LAr corresponds to a most probable (MPV) energy loss of 1.8 (1.5) MeV/cm.
Click the following links to retrieve the relevent plots:
When the total energy of an ionizing particle, $E_0$, is absorbed in the detector medium, the number of the
ion pairs $N_i$ is given as $E_0/W$, where W is an average energy required to form an ion pair (called $W$-value). The r.m.s. fluctuation of the number of ion pairs can be calculated to be $\sqrt{FN_i}$, where $F$ is the so-called "Fano-Factor" [2].
Considering the excitation (requiring $W_e$) and ionization (requiring $W_i$) in the atom, we have \[ E_0 = N_i \cdot W = N_i \cdot W_i + N_e \cdot W_e. \] Due to the actual much smaller $W_e$ than $W_i$, every "collision" will most likely cause excitation while the ionization must be combined with the excitation to dissipate precisely the total energy. Define $\sigma_x = \sigma(N_x)$ and $\sigma_e = \sqrt{N_e}$ ($N_e$ can be well approximated by a Poisson distribution), it can be derived that \[ \sigma_i \cdot W_i = \sigma_e \cdot W_e = \sqrt{N_e} \cdot W_e \] The fano factor then can be expressed as the last term in the following formula, \[ \sigma_i = \sqrt{N_i} \cdot \sqrt{\frac{W_e}{W_i}(\frac{W}{W_i}-1)}. \] In reality, three impacts are involved in the estimation of fano factor, which are ionization, excitation, and nuclear scattering, respectively. The contribution of nuclear scattering can be neglected.
The ultimate number of ionization electrons has to consider the recombination (discussed below) to the ions which serves as the survival probabilty of the ionization electrons liberated from the atoms. The excitation and recombination will contribute to the scintillation photons of liquid argon.
The recombination factor $R_c$ describes the efficiency in converting $dE/dx$ into $dQ/dx$, the visible energy available to be collected by the sense wires before attenuation. The ionized electron has about 5 eV energy with a distance about 0.5 nm from the ion. In about 2 ns, the ionized electron would go through 10,000 collisions before reaching thermal energy (~0.01 eV). At this point, the average separation distance is about 2.5 $\mu$m [3,4].
There are two common methods to model the recombination effect.
Birks Model:
\[
R_c = \frac{dQ/dx}{dE/dx} = \frac{A_{3t}}{1+k_{3t}/\varepsilon \times dE/dx}
\]
, where $A_{3t} = 0.8$, $k_{3t} = 0.0486 (g/MeV cm^2)(kV/cm)$ and $\varepsilon = E \rho$ is the product of electric field and density.
Modified Box model:
\[
R_c = \frac{dQ/dx}{dE/dx} = \frac{\ln(A+B/\varepsilon \times dE/dx)}{B/\varepsilon \times dE/dx}
\]
, where $A = 0.930$, $B = 0.212 (g/MeV cm^2)(kV/cm)$ and $\varepsilon = E \rho$ is the product of electric field and density.
Click the following links to retrieve the relevent plots:
There are three microscopic theories describing the recombination: Germinate, Bulk, and Columnar. In particular, the prediction from Columnar theory suggests that the recombination has a dependence on the track angle with respect to the drift direction. Such an effect has been studied in ArgoNeuT [3]. While there are some hints regarding the angular dependence, the effect seems to be much smaller than that predicted by Columnar theory (see Fig. 9 from Ref [5]). |
There is an argument I do not understand given in "Introduction to quantum electrodynamics" by Cohen-Tannoudji (page 111 for the French version of the book).
We are dealing with the
non-relativistic version of the QED Lagrangian.
It takes the following form:
$$ L = \sum_a \frac{1}{2} m_a \dot{\boldsymbol{r}}_a^2 + \int d^3r\left(\frac{\epsilon_0}{2}\boldsymbol{E}^2(\boldsymbol{r})-\frac{\epsilon_0}{2}\boldsymbol{B}^2(\boldsymbol{r}) + \boldsymbol{j(r)}\cdot\boldsymbol{A(r)} - \rho(\boldsymbol{r})U(\boldsymbol{r}) \right).\tag{B.5.a-b}$$
I remind that:
$$ \boldsymbol{E} = - \nabla U - \frac{\partial \boldsymbol{A}}{\partial t} \tag{B.3.a}$$ $$ \boldsymbol{B} = \nabla \wedge \boldsymbol{A}.\tag{B.3.b} $$
In this book, the author says that our Lagrangian has 8 dynamical variables: 3 generalised position for $A$ (its 3 components), 1: the potential $U$. And 4 more because of generalised velocities associated to those.
First thing I don't understand:
Why we don't take in account the spatial derivatives? In the Lagrangian density (if we put apart the free particle lagrangian) they also appear? Which would add more variables.
Then, the author says that the Maxwell equations have 6 degrees of freedom (the 3+3 components of the $E$ and $B$ fields).
Second thing I don't understand
I don't understand the comparison here. When he talks about the dependences in term of potentials, he has a Lagrangian and he counts the generalised positions and velocities dependences: the dynamical variables.
To make a clear comparison with the $E,B$ dependences, we would need to write a Lagrangian density in term of those variables, count the coordinates/velocities and make the conclusion.
Because probably that in this Lagrangian the velocities of the field would also appear (thus we would have more than 6 variables).
(Some context about this part of the book: he says all this to explain that there is a problem to fix that we have more variables in term of potential than in term of $E, B$ fields).
In summary: All my confusions here come from the fact: what do we exactly call "dynamic variables"? Isn't it the field dependence of the Lagrangian density: $\phi, \partial_t \phi, \partial_i \phi$? If so, why didn't he count the spatial derivative when he talked about potentials?
And how can he make the comparison with E and B without having written a Lagrangian in term of $E$ and $B$? |
Let $V$ a Hilbert space, $a:V\times V\rightarrow \mathbb{R}$ a bounded, symmetric and positive bilinear form and $f:V\rightarrow\mathbb{R}$ bounded.
Is well known that problem
$$\left\lbrace\begin{array}{l}\textrm{Find }u\textrm{ such that}\\ a(u,v)=f(v)\quad\forall v\in V\end{array}\right.$$
is equivalent to minimize the following (energy) functional:
$$J(v)=\dfrac{1}{2}a(v,v)-f(v).$$
If now I have some extra condition I get the problem
$$\left\lbrace\begin{array}{l}\textrm{min }J(v)\textrm{ with }v\in V\\ \textrm{ subjetc to }u=g\quad\forall v\in \tilde V\subset V\end{array}\right.$$
I'm thinking in how to get the corresponding Lagrange multipliers in
Stokes-like problems, for example, where the problem is
$$\left\lbrace\begin{array}{l}\textrm{min }J(u,p)\textrm{ with }(u,q)\in V\times Q\\ \textrm{ subjetc to }\int_\Omega p=0\quad\end{array}\right.$$
but when I try to use Lagrange multipliers (learned in my first course of calculus in several variables) I need to calculate things like:
$$\nabla f=\lambda\nabla g$$
but what is "$\nabla$" operator in my context (with bilinear forms)? In other words, How can I use Lagrange multipliers in this context? |
I'm concerned with equation 2.24 of http://arxiv.org/abs/1601.00482
The superconformal hypermultiplets in this paper have a conic hyperkahler target manifold and the authors want to gauge some isometries of this manifold. Letting the isometry group be $G$ and to have an associated Lie algebra $\mathfrak{g}$ generated by Killing vectors $k_I$, we can express this as $\mathcal{L}_{k_I} g=0$ where $g$ is the metric on the conic hyperkahler manifold.
Then, in order to not break SUSY, the $k_I$ must commute with the SUSY generators. Apparently this is equivalent to the Killing vectors being triholomorphic $\mathcal{L}_{K_I} J_\alpha=0$ where $J_\alpha$ are the triplet of complex structures. Does anyone know why this is the case?
Secondly, they say in 2.24 of this paper that the moment maps associated to these symmetries must satisfy the "equivariance condition". Unfortunately they don't offer any explanation of what this is or where it comes from. There is some discussion in other literature along the lines of "we can also derive the equivariance condition...." but they never say what it is or explain how they found it. The best I've found is in the Freedman/van Proeyen Supergravity book where in eqn (13.61), they seem to say it comes from requiring the moment maps to transform in the adjoint:
$$(k_I^\alpha \partial_\alpha + k_I^{\bar{\alpha}} \partial_{\bar{\alpha}} ) \mu_J = f_{IJ}^K \mu_K$$
They then use some identities to write this as (13.62):
$$k_I^\alpha g_{\alpha \bar{\beta}} k_J^{\bar{\beta}} - k_J^\alpha g_{\alpha \bar{\beta}} k_I^{\bar{\beta}} = i f_{IJ}^K \mu_K$$
Although I don't see how this looks anything like (2.24) of the attached paper.
If anyone can offer any help or thoughts on either of these issues I'd greatly appreciate it!This post imported from StackExchange Physics at 2016-06-26 09:46 (UTC), posted by SE-user user11128 |
oschvr a blog about software, programming, business, travelling and life. ← Problems Largest prime factor 24 days ago 02/05/2019 The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?
We neeed to find the largest prime factor of
600851475143.
We know that the largest prime factor of a number, its the number itself.
As a concept refresher, any
prime number is only divisible by 1 ant itself, thus, when dividing by a certain number, there should not be a reminder.
2 is the first prime number, and holds the honor of being the only even prime numnber.
We'll start by making a function to return a
bool if the former condition holds.
1 2 3 4 5 6 7 8 9
func isPrime(x int64) bool { var i int64 = 2 // We start with 2 as it is the first prime number for ; i < x; i++ { if x % i == 0 { return false } } return true }
Next, we need to test factors and see if they are prime numbers. Let's state:
$$ ab = N \space where \space 1 < a \leq b \leq N$$
$$N = ab \geq a^2 \Leftrightarrow a^2 \leq N \Rightarrow \sqrt[]{N}$$
This means that in order to test the primality of a composite number N, you check if this number is a product of two numbers. Thus we use square root.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
package main import ( "fmt" "math" "time" ) const n int64 = 600851475143 func isPrime(x int64) bool { var i int64 = 2 for ; i < x; i++ { if x%i == 0 { return false } } return true } func main() { start := time.Now() var i = int64(math.Sqrt(float64(n))) for ; i > 1; i-- { if n%i == 0 && isPrime(i) { fmt.Println("Execution time: ", time.Since(start)) fmt.Println("Largest Prime Factor: ", i) break } } }
Execution time:
7.990898ms
Answer
Largest Prime Factor:
6857 |
Advances in Differential Equations Adv. Differential Equations Volume 1, Number 2 (1996), 265-281. Almost periodicity enforced by Coulomb friction Abstract
We describe the influence of Coulomb friction $\mu$ sgn $ \dot x$ on the behavior of the linear oscillator given by $\ddot x+x=\varphi(t)$, where $\varphi$ is continuous and almost periodic. Depending on $\varphi$, we characterize the range of $\mu>0$ such that nontrivial almost periodic motions exist. We also show that dissipation caused by Coulomb friction may be too weak to ensure uniqueness of such motions, a phenomenon which appears already in case $\varphi$ is $2k\pi$-periodic with $k>1$. Nevertheless, we get a rather complete picture of the asymptotic behavior of such a system, but also have some interesting open questions, for example concerning the shape of the almost periodic solutions.
Article information Source Adv. Differential Equations, Volume 1, Number 2 (1996), 265-281. Dates First available in Project Euclid: 25 April 2013 Permanent link to this document https://projecteuclid.org/euclid.ade/1366896240 Mathematical Reviews number (MathSciNet) MR1364004 Zentralblatt MATH identifier 0838.34016 Subjects Primary: 34C25: Periodic solutions Secondary: 34A60: Differential inclusions [See also 49J21, 49K21] 70K20: Stability 70K30: Nonlinear resonances Citation
Deimling, Klaus; Hetzer, Georg; Shen, Wen Xian. Almost periodicity enforced by Coulomb friction. Adv. Differential Equations 1 (1996), no. 2, 265--281. https://projecteuclid.org/euclid.ade/1366896240 |
(This is a cross-post from the Theoretical Computer Science Stack Exchange.)
For the purposes of this question, a
cut in a graph $G$ is the edge-set $\delta (S)\subseteq E(G)$ between some vertex-set $S$ and its complement. A max cut is one with at least as many edges as any other cut. Finding a max cut is NP-hard, but a greedy algorithm (e.g.) can approximate a max cut, finding a cut with at least half as many edges as possible.
Equivalently, a cut $\delta (S)\subseteq E(G)$ is a max cut if and only if$$| \delta (S)\cap \delta (T) |\geq \frac{1}{2}|\delta (T)| \qquad \forall\text{ cuts } \delta (T)\subseteq E(G).$$
Proof: It should be clear that the standard definition implies this one. To show that these inequalities imply $\delta (S)$ is at least as big as any other cut $\delta (S')$, observe that $$|\delta (S)| - |\delta (S')| = |\delta (S) \cap \delta (S\Delta S')| - |\delta (S') \cap \delta (S\Delta S')|.$$ Because the two edge-sets appearing in the right-hand side partition the edges of the cut $\delta (S\Delta S')$, applying the above inequality with $S\Delta S'$ in the role of $T$ gives $|\delta (S)|-|\delta (S')|\geq 0$. $\square$
I'd like to know whether max cuts can be easily approximated in the sense of my second definition. Specifically:
Question:Is there a polynomial-time algorithm to find a cut $\delta (S)\subseteq E(G)$ with $$| \delta (S)\cap \delta (T) |\geq \epsilon |\delta (T)| \qquad \forall\text{ cuts } \delta (T)\subseteq E(G)$$ for some $\epsilon > 0$? |
LLVM optimizes power sums, such as
int sum(int count) { int result = 0; for (int j = 0; j < count; ++j) result += j*j; return result; }to code that calculates the result without a loop (godbolt) sum(int): test edi, edi jle .LBB0_1 lea eax, [rdi - 1] lea ecx, [rdi - 2] imul rcx, rax lea eax, [rdi - 3] imul rax, rcx shr rax imul eax, eax, 1431655766 add eax, edi shr rcx lea ecx, [rcx + 2*rcx] lea eax, [rax + rcx] add eax, -1 ret .LBB0_1: xor eax, eax retIt handles more complex cases too (godbolt) – that is, the optimization is not just a simple pattern matcher. This post will show how the optimization is done. Loop analysis – scalar evolution
There are many cases where compilers need to track how values are updated within a loop. For example, the loop vectorizer needs to check that the pointers are moved to the adjacent element in the next iteration, and check that no other pointer indexing may alias the range we are vectorizing.
Both GCC and LLVM does this in the same way in their scalar evolution passes, where each variable at iteration \(i\) (we start enumerating iterations from \(0\)) is represented as a function \(f_0(i)\) defined as a linear recurrence of the form
\[f_j(i) = \begin{cases} \phi_j & \text{if $i = 0$} \\ f_j(i-1) \odot_{j+1} f_{j+1}(i-1) & \text{if $i > 0$} \end{cases}\] where \(\odot\in\{+,*\}\). \begin{cases} 0 & \text{if $i = 0$} \\ f(i-1) + 1 & \text{if $i > 0$} \end{cases}\] \begin{cases} 2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\ f_2(i-1) + 6 & \text{if $i > 0$} \end{cases}\\ f_1(i) & = \begin{cases} k-1 & \text{if $i = 0$} \\ f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$} \end{cases}\\ f(i) = f_0(i) & = \begin{cases} 7 & \text{if $i = 0$} \\ f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$} \end{cases}\end{align}\] One optimization we can see directly from these functions is that the value can be calculated by just three additions within the loop \begin{cases} 2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\ f_2(i-1) + 6 & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{2,+,6\}$}\\ f_1(i) & = \begin{cases} k-1 & \text{if $i = 0$} \\ f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{k-1,+,f_2\}$}\\ f(i) = f_0(i) & = \begin{cases} 7 & \text{if $i = 0$} \\ f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$} \end{cases} \phantom{xx}\text{is written as $\{7,+,f_1\}$}\end{align}\] These can be chained, so \(f(i)\) can be written as a \{\phi_0, +, \phi_1\} + \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 + \psi_0, +, \phi_1 + \psi_1\} \\ \{\phi_0, +, \phi_1\}* \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 * \psi_0, +, \psi_1 * \{\phi_0, +, \phi_1\} + \phi_1 * \{\psi_0, +, \psi_1\} + \phi_1*\psi_1\} \\ \{\phi_0, +, \phi_1,+,0\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0, +, \phi_1\}\end{align}\] So for the loop from the & = \{0 * 0, +, 1 * \{0, +,1\} + 1 * \{0, +, 1\} + 1*1\} \\ & = \{0, +, 1,+,2\}\end{align}\] Similar calculations for
\[f_j(i) =
\begin{cases}
\phi_j & \text{if $i = 0$} \\
f_j(i-1) \odot_{j+1} f_{j+1}(i-1) & \text{if $i > 0$}
\end{cases}\] where \(\odot\in\{+,*\}\).
Example 1The simplest case is a loop such as void foo(int m, int *p) { for (int j = 0; j < m; j++) *p++ = j; }The loop writes \(0\) to
*p++ in the first iteration, \(1\) in the second, etc. So we can express the value written at iteration \(i\) as\[f(i) =
\begin{cases}
0 & \text{if $i = 0$} \\
f(i-1) + 1 & \text{if $i > 0$}
\end{cases}\]
Example 2Polynomials in the iteration variable can also be expressed in this form. void foo(int m, int k, int *p) { for (int j = 0; < m; j++) *p++ = j*j*j - 2*j*j + k*j + 7; }We will see below how to build the functions, but the result for the value stored in this loop is \[\begin{align}f_2(i) & =
\begin{cases}
2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\
f_2(i-1) + 6 & \text{if $i > 0$}
\end{cases}\\
f_1(i) & =
\begin{cases}
k-1 & \text{if $i = 0$} \\
f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases}\\
f(i) = f_0(i) & =
\begin{cases}
7 & \text{if $i = 0$} \\
f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases}\end{align}\] One optimization we can see directly from these functions is that the value can be calculated by just three additions within the loop
void foo(int m, int k, int *p) { int t0 = 7; int t1 = k-1; int t2 = 2; for (int j = 0; j < m; j++) { *p++ = t0; t0 = t0 + t1; t1 = t1 + t2; t2 = t2 + 6; } }which is a useful optimization for architectures were multiplication is expensive. This kind of code is, however, uncommon, so most compilers do not do this optimzation – but they usually do this for simpler cases, such as void foo(int m, int k, int *p) { for (int j = 0; < m; j++) *p++ = k*j + 7; }as constructs of the form
k*j+7 are common in address calculations.
Chains of recurrencesIt is cumbersome to write the recursive functions all the time, so the functions are usually written in the form \(\{\phi_j, \odot_{j+1}, f_{j+1}\}\). For example \[\begin{align}f_2(i) & =
\begin{cases}
2\phantom{f_0(i-1) + f_1(i-1)} & \text{if $i = 0$} \\
f_2(i-1) + 6 & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{2,+,6\}$}\\
f_1(i) & =
\begin{cases}
k-1 & \text{if $i = 0$} \\
f_1(i-1) + f_2(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{k-1,+,f_2\}$}\\
f(i) = f_0(i) & =
\begin{cases}
7 & \text{if $i = 0$} \\
f_0(i-1) + f_1(i-1)\phantom{2} & \text{if $i > 0$}
\end{cases} \phantom{xx}\text{is written as $\{7,+,f_1\}$}\end{align}\] These can be chained, so \(f(i)\) can be written as a
chain of recurrences(CR) \(\{7,+,\{k-1,+,\{2,+,6\}\}\}\). The inner curly braces are redundant, so the CR is usually written as a tuple \(\{7,+,k-1,+,2,+,6\}\). Building the chains of recurrencesThe chains of recurrences are built by iterating over the code and calculating the CR for the result of each operation (or marking it as unknown if we cannot handle the operation), using simplification rules such as\[\begin{align}c * \{\phi_0, +, \phi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{c * \phi_0, +, c * \phi_1\} \\
\{\phi_0, +, \phi_1\} + \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 + \psi_0, +, \phi_1 + \psi_1\} \\
\{\phi_0, +, \phi_1\}* \{\psi_0, +, \psi_1\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0 * \psi_0, +, \psi_1 * \{\phi_0, +, \phi_1\} + \phi_1 * \{\psi_0, +, \psi_1\} + \phi_1*\psi_1\} \\
\{\phi_0, +, \phi_1,+,0\} & \phantom{xx} \Rightarrow \phantom{xx} \{\phi_0, +, \phi_1\}\end{align}\] So for the loop from the
sum function
for (int j = 0; j < count; ++j) result += j*j;we start with
j which we know has the CR \(\{0,+,1\}\) per Example 1. This is then used as
j*j when calculating
result, so we calculate the CR for
j*j using the simplification rules as \[\begin{align}j*j& = \{0,+,1\} * \{0,+,1\} \\
& = \{0 * 0, +, 1 * \{0, +,1\} + 1 * \{0, +, 1\} + 1*1\} \\
& = \{0, +, 1,+,2\}\end{align}\] Similar calculations for
result gives us the CR \(\{0,+,0,+,1,+,2\}\) for the value at the beginning of the iteration, and \(\{0,+,1,+,3,+,2\}\) after adding
j*j.
Doing the optimization
The optimization is done during induction variable simplification, and LLVM has transformed the function to a form more convenient for analysis and optimization
One nice property of the chains of recurrences is that it is easy to calculate the value at a specific iteration – if we have a CR \(\{\phi_0,+,\phi_1,+,\ldots,+,\phi_n\}\), then the value at iteration \(i\) can be calculated as \[\begin{align}f(i) & = \sum_{j=0}^{n}\phi_j{i \choose j} \\ & = \phi_0 + \phi_1i + \phi_2{i(i-1)\over 2!} + \ldots + \phi_n{i(i-1)\cdots(i-n+1)\over n!}\end{align}\] Inserting the values for the CR \(\{0,+,1,+,3,+,2\}\) describing One problem with this optimization is that it is hard for developers to make the compiler generate a loop for this if they know that the majority of values used in reality are small enough for the loop to be the fastest option. GCC does, therefore, not replace the final value of a loop if the expression is expensive int sum(int count) { int result = 0; if (count > 0) { int j = 0; do { result = result + j*j; ++j; } while (j < count); } return result; }or as it looks like in the LLVM IR define i32 @sum(i32) { %2 = icmp sgt i32 %0, 0 br i1 %2, label %3, label %6 ; <label>:3: br label %8 ; <label>:4: %5 = phi i32 [ %12, %8 ] br label %6 ; <label>:6: %7 = phi i32 [ 0, %1 ], [ %5, %4 ] ret i32 %7 ; <label>:8: %9 = phi i32 [ %13, %8 ], [ 0, %3 ] ; {0,+,1} %10 = phi i32 [ %12, %8 ], [ 0, %3 ] ; {0,+,0,+,1,+,2} %11 = mul nsw i32 %9, %9 ; {0,+,1,+,2} %12 = add nuw nsw i32 %11, %10 ; {0,+,1,+,3,+,2} %13 = add nuw nsw i32 %9, 1 ; {1,+,1} %14 = icmp slt i32 %13, %0 br i1 %14, label %8, label %4 }The compiler can see that the function returns
0 if
count <= 0, otherwise it returns the value of
result at loop iteration
count-1.
One nice property of the chains of recurrences is that it is easy to calculate the value at a specific iteration – if we have a CR \(\{\phi_0,+,\phi_1,+,\ldots,+,\phi_n\}\), then the value at iteration \(i\) can be calculated as \[\begin{align}f(i) & = \sum_{j=0}^{n}\phi_j{i \choose j} \\ & = \phi_0 + \phi_1i + \phi_2{i(i-1)\over 2!} + \ldots + \phi_n{i(i-1)\cdots(i-n+1)\over n!}\end{align}\] Inserting the values for the CR \(\{0,+,1,+,3,+,2\}\) describing
result gives us \[f(i) = i + {3i(i-1)\over 2} + {i(i-1)(i-2) \over 3}\] The compiler now only need to insert code that calculates this with \(i =\)
count-1, after the loop
result = count-1 + 3*(count-1)*(count-2)/2 + (count-1)*(count-2)(count-3)/3;but it need to take some care to calculate in the correct precision (as temporary values may not fit in 32-bit integers). Integer division is slow, so it is also doing tricks to replace the divisions with multiplication and shift instructions. The result is the LLVM IR %4 = add i32 %0, -1 %5 = zext i32 %4 to i33 %6 = add i32 %0, -2 %7 = zext i32 %6 to i33 %8 = mul i33 %5, %7 %9 = add i32 %0, -3 %10 = zext i32 %9 to i33 %11 = mul i33 %8, %10 %12 = lshr i33 %11, 1 %13 = trunc i33 %12 to i32 %14 = mul i32 %13, 1431655766 %15 = add i32 %14, %0 %16 = lshr i33 %8, 1 %17 = trunc i33 %16 to i32 %18 = mul i32 %17, 3 %19 = add i32 %15, %18 %20 = add i32 %19, -1Inserting this makes the loop become dead, so it is later removed by dead code elimination, and we eventually end up with the code sum(int): test edi, edi jle .LBB0_1 lea eax, [rdi - 1] lea ecx, [rdi - 2] imul rcx, rax lea eax, [rdi - 3] imul rax, rcx shr rax imul eax, eax, 1431655766 add eax, edi shr rcx lea ecx, [rcx + 2*rcx] lea eax, [rax + rcx] add eax, -1 ret .LBB0_1: xor eax, eax ret PerformanceThis optimization is not always profitable. For example, int sum(int count) { int result = 0; for (int j = 0; j < count; ++j) result += j*j*j*j*j*j; return result; }is calculated by three 32-bit multiplications and one addition within the loop, while the optimized version needs six 64-bit multiplications, five 32-bit multiplications, and a slew of other instructions (godbolt), so the optimized version is slower for small values of
count.I benchmarked on my PC, and
count must be larger than
5 for the optimized version to be faster than the loop. Smaller CPUs with, for example, more expensive 64-bit multiplication, will need a higher
count for the optimization to help. And CPUs not having instructions for 64-bit multiplication (godbolt) will need a
muchhigher
count.
One problem with this optimization is that it is hard for developers to make the compiler generate a loop for this if they know that the majority of values used in reality are small enough for the loop to be the fastest option. GCC does, therefore, not replace the final value of a loop if the expression is expensive
/* Do not emit expensive expressions. The rationale is that when someone writes a code like while (n > 45) n -= 45; he probably knows that n is not large, and does not want it to be turned into n %= 45. */ || expression_expensive_p (def))So GCC not doing this optimization is a feature – not a bug. Further readingsChains of recurrences: Olaf Bachmann, Paul S. Wang, Eugene V. Zima. “Chains of recurrences – a method to expedite the evaluation of closed-form functions” Eugene V. Zima. “On computational properties of chains of recurrences” Robert A. van Engelen. “Symbolic Evaluation of Chains of Recurrences for Loop Optimization” Robert A. van Engelen. “Efficient Symbolic Analysis for Optimizing Compilers” Torbjörn Granlund, Peter L. Montgomery. “Division by Invariant Integers using Multiplication” Updated: The original post incorrectly called the power sums “geometric sums”. |
I was thinking about the meaning of location-scale family. My understanding is that for every $X$ member of a location scale family with parameters $a$ location and $b$ scale, then the distribution of $Z =(X-a)/b$ does not depend of any parameters and it's the same for every $X$ belonging to that family.
So my question is could you provide an example where two random from the same distribution family are standardized but that does not results in a Random Variable with the same distribution?
Say $X$ and $Y$ come from the same distribution family (where with family I mean for example both Normal or both Gamma and so on ..). Define:
$Z_1 = \dfrac{X-\mu}{\sigma}$
$Z_2 = \dfrac{Y-\mu}{\sigma}$
we know that both $Z_1$ and $Z_2$ have the same expectation and variance, $\mu_Z =0, \sigma^2_Z =1$.
But can they have different higher moments?
My attempt to answer this question is that if the distribution of $X$ and $Y$ depends on more than 2 parameters than it could be. And I am thinking about the generalized $t-student$ that has 3 parameters.
But if the number of parameters is $\le2$ and $X$ and $Y$ come from the same distribution family with the same expectation and variance, then does it mean that $Z_1$ and $Z_2$ has the same distribution (higher moments)?
Thank you |
@Jamal showed a good way to calculate it, and that they can be somewhat intertwined. Still, for the cosmological solution one can easily see what causes the curvature, and what is curved. A good way to understand it is to solve the equations for general relativity. If you do it for d=4, with one of the dimensions timelike, you get the Robertson Walker metric, and for the modeled equations of state (which account for matter, radiation and if one introduces the cosmological constant, dark energy) one gets the full FLRW solution.
The metric solution can be written as, with k = 0, 1 or -1,
$$ds^2 = dt^2 - a(t)^2[dr^2/(1-kr^2) + d\Sigma^2]$$
with the item in square parenthesis the metric of a constant curvature k spatial section, and the last term is simply the angular components
$$d\theta^2 + sin^2(\theta)d\phi^2$$
You can calculate curvatures, but it is clear the spatial section has intrinsic curvature k, and the spatial slices are growing with a linear scale factor a(t) in each spatial dimension.
The Ricci scalar R, which is an invariant measure of the 4d curvature is dependent on k and a(t). See the reference to Wikipedia below for the exact dependence. This says there is a spacetime curvature, and it is true even if k = 0, which gives flat spatial slices.
So yes, space may be curved (cosmological data has k =0 to within a few percent, i.e. flat space) but spacetime definitely is curved if a(t) changes, which we know it does.
See wiki at https://en.m.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric
This is intrinsic curvature, for the spacetime, and for the spatial slices you can see where is the contribution, if k is not zero, and that otherwise they have zero curvature. The wiki article gives you the non-zero terms for the 4d Ricci components, and in fact $R_{tt}$ is proportional to the second derivative of a(t). Other Ricci components also depend on derivatives of a(t).
You can do all this without invoking any other dimensions. |
I am referencing B. Schutz, A First Course in General Relativity. 2009, p. 256-258.
Note first that the line element on a 2-sphere with radius of curvature $r$ is $dl^2=r^2(d\theta^2+\sin^2\theta d\phi^2)$. Since we want a metric that is spherically symmetric about a point in space, say $r'=0$, we must not be able to tell the difference (in terms of intrinsic geometry) between points located at the same distance $r'$ away from the centre of symmetry and at the same time $t$. Therefore every point in space-time is located on a spatial 2-sphere whose radius of curvature may depend on coordinates $r'$ and $t$, i.e. $r=r(r',t)$. The line element on such a 2-sphere is then $dl^2=r^2(d\theta^2+\sin^2\theta d\phi^2)$, where we have chosen to label each sphere by its radius of curvature $r$ instead of the actual radial distance $r'$ of its points away from the centre of symmetry.
Now consider two spheres corresponding to the same time $t$ but different distances $r'$ and $r'+dr'$. A priori, the coordinates on the two spheres might be defined with respect to different poles where $\theta=0$ and different half-planes where $\phi=0$. However, because the spheres are centred about the same point $r'=0$, we can always choose these references to be the same such that the curves $\theta=const.$, $\phi=const.$ (implicity, we also have $t=const.$) are orthogonal to all spheres at different distances $r'$. By definition, a tangent to these curves is the coordinate basis vector $\textbf{e}_r$. Therefore we have $g_{\theta r}=\textbf{e}_{\theta}\cdot\textbf{e}_r=0$ and $g_{\phi r}=\textbf{e}_{\phi}\cdot\textbf{e}_r=0$. We have thus reduced the most general metric in spherical coordinates, with all the possible cross terms, to the slightly simpler form: $$ds^2=g_{00}dt^2+2g_{0r}dtdr+2g_{0\theta}dtd\theta+2g_{0\phi}dtd\phi+g_{rr}dr^2+r^2d\Omega^2,$$ where I have defined $d\Omega^2\equiv d\theta^2+\sin^2\theta d\phi^2$.
Similarly, we argue that the curves $r=const.$, $\theta=const.$, $\phi=const.$, with tangent vector $\textbf{e}_t$ must be orthogonal to the two-spheres. Otherwise, $\textbf{e}_t$ would have components in the directions of $\textbf{e}_\theta$ and $\textbf{e}_\phi$. There would then be a preferred spatial direction on the 2-sphere, namely that direction parallel to the projection of $\textbf{e}_t$ on the sphere. But this is forbidden by spherical symmetry, therefore $g_{0\theta}=\textbf{e}_t\cdot\textbf{e}_\theta=0$ and $g_{0\phi}=\textbf{e}_t\cdot\textbf{e}_\phi=0$. We have now reduced the metric to: $$ds^2=g_{00}dt^2+2g_{0r}dtdr+g_{rr}dr^2+r^2d\Omega^2.$$ Lastly, a static spacetime is one where the metric is unchanged by a transformation $t \rightarrow -t$. This implies that $g_{0r}=-g_{0r}=0$. Therefore we get a line element that only contains $dt^2$, $dr^2$ and $r^2\Omega^2$, as required by the question. |
Let me first explain what a conjugate prior is. I will then explain the Bayesian analyses using your specific example.Bayesian statistics involve the following steps:Define the prior distribution that incorporates your subjective beliefs about a parameter (in your example the parameter of interest is the proportion of left-handers). The prior can be "...
[Warning: as a card-carrying member of the Objective Bayes Section of ISBA, my views are not representative of all Bayesian statisticians!, quite the opposite...]In summary, there is no such thing as a prior with "truly no information".Indeed, the "uninformative" prior is sadly a misnomer. Any prior distribution contains some specification that is akin ...
The answer depends on whether you are assuming the symmetric or asymmetric dirichlet distribution (or, more technically, whether the base measure is uniform). Unless something else is specified, most implementations of LDA assume the distribution is symmetric.For the symmetric distribution, a high alpha-value means that each document is likely to contain a ...
Suppose that you and a friend are analyzing the same set of data using a normal model. You adopt the usual parameterization of the normal model using the mean and the variance as parameters, but your friend prefers to parameterize the normal model with the coefficient of variation and the precision as parameters (which is perfectly "legal"). If both of you ...
It is not that easy. Information in your data overwhelms prior information not only your sample size is large, but when your data provides enough information to overwhelm the prior information. Uninformative priors get easily persuaded by data, while strongly informative ones may be more resistant. In extreme case, with ill-defined priors, your data may not ...
A half-Cauchy is one of the symmetric halves of the Cauchy distribution (if unspecified, it is the right half that's intended):Since the area of the right half of a Cauchy is $\frac12$ the density must then be doubled. Hence the 2 in your pdf (though it's missing a $\frac{1}{\pi}$ as whuber noted in comments).The half-Cauchy has many properties; some are ...
First of all, there is no such a thing as uninformative prior. Below you can see posterior distributions resulting from five different "uninformative" priors (described below the plot) given different data. As you can clearly see, the choice of "uninformative" priors affected the posterior distribution, especially in cases where the data itself did not ...
Let me complete Zen's answer. I don't very like the notion of "representing ignorance". The important thing is not the Jeffreys prior but the Jeffreys posterior. This posterior aims to reflect as best as possible the information about the parameters brought by the data. The invariance property is naturally required for the two following points. Consider for ...
The Jeffreys prior coincides with the Bernardo reference prior for one-dimensional parameter space (and "regular" models). Roughly speaking, this is the prior for which the Kullback-Leibler divergence between the prior and the posterior is maximal. This quantity represents the amount of information brought by the data. This is why the prior is considered to ...
For simplicity let's just consider a single observation of a variable $Y$ such that$$Y|\mu, \sigma^2 \sim N(\mu, \sigma^2),$$$\mu \sim \mbox{Laplace}(\lambda)$and the improper prior$f(\sigma) \propto \mathbb{1}_{\sigma>0}$.Then the joint density of $Y, \mu, \sigma^2$ is proportional to$$f(Y, \mu, \sigma^2 | \lambda) \propto \frac{1}{\sigma}\...
Yes. The posterior distribution for a parameter $\theta$, given a data set ${\bf X}$ can be written as$$ p(\theta | {\bf X}) \propto \underbrace{p({\bf X} | \theta)}_{{\rm likelihood}} \cdot \underbrace{p(\theta)}_{{\rm prior}} $$or, as is more commonly displayed on the log scale,$$ \log( p(\theta | {\bf X}) ) = c + L(\theta;{\bf X}) + \log(p(\theta)...
Two reasons one may go with a Bayesian approach even if you're using highly non-informative priors:Convergence problems. There are some distributions (binomial, negative binomial and generalized gamma are the ones I'm most familiar with) that have convergence issues a non-trivial amount of the time. You can use a "Bayesian" framework - and particular ...
This is obvious by inspection of the quantity the LASSO is optimizing.Take the prior for $\beta_i$ to be independent Laplace with mean zero and some scale $\tau$.So $p(\beta|\tau) \propto e^{-\frac{1}{2\tau} \sum_i|\beta_i|}$.The model for the data is the usual regression assumption $y \stackrel{\text{iid}}{\sim}N(X\beta,\sigma^2)$.$f(\mathbf{y}|\...
Notice that:\begin{equation}\frac{\alpha\cdot\beta}{(\alpha+\beta)^2}=(\frac{\alpha}{\alpha+\beta})\cdot(1-\frac{\alpha}{\alpha+\beta})\end{equation}This means the variance can therefore be expressed in terms of the mean as\begin{equation}\sigma^2=\frac{\mu\cdot(1-\mu)}{\alpha+\beta+1} \\\end{equation}If you want a mean of $.27$ and a standard ...
The relation of Laplace distribution prior with median (or L1 norm) was found by Laplace himself, who found that using such prior you estimate median rather than mean as with Normal distribution (see Stingler, 1986 or Wikipedia). This means that regression with Laplace errors distribution estimates the median (like e.g. quantile regression), while Normal ...
Many frequentist confidence intervals (CIs) are based on the likelihood function.If the prior distribution is truly non-informative, then the a Bayesianposterior has essentially the same information as the likelihood function.Consequently, in practice, a Bayesian probability interval (or credible interval)may be very similar numerically to a frequentist ...
Here is an attempt to illustrate the last paragraph in Macro's excellent (+1) answer. It shows two priors for the parameter $p$ in the ${\rm Binomial}(n,p)$ distribution. For a few different $n$, the posterior distributions are shown when $x=n/2$ has been observed. As $n$ grows, both posteriors become more and more concentrated around $1/2$.For $n=2$ the ...
You should definitely invest some time in learning the bases of Bayesian statistics and MCMC methods from textbooks or on-line courses.The title and the wording of the question seem to indicate some confusion between the prior modelling [which pertains to the statistical model] and the MCMC implementation [which pertains to the computational resolution]....
From (an earlier version of) the Stan reference manual:Not specifying a prior is equivalent to specifying a uniform prior.A uniform prior is only proper if the parameter is bounded[...]Improper priors are also allowed in Stan programs; they arise from unconstrained parameters without sampling statements. In some cases, an improper prior may ...
Generally, informative priors are typically viewed as your information about parameters (or hypotheses) before seeing the data. So any data-based prior is violating the likelihood principle since evidence from the sample is coming through the likelihood function and the prior.
In frequentist decision theory, there exist complete class results that characterise admissible procedures as Bayes procedures or as limits of Bayes procedures. For instance, Stein necessary and sufficientcondition (Stein. 1955; Farrell, 1968b) states that, under the followingassumptionsthe sampling density $f(x|\theta)$ is continuous in $\theta$ and ...
An appealing property of formal noninformative priors is the "frequentist-matching property" : it means that a posterior 95%-credibility interval is also (at least, approximately) a 95%-confidence interval in the frequentist sense.This property holds for Bernardo's reference prior although the fundations of these noninformative priors are not oriented ...
Although the results are going to be very similar, their interpretations differ.Confidence intervals imply the notion of repeating an experiment many times and being able to capture the true parameter 95% of times. But you cannot say you have a 95% chance of capturing it.Credible intervals (Bayesian), on the other hand, allow you to say that there is a ...
Improper priors are $\sigma$-finite non-negative measures $\text{d}\pi$ on the parameter space $\Theta$ such that$$\int_\Theta \text{d}\pi(\theta) = +\infty$$As such they generalise the notion of a prior distribution, which is a probability distribution on the parameter space $\Theta$ such that$$\int_\Theta \text{d}\pi(\theta) =1$$They are useful in several ...
First of all, there is no such a thing as "uninformative priors" (we rather talk about "weakly informative" priors), each prior brings some kind of assumptions into your model. On another hand, the more information does your data provides, the less influential your prior becomes.But taking this aside, with "uninformative" priors the point estimates from ...
To give a more narrow response than the excellent ones that have already been posted, and focus on the advantage in interpretation - the Bayesian interpretation of a, e.g., "95% credible interval" is that the probability that the true parameter value lies within the interval equals 95%. One of the two common frequentist interpretations of a, e.g., "95% ...
Simply put, a flat/non-informative prior is used when one has little/no knowledge about the data and hence it has the least effect on outcomes of your analysis (i.e. posterior inference).Conjugate distributions are those whose prior and posterior distributions are the same, and the prior is called the conjugate prior. It is favoured for its algebraic ...
As stated in comment, the prior distribution represents prior beliefs about the distribution of the parameters.When prior beliefs are actually available, you can:convert them in terms of moments (e.g. mean and variance) to fit a common distribution to these moments (e.g. Gaussian if your parameter lies to the real line, Gamma if it lies to $R^+$).use ... |
I'm working through Dr. Pete Clark's convergence notes here: http://math.uga.edu/~pete/convergence.pdfand I've been thinking about Exercise 3.2.2 (a) and I am completely stumped.The exercise says ...
In the definition of Riemann integral or Darboux-integral we first study partitions (or tagged partition) of the given interval determined by finitely many points. To each partition and a function $f$ ...
Let $\{(X_\nu,\mu_\nu):\nu\in\Lambda\}$ be a family of measurable spaces. Is it true that$\bigoplus_2\{L_2(X_\nu,\mu_\nu):\nu\in\Lambda\}$ isometrically isomorphic to $L_2\left(\bigsqcup\{(X_\nu,\mu_\...
I'm trying to understand the chapter on Elliptic functions in Stein's Complex Analysis. In particular, I am interested in the construction of Weierstrass's $\wp$ function.Let $\Lambda = \{n + m\tau :...
I was fascinated when I heard that the most intuitive laws of arithmetic are no longer necessarily valid when it comes to the sum of an infinite sequence, which be denoted by$$S = \sum_{n=0,1,2,...}... |
For a linear system
$$ M \dot{u} = Au \qquad \textrm{or} \qquad \dot{u} = L u $$
The
generalized eigenvalue problem is$$A e = \lambda M e$$
We can use the
time-stepper approach which essentially computes the eigenvalues of the operator
$$ E = \exp(L T) $$
If $\mu$ is an eigenvalue of $E$ then it is related to the original eigenvalue by
$$\mu = \exp(\lambda T)$$and $T$ has to be chosen appropriately. If I have a code $E_{\Delta} \approx E$ which solves the initial value problem, then I can couple this to ARPACK to compute the eigenvalues. I would use the
standard eigenvalue solver of ARPACK on the operator $E_\Delta$.
But I would also like to use the inner product induced by $M$ in the Arnoldi process so that the eigenvectors have unit norm in the inner product of $M$. How can I achieve this with ARPACK ? |
Inverted logic can be unnatural. Let's move over to quantified logic:
$$\forall x:({duck}(x)\land {quacks}(x))\lor ({dog}(x)\land {barks}(x))\lor(\lnot {duck}(x)\land\lnot{dog}(x))$$
"Everything is either a duck (and quacks), or a dog (and barks) or else it is neither duck nor dog."
If write down the dual, and then use DeMorgan's on it to flip the logic, we get something unnatural:
Dual (so far so good):
$$\lnot\exists x:\lnot((({duck}(x)\land {quacks}(x))\lor ({dog}(x)\land {barks}(x))\lor(\lnot {duck}(x)\land\lnot{dog}(x)))$$
DeMorgan's, step 1:
$$\lnot\exists x:\lnot(({duck}(x)\land {quacks}(x))\land\lnot({dog}(x)\land {barks}(x)\land\lnot(\lnot {duck}(x)\land\lnot{dog}(x))$$
step 2:
$$\lnot\exists x:(({\lnot duck}(x)\lor {\lnot quacks}(x))\land({\lnot dog}(x)\lor {\lnot barks}(x)\land({duck}(x)\lor{dog}(x))$$
"There does not exist a thing which, simultaneously:
is either a non-quacker or a non-duck; and is either a non-barker or a non-dog; and is a dug duck or a dog."
Say what? :)
Sum-of-products goes hand in hand with divide-and-conquer. A sum-of-products representation of a proposition divides it into all the cases which
independently make it true. Proposition P is true if such and such; or some situation; or if that other case. Division into independent cases assists clarity in reasoning.
Furthermore, in predicate logic and related reasoning, we usually deal with positives, like "duck", and less with negatives like "non-duck". "non-duck" is not a class of object. Things are classified using positive attributes that they do have, not what they don't have. The space of things which are "non-duck" is unbounded. Reasoning with such negatives is confusing.
In propositional logic, as in zeroth order logic without quantifiers, like what we deal with in logic circuits, we can write down the complete truth table. It may turn out that the negative space of a function is in fact simpler to characterize.
For instance a boolean formula over four variables has only a 16 row table. Suppose there are three rows for which it is true, and it is false everywhere else. Then a simple formula is produced by giving those three combinations of four variables, and combine them with
or.
But suppose that the formula is only false in three rows. Then it may be more convenient and natural to characterize these exceptions, and express it that way: the formula is true when the variables are not in this combination,
and not in this other combination, and not in this third combination. The not operators can then distribute into the combinations, yielding a product over sums.
Positive example:
A B C D P
0 0 0 0 0
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1 *
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1 * Sum of products:
1 0 0 0 0 P = A'BC'D' + A'BCD + ABC'D
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 1 *
1 1 1 0 0
1 1 1 1 0
Negative example:
A B C D P
0 0 0 0 1
0 0 0 1 1
0 0 1 0 1
0 0 1 1 1
0 1 0 0 0 *
0 1 0 1 1
0 1 1 0 1
0 1 1 1 0 * Product of sums:
1 0 0 0 1 P = (A'BC'D' + A'BCD + ABC'D)'
1 0 0 1 1 P = (A'BC'D')'(A'BCD)'(ABC'D)'
1 0 1 0 1 P = (A + B' + C + D)(A + B' + C' + D')(A' + B' + C + D')
1 0 1 1 1
1 1 0 0 1 Sum of products:
1 1 0 1 0 * A'B'C'D' + A'B'C'D + A'B'CD' ... (10 more terms)
1 1 1 0 1
1 1 1 1 1
Even so, in spite of its simplicity, it is somewhat hard to understand the third formula (product-of-sums) compared to the second (product-of-negated-products). However, the alternative unsimplified sum of 13 products is also hard to understand, due to the large number of terms. |
Morias
\(\mathrm{mori}(z)=\displaystyle \frac{ J_0 (L_1 z)}{1-z^2}\)
where \(L_1\approx 2.4\) is first zero of the Bessel function, \(J_0(L_1)\!=\!0~\).
\(u\!+\!\mathrm i v=\mathrm{mori}(x\!+\!\mathrm i y)\)
For integer \(m>0\) and \(|z|\gg 1\), function mori\((z)\) can be approximated with
\(\displaystyle \mathrm{mori}(z) \approx \mathrm{morias}_m(z) \) \(\displaystyle= - \sqrt{\frac{2}{\pi L_1}} z^{-5/2} \, G_m(z^2)\, \cos\left( - \frac{\pi}{4} + L_1 z\, F_m(z^2) \right) \)
where
\(G_m(x) = \sum_{n=0}^{m} g_n x^{-n}\) \(F_m(x) = \sum_{n=0}^{m} f_n x^{-n}\)
The coefficients \(f\) and \(g\) can be calculated from the asymptotic expansion of the Hankel function;
\(\begin{array}{l} f_0=1\\ f_1=- 0.021614383628830615865\\ f_2=0.0019465899152260872595\\ f_3=-0.0010834984344719114778\\ f_4=0.001464410164512283719\\ f_5=-0.003628876399615993660\\ f_6=0.01431760830195380729\\ f_7=-0.0824438982874790057\\ f_8=0.652747801052423657\\ f_9=-6.80376838070624330\\ f_{10}=90.322658904953727\\ f_{11}=-1487.942880868968812\\ f_{12}=29785.50901270392374\\ \end{array}\) \(~\) \(\begin{array}{l} g_0=1 \\ g_1=0.989192808185584692068 \\ g_2=0.99228788615079417081 \\ g_3= 0.989481317221334367489\\ g_4= 0.994709980602617872387\\ g_5= 0.97818700495778240956\\ g_6=1.0575251177784290263\\ g_7= 0.5188843197279991625\\ g_8=5.432808917007474985\\ g_9=-52.5640507009104629\\ g_{10}=807.429675670594971\\ g_{11}= -14844.4023379433794\\ g_{12}=328204.367306340176 \end{array}\)
For \(m\!=\!11\), the complex map of function \(\mathrm{morias}_m\) is shown in figure at right. At this value of \(m\), the last line in the table above is not used, as the following increase of number of terms, taken into account, does not improve the precision of approximation with the complex double arithmetics in the examples and numerical tests.
Overlapping of the first two maps is shown in the third image at right.
Function \(\mathrm{morias}_{11}(z)\) approximates \(\mathrm{mori}(z)\) at \(|z|\!>\!2\), and, for \(|z|\!>\!4\), it provides the precise approximation; the error of this approximation is small compared to the rounding error for the complex double variables.
Application
Function morias seems to be useful for the efficient evaluation of function naga that refers of the loss of the pinhole waveguide. However, for the contour integral, morias of the complex argument should be implemented.
For the numerical testing of the asymptotic formulas of the complex argument, it is neserably to have function implemented in a "dingle piece". Then, for the commercial implementation as a built-in function of some language or software, the appropriate combination of morias and expansion of function mori for small and moderate values of the argument can be arranged.
References |
Give me a descent direction and a step length to move and I will find the optimum.
To recap the optimization problem from the previous post, in an unconstrained problem, the objective function is:
\(\min f(x) \tag{1}\)
We use iterative approaches to move \(x^k\) to \(x^{k+1}\) such that
\(f(x^{k+1}) < f(x^k) \tag{2}\)
The problem now is how to move \(x\).
Descent direction
The linear search algorithm [1] chooses a direction \(p^k \in R^n\) and searches along this direction from the current iterate \(x^k \in R^n\) for a new iterate \(x^{k+1} \in R^n\) with a lower function value:
\(x^{k+1} = x^k + \alpha p^k \tag{3}\)
The distance to move along \(p^k\) is called step length \(\alpha^k \in R\), which can be found by solving a 1-D minimization problem:
\(\min_{\alpha > 0} f(x^k + \alpha p^k) \tag{4}\)
If \(f(x)\) is continuously differentiable with gradient \(\nabla f(x)\), using Taylor theorem, we have:
\(f(x^k + \alpha p^k) = f(x^k) + \alpha^k \nabla f(x^k)^T p^k + o\|\alpha^k p^k \|\tag{5} \)
We would like to have a descent direction for \(p^k\), which means
\(\nabla f(x^k)^Tp^k < 0 \tag{6.1}\)
\(\nabla f(x^k) = (\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, … \frac{\partial f}{\partial x_n})^T \tag{6.2}\)
\(\nabla f(x^k) \in R^n\) is a n-dimension vector, and \(\nabla f(x^k)^Tp^k \) is a scalar.
Most search algorithms have the solution format as below:
\(p^k = -(B^k)^{-1} \nabla f(x^k) \tag{7} \)
\(B^k \in R^{n \times n}\) is a symmetric and nonsingular square matrix. From previous post on matrix, we learned that a symmetric matrix can be decomposed into triangular matrices \(A = LL^T\) using Cholesky decomposition.
Steepest descent
In equation 5, at \(x^k\), function \(f\) changes along direction \(p^k\) with a rate of the step length \(\alpha^k\). Thus, \(p^k\) is the most rapid decrease direction:
\(p^k = \min_p \nabla f(x^k)^T p, s.t.\|p\| = 1 \tag{8} \)
Note that here \(p^k \in R^n\) is a direction vector, and \(\nabla f(x^k)\) represents the gradient of \(f\) at \(x^k\). Thus
\(\nabla f(x^k)^T p = \|\nabla f(x^k)\|\|p\| cos \theta \tag{9} \)
Equations 9 has the minimum value when \(cos \theta = -1, \theta = \pi\). This means \(p^k\) takes the opposite direction of the function gradient.
\(p^k = – \frac {\nabla f(x^k)^T}{\| \nabla f(x^k)\|} \tag{10} \)
As shown in the figure, in 1-D case, \(p^k\) represents the tangential of function \(f\) and takes opposite direction from the gradient at \(x^k\). In 2-D case, function \(f\) is usually depicted by its contours and \(p^k\) is orthogonal to the contours [5].
The steepest descent algorithm uses
\(p^k = – \nabla f(x^k)^T \tag{11}\)
This means in equation 7, we have:
\(B = I \tag{12}\)
The optimal step length: search or not search
If you recall the implementation of logistic regression, we use a fixed value of step length of 1. Using too large a step length could lead to non-convergence, and too small a step length could lead to slow convergence. In theory, we can perform a 1-D search such that
\(\min_{\alpha >0} \phi (\alpha) \tag{13.1} \)
\(\phi (\alpha) = f(x^k + \alpha p^k) \tag{13.2}\)
Solving equation 13.1, we achieve the maximum benefit (decrease) of moving along \(p^k\).
We can use iterative golden-section search and interpolation to search the minimum [2]. In practice, however, a precise 1-D search can be expensive and slow, and most of the time we only need an approximate to achieve sufficient reduction of \(f\). There are several approaches to approximate step length [1], which I won’t cover here.
The implementation of iterative gradient descent methods usually does not perform search for the optimal step length at each step. Alternatively, we may treat step length as a hyperparameter to tune in model training and use a fixed value for each model, or we may adaptively modulate step length at each step.
Configuring learning rate is critical in training complex models such deep neural network, and latest development such as momentum, adaptive learning rate, learning rate schedule has been shown to improve the optimization process [3][4]. I will discuss more about learning rate later in deep learning posts.
Algorithm
Termination condition is defined by tolerance \(\epsilon >0\), see previous post.
initiate \(x^0\) while termination condition not met: compute \(p^k = – \nabla f(x^k)\) compute \(\alpha^k = \min_{\alpha >0} f(x^k + \alpha p^k)\) or use a fixed value of \(\alpha\) \(x^{k+1} = x^k + \alpha^k p^k\) \(k = k+ 1\) return \(x^k, f(x^k)\)
When using precise search for step length \(\alpha^k\), from equation 13.1, we get
\(\nabla \phi (\alpha)= 0 \tag {14.1}\)
\(\nabla \phi (\alpha) = \nabla f(x^k + \alpha p^k)^T p^k = 0 \tag {14.2}\)
In the next iteration,
\(p^{k+1} = -\nabla f(x^{k+1}) = – \nabla f(x^k + \alpha p^k) \tag{15}\)
Therefore, from equation 14.2 and 15, we have
\((p^{k+1})^T p^k =0 \tag{16.1}\)
\(\nabla f(x^{k+1})^T \nabla f(x^{k}) =0 \tag{16.2}\)
As a result of this zig-zag search manner, steepest descent achieves most rapid local descent, it may not achieve the most rapid global descent.
Newton method
If \(f(x)\) is continuously twice differentiable, using Taylor theorem, we have:
\(f(x^k + p^k) = f(x^k) + \nabla f(x^k)^T p^k + \frac{1}{2}(p^k)^T \nabla ^2 f(x^k)p^k + o\|p^k \|^2\tag{17} \)
\(p^k = \min_p m(p) \tag{18.1}\)
\(m(p^k) = f(x^k + p^k) \tag{18.2}\)
\(\nabla m(p^k) = \nabla f(x^k) + \nabla^2 f(x^k) p^k = 0 \tag{19}\)
\(\nabla ^2 f(x^k) \in R^{n \times n}\) is a symmetric square matrix of second derivatives of \(f\). If \(\nabla ^2 f(x^k)\) is non-singular, from equation 19, we can derive:
\(p^k = – (\nabla^2 f(x^k))^{-1}\nabla f(x^k) \tag{20} \)
We can write first and second derivatives as:
\(g = \nabla f(x), H = \nabla^2 f(x) \tag{21}\)
\(p^k\) is a descent direction, only if Hessian matrix \(H\) is positive definite and satisfies \(x^THx > 0\). Combining equation 20 and 21, we have
\(\nabla f(x^k)^T p^k \\= – \nabla f(x^k)^T (\nabla^2 f(x^k))^{-1} \nabla f(x^k)\\ = – g^T H^{-1} g < 0\tag{22}\)
This means in equation 7, we have:
\(B = \nabla^2 f(x^k) \tag{23}\)
Algorithm initiate \(x^0\) while termination condition not met: compute \(p^k = – (\nabla^2 f(x^k))^{-1}\nabla f(x^k) \) \(x^{k+1} = x^k + p^k \) \(k = k+1\) return \(x^k, f(x^k)\)
Note the step length is set as 1 for Newton Method. It is also possible to perform 1-D search for the optimal step length as done in steepest descent.
Shown in the figure at the beginning of this post are steepest descent (white lines) and Newton method (red lines) [6]. Compared with steepest descent, Newton method converges faster when the initial selection of \(x^0\) is close enough to the minimum \(x^{*}\). However, computation of Hessian matrix and its inverse can be expensive and slow. In addition, if \(x^0\) is too far away from \(x^{*}\), \(x^k\) may not converge.
In practice, steepest descent is more commonly used in function optimization given its advantage in computation.
Other not-so-common methods Quasi-Newton
To avoid computation of Hessian matrix, we can replace the true Hessian with an approximation \(B^k\), which makes use of the first gradient to gain knowledge of the second derivative [1].
Conjugate gradient
The conjugate gradient method is used to solve linear optimization problems when the objective function has a quadratic format:
\(f(x) = \frac{1}{2}x^TAx + b^Tx \tag{24}\)
From equation 16.1 in steepest descent, we know that \(p^k, p^{k+1}\) are orthogonal. We may also construct a series of directions \(p^0, p^1, … p^{n-1}\) such that \(p_i, p_j\) are conjugate with respect to a positive definite symmetric matrix \(A\):
\(p_i^TAp_j = 0, i \neq j \tag{25.1}\)
\(p_i^TAp_i > 0 \tag{25.2}\)
\(p^0, p^1, … p^{n-1}\) are also linearly independent.
The first derivative of equation 24 is
\(\nabla f(x) = Ax + b \tag{26} \)
With limited steps \(n\), we can reach convergence.
\(\nabla f(x^n) = Ax^n + b = 0 \tag{27} \)
The construction of conjugate coordinates is flexible. Usually, we can initiate \(p^0\) based on gradient, and construct subsequent \(p^{k+1}\) using \(p^k\) and first derivatives at both \(x^k\) and \(x^{k+1}\), which I won’t cover here.
Algorithm
Termination condition: \(\nabla f(x^k) = 0\)
initiate \(x^0, k = 0\) compute \(p^0 = – \nabla f(x^0)\) while termination condition not met: compute \(\alpha^k = \min_{\alpha >0} f(x^k + \alpha p^k)\) or use a fixed value of \(\alpha\) \(x^{k+1} = x^k + \alpha^k p^k\) \(p^{k+1} = – \nabla f(x^{k+1}) + \upsilon ^k p^k\), \(\upsilon^k = \frac{\|\nabla f(x^{k+1}) \|^2} {\|\nabla f(x^{k}) \|^2} \) \(k = k+1\) return \(x^{k+1}, f(x^{k+1})\) What about gradient-free functions?
All of the methods discussed so far compute first derivative of function \(f\) (gradient). What if \(f\) does not have a explicit format and we do not know its gradient?
Gradient-free functions are essentially a black box and there are several commonly used methods for black-box optimization including grid search, genetic algorithm, simulated annealing, and Bayesian Optimization. I will discuss Bayesian Optimization in later post.
References
[1] Numerical Optimization, Jorge Nocedal, Stephen J. Wright
[2] https://github.com/wzhe06/Ad-papers/blob/master/Optimization%20Method/%E9%9D%9E%E7%BA%BF%E6%80%A7%E8%A7%84%E5%88%92.doc
[3] learning rate https://machinelearningmastery.com/learning-rate-for-deep-learning-neural-networks/
[4] momentum https://distill.pub/2017/momentum/
[5] wikipedia https://en.wikipedia.org/wiki/Gradient_descent
[6] wikipedia https://it.wikipedia.org/wiki/Metodo_del_gradiente_coniugato |
First of all, by a simple geometry principle:$\triangle CED$ and $\triangle AED$ have the same base $|ED|$ and thearea ratio between $\triangle CEF$ and $\triangle CFD$ has to be thesame as the length ratio between $|EF|$ and $|FD|$ so, similarly,you can tell the area ratio between $\triangle AEF$ and $\triangleAFD$ as $2x$ and $3x$.Notice from ...
It's possible if Professor Erasmus can reflect triangles when creating isosceles triangles. If we imagine them as being cut out of paper, this would be by flipping over the paper.This is an isosceles triangle if we set $x=\sqrt2-1$, making the bottom and right edges both have length $\sqrt2$.The two leftmost triangles form a symmetric isoceles triangles, ...
Using non-overlapping sticks I get 36.I'll give measurements and positions to make this easy to picture. Imagine we're drawing this on graph paper. Make one horizontal stick of width eight units. Pick a point somewhere above it, and attach sticks from it to each of 9 points on the line, all one unit apart.Now how many triangles are there?say the ...
Now that we have two correct answers, I figured I'd present my own approach. It's similar to Paul's but doesn't work with the ratios of the side lengths, but instead directly with the ratios of the areas.I'm splitting the solution into several spoiler paragraphs so that they can be used as individual hints towards the solution.As a corollary, note that ...
No, it's not possible.For any vertex, the three edges coming out of it must have distinct length, since otherwise the triangle must be isosceles as well as right, which forces an equilateral triangle for the opposite face. Therefore, the three angles at each vertex must correspond to the three angles of the right triangle; call them $\alpha$, $\beta$, and $...
It is possible to do better than a hexagon, if an irregular polygon is acceptable.It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. By taking $\sqrt{3}$ by $1$ squares, each made from two triangles, you could ...
Two 5-point stars should do the trick! This solution uses 10 'sticks' (each stick spans vertex-to-vertex) of exactly the same length in 2D.As someone commented briefly (sadly, it was deleted too quickly for me to reference), we get 10 triangles even with a single 5-point star: 5 large, 5 small, and the large triangles overlap the small triangles.Note: ...
Here is one possible triangulation.Note that all triangulations (that do not introduce extra vertices) will have the same number of triangles, which depends only on the number of sides in the polygon.A sketch of the proof: A polygon with three sides is already a triangle, so the minimal number of triangles needed to triangulate it is exactly $1$. For a ...
If the triangles must have a named(A to H) point at each corner then I think the answer is actually:Count:The triangles in one half only(left half in this case):(B,E,D) (B,E,G) (B,G,D)(A,D,E) (A,D,G) (A,G,E)(A,B,E) (A,D,B)so 16 for both halvesplus the triangles which go over both halves(A,H,G) (A,C,B)(D,G,H) (D,E,F)(D,...
The professor has blundered. The triangle he claims to have constructed cannot exist in Euclidean space.(I apologize for the lack of pictures. I know they would make my explanation/proof easier to follow, but I do not currently have access to any tools to create or upload them.)Let's start by considering an isosceles triangle, $ABC$, that has been ...
I count 40 different triangles in the pictureTo check this I write them down in a sorted list of triples. Each triple isnormalized in that way that the components of the triple are alphabecticalyordered.1. A B C2. A B D3. A B E4. A B J5. A C D6. A C F7. A C J8. A D E9. A D F10. A D G11. A ... |
I am not very familiar with the common discretization schemes for PDEs. I know that Crank-Nicolson is popular scheme for discretizing the diffusion equation.
Is also a good choice for the advection term?
I am interesting in solving the Reaction-Diffusion-Advection equation,
$\frac{\partial u}{\partial t} + \nabla \cdot \left( \boldsymbol{v} u - D\nabla u \right) = f$
where $D$ is the diffusion coefficient of substance $u$ and $\boldsymbol{v}$ is the velocity.
For my specific application the equation can be written in the form,
$\frac{\partial u}{\partial t} = \underbrace{D\frac{\partial^2 u}{\partial x^2}}_{\textrm{Diffusion}} + \underbrace{\boldsymbol{v}\frac{\partial u}{\partial x}}_{\textrm{Advection (convection)}} + \underbrace{f(x,t)}_{\textrm{Reaction}}$
Here is the Crank-Nicolson scheme I have applied,
$\frac{u_{j}^{n+1} - u_{j}^{n}}{\Delta t} = D \left[ \frac{1 - \beta}{(\Delta x)^2} \left( u_{j-1}^{n} - 2u_{j}^{n} + u_{j+1}^{n} \right) + \frac{\beta}{(\Delta x)^2} \left( u_{j-1}^{n+1} - 2u_{j}^{n+1} + u_{j+1}^{n+1} \right) \right] + \boldsymbol{v} \left[ \frac{1-\alpha}{2\Delta x} \left( u_{j+1}^{n} - u_{j-1}^{n} \right) + \frac{\alpha}{2\Delta x} \left( u_{j+1}^{n+1} - u_{j-1}^{n+1} \right) \right] + f(x,t)$
Notice the $\alpha$ and the $\beta$ terms. This enables scheme to move between:
$\beta=\alpha=1/2$ Crank-Niscolson, $\beta=\alpha=1$ it is fully implicit $\beta=\alpha=0$ it is fully explicit
The values can be different, which allows the diffusion term to be Crank-Nicolson and the advection term to be something else. What is the most stable approach, what would you recommend? |
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues?
Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson...
Hmm, it seems we cannot just superimpose gravitational waves to create standing waves
The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line
[The Cube] Regarding The Cube, I am thinking about an energy level diagram like this
where the infinitely degenerate level is the lowest energy level when the environment is also taken account of
The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume
Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings
@Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer).
Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it?
Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks.
I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh...
@0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P)
Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio...
the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above
If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta$$$$\vec{A}\cdot\...
@ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there.
@CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer |
I've tried to explain the solution using as little math as possible, and to give some intuition as to what makes it tick. Nonetheless there will be a little mathematical notation at the end.
First steps: going beyond the obvious solution in the simplest case (N=2)
The statement of the puzzle as presented here doesn't make this very clear, but the puzzle relies on the prisoners not knowing anything about which name is located in which box (until they get into the room, after which they cannot communicate anymore). If every prisoner checks 50 boxes at random, then each prisoner has a ½ chance of finding his own name. If all the prisoners choose a set of boxes at random, independently, then the probability that they all find their own name is ½ × … × ½ = 1/2
100 — infinitesimal.
Making independent choices is a waste, though. If anyone gets it wrong, the situation isn't worse than if everybody gets it wrong. Rather than make independent choices, they can make correlated choices; the idea is to try to arrange that either everybody gets it right, or many get it wrong.
Let's consider the simpler case when there are two prisoners. If they both choose at random, then they have ½ × ½ = ¼ chance of surviving. But there's an obvious waste: suppose prisoner #1 opens the left-hand box and finds his name: then prisoner #2 will not find his name in the left-hand box. So the prisoners can decide that #1 will look at the box on the left and #2 looks at the box on the right: that way, either they both get it right or both get it wrong, and they have a ½ chance of survival.
Incidentally, note that another assumption that wasn't clearly stated here is that the prisoners get to formulate their strategy in secret. If the warden knows which prisoner chose which box, he can arrange for the prisoners to fail by putting prisoner #1's name in the right-hand box.
The next step: N=4
The obvious way to generalize this to more prisoners is to assign each prisoner a fixed set of boxes that he'll open. However, I won't pursue this further, because it doesn't take advantage of an important ability: after a prisoner has opened a first box, he can base his decision on which box to open next on the content of the first box, and so on.
Consider the case with 4 prisoners and 4 boxes. I'll use numbers for the prisoners' names, and assume that the boxes are numbered as well. Intuitively, it is preferable for each prisoner to pick a different box to open first, since otherwise some common choices are wasted. So prisoner #1 opens box #1 and finds a name (number). Now what? If he finds his own name (#1) (¼ chance), of course, he can stop. If he finds some different name (say 2) (¾ chance), what information does this provide? Well, since each box contains a different name, prisoner #1 now knows that box #2 does not contain 2, so prisoner #2 will not be lucky the first time either. Furthermore, the strategy should favor arranging for prisoner #2 to pick box #1 next.
To simplify the analysis, I'll only look at cases where all prisoners follow the same strategy. (I don't have an intuitive argument as to why breaking the symmetry wouldn't be advantageous.) Either they all open the box whose number they found in the first box, or they all open a different box.
If #1 opens box #2 and finds his name there, then either boxes #3 and #4 contain 3 and 4 respectively, or they contain 4 and 3. Either way, with the strategy of using the name in the first box, if one prisoner is lucky the second time then every prisoner is lucky! If #1 opens box #3 instead and finds his name there, then there is a ½ chance that prisoner #2 will find his name in box #2, and a ½ chance that he'll find his name in box #4. But what about prisoner #3? He finds the name of prisoner #1 in box #3, which doesn't give any clue as to where 3 might be instead.
So let's concentrate on the strategy where each prisoner opens the second box whose number is what he found in the first box. What arrangement of numbers in boxes make it work?
There are 4 ways to choose which box contains the number 1. The number 2 can go into any of the 3 remaining boxes. The number 3 can go in either of the 2 remaining boxes, and the number 4 must go into the one remaining box. So there are 4×3×2 = 24 different arrangements. The following arrangements lead to success because each number is either in its own box or swapped with another number:
1234 1243 1324 1432 2134 2143 3214 3412 4231 4321
That's 10 successful arrangements out of 24. The chance of success isn't very far from the theoretical maximum of ½, which is encouraging.
Note that in order for the chance of success to be 10/24, we need to know that the arrangements have an equal chance of being chosen. If the warden is nasty and arranges the numbers as 2341, the prisoners are sure to lose. This is where the fact that the prisoners choose a strategy in secret comes in. In my analysis, I used numbers for prisoners — but fact the prisoners are names, not numbers, and they can pick a random assignment of names to numbers as part of their secret strategy (in fact, this assignment is the only secret part, since the warden may have looked up the solution of the puzzle).
General analysis
Let's explore a strategy that generalizes what we explored for 4 boxes: each prisoner opens the box with his own number, then the box whose number is contained in the first box, and so on. Consider the sequence of numbers that a certain prisoner encounters: $x_0$ (the inital box numbered with the prisoner's own number), $x_1$ (number contained in box $x_0$), $x_2$ (number contained in box $x_1$), … Since each number is contained in only one box, this sequence cannot contain any repeated element as long as it doesn't loop back to $x_0$. Eventually the sequence has to loop back to $x_0$ since it will run out of numbers. At that point, the prisoner has found his own name. The critical problem for the prisoner is whether the loop completes before or after the prisoner has opened the maximum of 50 boxes.
From now on, let me use the proper mathematical vocabulary. A way to arrange distinct numbers into as many boxes is called a
permutation. Opening box number $k$ and looking at the number that it contains is called applying that permutation. Repeated applications of a permutation eventually runs into a loop; such a loop is called a cycle. The prisoners succeed if all of the cycles for the permutation have a length of at most 50.
Let's call a cycle
long if it contains 51 or more elements. Observe that there cannot be more than one long cycle (if one cycle has at least 51 elements, then there are only 49 or fewer elements to share between the other cycles). So we can count the losing configurations by adding up the permutations of 100 elements that have a cycle of length 51, 52, …, 100.
Lemma: there are $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot (n-1) \cdot n$ distinct permutations of $n$ elements. Proof: there are $n$ ways to pick the image of the first element, $n-1$ remaining ways to pick the image of the second elements, etc., down to a single way to pick the image of the last element.
Now let's count the number of permutations that have a cycle of length $c$ (for $c \ge 51$, so that there's a single such cycle). We're actually going to count each permutation $c$ times, once for each element of the cycle. Pick the first element in the cycle: there are 100 possibilities. There are 99 possibilities for the second element, and so on, until we've picked $c$ elements. So far, that's $100 \times 99 \times \ldots \times (100-c+1)$ possibilities. There are $100-c$ remaining elements, and they can be permuted in any way, so there are $(100-c)!$ possibilities as per the lemma above. That's a total of $(100 \times 99 \times \ldots \times (100-c+1)) \times ((100-c) \times \ldots \times 2 \times 1)$ possibilities, which nicely collapses to $100!$. Recall that we counted each permutation $c$ times, since we counted it once per element in the cycle. So the number of permutations with a cycle of length $c$ is $100! / c$.
The number of permutations with a long cycle is thus$$ \frac{100!}{100} + \frac{100!}{99} + \ldots + \frac{100!}{51} $$That's out of a total of $100!$ permutation. The proportion of failing permutations is thus$$ \frac{1}{100} + \frac{1}{99} + \ldots + \frac{1}{51} $$Numerically, this is about 0.6882, i.e. the chance of success is about 31.18%, a little over the requisite 30%.
In general, the proportion of failing permutations for $N$ prisoners is $H_N - H_{N/2}$ where$$ H_n = 1 + \frac{1}{2} + \ldots + \frac{1}{n} $$is called the $n$th harmonic number. For large values of $n$, $H_n \approx \ln n + C$ for some number C, and the series $H_N - H_{N/2}$ converges to $\ln 2 \approx 0.6931$ from below. (I will not provide an elementary proof of that). This gives a lower limit to the chance of success for large numbers of prisoners: 30.68% is achievable. |
Sample Quantiles
The generic function
quantile produces sample quantiles corresponding to the given probabilities. The smallest observation corresponds to a probability of 0 and the largest to a probability of 1.
Keywords univar Usage
quantile(x, …)
# S3 method for defaultquantile(x, probs = seq(0, 1, 0.25), na.rm = FALSE, names = TRUE, type = 7, …)
Arguments x
numeric vector whose sample quantiles are wanted, or an object of a class for which a method has been defined (see also ‘details’).
NAand
NaNvalues are not allowed in numeric vectors unless
na.rmis
TRUE.
probs
numeric vector of probabilities with values in \([0,1]\). (Values up to
2e-14outside that range are accepted and moved to the nearby endpoint.)
na.rm
logical; if true, any
NAand
NaN's are removed from
xbefore the quantiles are computed.
names
logical; if true, the result has a
namesattribute. Set to
FALSEfor speedup with many
probs.
type
an integer between 1 and 9 selecting one of the nine quantile algorithms detailed below to be used.
…
further arguments passed to or from other methods.
Details
A vector of length
length(probs) is returned; if
names = TRUE, it has a
names attribute.
The default method works with classed objects sufficiently like numeric vectors that
sort and (not needed by types 1 and 3) addition of elements and multiplication by a number work correctly. Note that as this is in a namespace, the copy of
sort in base will be used, not some S4 generic of that name. Also note that that is no check on the ‘correctly’, and so e.g.
quantile can be applied to complex vectors which (apart from ties) will be ordered on their real parts.
Types
quantile returns estimates of underlying distribution quantiles based on one or two order statistics from the supplied elements in
x at probabilities in
probs. One of the nine quantile algorithms discussed in Hyndman and Fan (1996), selected by
type, is employed.
All sample quantiles are defined as weighted averages of consecutive order statistics. Sample quantiles of type \(i\) are defined by: $$Q_{i}(p) = (1 - \gamma)x_{j} + \gamma x_{j+1}$$ where \(1 \le i \le 9\), \(\frac{j - m}{n} \le p < \frac{j - m + 1}{n}\), \(x_{j}\) is the \(j\)th order statistic, \(n\) is the sample size, the value of \(\gamma\) is a function of \(j = \lfloor np + m\rfloor\) and \(g = np + m - j\), and \(m\) is a constant determined by the sample quantile type.
Discontinuous sample quantile types 1, 2, and 3
For types 1, 2 and 3, \(Q_i(p)\) is a discontinuous function of \(p\), with \(m = 0\) when \(i = 1\) and \(i = 2\), and \(m = -1/2\) when \(i = 3\).
Type 1
Inverse of empirical distribution function. \(\gamma = 0\) if \(g = 0\), and 1 otherwise.
Type 2
Similar to type 1 but with averaging at discontinuities. \(\gamma = 0.5\) if \(g = 0\), and 1 otherwise.
Type 3
SAS definition: nearest even order statistic. \(\gamma = 0\) if \(g = 0\) and \(j\) is even, and 1 otherwise.
Continuous sample quantile types 4 through 9
For types 4 through 9, \(Q_i(p)\) is a continuous function of \(p\), with \(\gamma = g\) and \(m\) given below. The sample quantiles can be obtained equivalently by linear interpolation between the points \((p_k,x_k)\) where \(x_k\) is the \(k\)th order statistic. Specific expressions for \(p_k\) are given below.
Type 4
\(m = 0\). \(p_k = \frac{k}{n}\). That is, linear interpolation of the empirical cdf.
Type 5
\(m = 1/2\). \(p_k = \frac{k - 0.5}{n}\). That is a piecewise linear function where the knots are the values midway through the steps of the empirical cdf. This is popular amongst hydrologists.
Type 6
\(m = p\). \(p_k = \frac{k}{n + 1}\). Thus \(p_k = \mbox{E}[F(x_{k})]\). This is used by Minitab and by SPSS.
Type 7
\(m = 1-p\). \(p_k = \frac{k - 1}{n - 1}\). In this case, \(p_k = \mbox{mode}[F(x_{k})]\). This is used by S.
Type 8
\(m = (p+1)/3\). \(p_k = \frac{k - 1/3}{n + 1/3}\). Then \(p_k \approx \mbox{median}[F(x_{k})]\). The resulting quantile estimates are approximately median-unbiased regardless of the distribution of
x.
Type 9
\(m = p/4 + 3/8\). \(p_k = \frac{k - 3/8}{n + 1/4}\). The resulting quantile estimates are approximately unbiased for the expected order statistics if
xis normally distributed.
Further details are provided in Hyndman and Fan (1996) who recommended type 8. The default method is type 7, as used by S and by R < 2.0.0.
References
Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988)
The New S Language. Wadsworth & Brooks/Cole.
Hyndman, R. J. and Fan, Y. (1996) Sample quantiles in statistical packages,
American Statistician 50, 361--365. 10.2307/2684934. See Also Aliases quantile quantile.default Examples
library(stats)
# NOT RUN {quantile(x <- rnorm(1001)) # Extremes & Quartiles by defaultquantile(x, probs = c(0.1, 0.5, 1, 2, 5, 10, 50, NA)/100)### Compare different typesquantAll <- function(x, prob, ...) t(vapply(1:9, function(typ) quantile(x, prob=prob, type = typ, ...), quantile(x, prob, type=1)))p <- c(0.1, 0.5, 1, 2, 5, 10, 50)/100signif(quantAll(x, p), 4)## for complex numbers:z <- complex(re=x, im = -10*x)signif(quantAll(z, p), 4)# }
Documentation reproduced from package stats, version 3.5.2, License: Part of R 3.5.2 |
No, I don't think auto-regulation explain much in the population sizes of predators. Group selection may explain such auto-regulation but I don't think it is of any considerable importance for this discussion.The short answer is, as @shigeta said[predators] tend to starve to death as they are too many!To have a better understanding of what @shigeta ...
A quick back-of-the-envelope answer to the number of generations that have passed since the estimated human-chimp split would be to divide the the split, approximately 7 million years ago (Langergraber et al. 2012), by the human generation time. The human generation time can be tricky to estimate, but 20 years is often used. However, the average number is ...
The conservation biology literature has a great deal of information, particularly with reference to developing species survival plans (e.g., Traill et al. [2007] report a minimum effective population size of ~4,000 will give a 99% persistence probability of 40 generations).Because the question specifically mentions human populations, I'll focus my answer ...
Remi.b's answer is an excellent one, and this should be taken as a supplement to it:It's possible your simulation is correctThe Lotka-Volterra equations are what is known as a deterministic model, and it describes the behavior of predator-prey systems (in a somewhat simplified fashion) in large populations. Small populations are subject to what is known ...
Let's see!I took the most recent WHO data from here and did a quick an dirty analysis in R. Here is the histogram as well as a normal distribution with the same mean and standard deviation as the actual data:Does not look very normally distributed. In fact, the shapiro test confirms this impression:Shapiro-Wilk normality testdata: df$life_expectancy...
it is impossible to know the exact number so here is my gross ballpark estimate of an upper bound - i.e. the maximum number of organisms that could have lived on earth in the extreme best case scenario. in practice it is probably much less, but this is to get an idea of what kind of numbers we are dealing with.The earth's volume is about 1.08321 * 10^12 ...
This article discusses the origin of the terms. They come directly from the equation used to describe population dynamics. As Canadianer mentioned the Wiki also covers it quite well."r" stands for "rate" {growth rate}, r strategists have a high r value and a low K value. They grow fast but most die."K" stands for Kapazitätsgrenzen which is german for ...
One of the possible adjustments of these mathematical models is to introduce a "place to hide", making some (small) percent of the prey population not accessible (or much more difficult to access) for predators. After the number of predators decreases from starvation, prey individuals are relatively safer outside the "place to hide" and can grow over this ...
Leonardo's already given you an excellent answer, but I thought I'd add my perspective. I'm a mathematical epidemiologist, so I'd at least like to believe these types of models are useful.For me, there are a number of things population dynamics models are especially useful for:Highlighting data requirements. Yes, models need data, as you've mentioned. ...
It is certainly possible to use general relationships to predict human population density or abundance. The relationship between population density/abundance and body size is an old topic in ecology, that fall within the field of allometrics (how different features of organisms scale with body size). A closely related allometric relationship is Kleiber's law,...
Mark-recapture is the most frequently used method for small mammals. It's best when combined with uncertainty estimates and population dynamics models (e.g. projection matrix).The fluctuations of small rodent populations have long fascinated scientists, and various models have been developed. Logistic regression models can be used to estimate likelihood ...
I think it does make sense - with a population density for finland that is so low, the disease with such a low beta cannot communicate to enough people to propagate.The number of people who have this disease will be fewer each week. I think this makes sense because at 16 / km^2, you can expect that practically nobody will ever see each other.This is ...
Two previous answers listed many applications of population dynamics models. I want to add that they are also important for conservation of endangered species. For example classical stage-class model (Crouse et al 1987, free copy) indicate that the most effective way to protect sea turtles is reducing mortality of large juveniles.Moreover, you don't have ...
There is one book that will perfectly suits your needs:A biologist's guide to Mathematical Modeling in Ecology and Evolution, by Sally OttoIt is a very good book that is very easy to understand and in the meantime goes pretty far (It ends with the use of diffusion equation in Evolutionary Biology). I highly recommend it.It covers:How to create a ...
You can make the continuous approximation when the population size is large. As mentioned by arboviral, there are algorithms that allow you to perform stochastic simulations with discrete variables. However, these are computationally much more intensive than integration of ODEs. Moreover, analytical solutions for the master-equations (time evolution of ...
I commented on this question, but the OP's response prompted me to think again. Here is the graph from the document that the OP linked to:Clearly what is confusing is that the parameter referred to as 'antibody level' rises quickly, but not as a step. In terms of the x-axis, vaguely labelled as weeks, it looks as if the level of antibody continues to rise ...
We call it Colony Collapse Disorder (CCD).ConsequencesThe phenomenon is observed worldwide and is pretty serious. Northern Ireland lost 50% of its beehives for example. Between 1997 and 2003, 10 millions beehives were lost. Many cultivated crops are pollinated by bees and we don't quite have an alternative today. In 2005 a study showed that the worth of ...
As Alan Boyd says, the relatively slow rise is due to gradual uptake of the injected antibody.If you deliver the antibodies by intravenous injection or another mode that allows rapid uptake (I use intraperitoneal injection in mice) then the antibody levels peak rapidly, less than a day and probably a couple hours, and then drop off; there is no continued ...
The chaotic behaviour you are referring to (at least the one described in your link in the comments) is a property of the discrete version of the logistic equation, where you get chaotic dynamics at growth rates above ~3.55 (see the logistic map). The behaviour of this equation has been described in a classic paper by Robert May (1976). As you increase ...
These equations describe how the haplotype frequencies will change over time due to a combination of recombination and natural selection.Before I proceed, I need to change your four $\delta X_i$ formulas above. Lewontin and Kojima (1960) writes the equations as:$$\Delta X_i = \frac{X_i(w_i - \bar w) \pm Drw_{14}}{\bar w}$$where the minus sign is used ...
Your calculations are the following. Assuming non-overlapping generations, the number of ancestors you have in the last $t$ generation is given by:$$\sum_{i=1}^t 2^t$$This sounds correct. But there are some very strong assumptions:Generations are non-overlapping. A more realistic model would need to consider $t$ as a continuous variable a give a ...
First of all, here is a program which simulates the evolution of the G-matrix over multiple generations, it's a few years old (they seem to have stopped developing it) and I've only played with it briefly. This could solve how to model the evolution of the G-matrix.Fisher's fundamental theorem is a great place to start off with the theory of this:The ...
First, Allee effects (also positive density dependence) can be modelled in several different ways, and the equation you give is one example. The terms weak and strong Allee effects are in my experience used in a couple of different ways. Most often, strong density dependence is used to denote Allee effects where the per capita population growth rate can ...
Here is a tree based on mitochondrial DNA variations in human populations.van Oven M, Kayser M. Hum Mutat. 2009 Feb;30(2):E386-94.Updated comprehensive phylogenetic tree of global human mitochondrial DNA variation.Looking at genetic distance between populations via mitochondria DNA, all nonafricans are descended from a founder in one mtDNA group (L3). ...
You need to add Bell curves to your simulation. The most important curve to simulate is the nutritional quality of the prey though there are plenty more thing to curve like speed and virility for prey and predators both. Nature uses lots of Bell curves so they must be good for something, such as softening the harsh effects of pure exponential growth. I ...
As you have phrased it, the question can be understood in two ways.The population size at t+1 is 350, after births, deaths and migration have taken place.The population size is 350 after deaths, but you also have to take births and migration into account to calculate population size at t+1For alternative 1 the growth rate is:$pgr = \frac{(230+21)-(...
Classification of equilibrium points is done on the basis of the eigenvalues.If the two eigenvalues have no real parts, it is a hyperbolic fixed point and represents undamped oscillation.If both have a negative real part, it is a stable fixed point. If any of the eigenvalues has an imaginary part then it represents damped oscillations (in that case the ...
I guess you meant the population size stability.It is considered that the biosystems will increase their capacity of adaptation when evolving in very fluctuating environments. I believe the population stability is embedded in the adaptability of individuals.There is a measurement about it, evolvability, when the environment changes, the faster the ... |
According to Kennedy's
Robust op-amp realization of Chua's circuit(1992), the differential equations satisfied by several physical quantities in Chua's circuit are
$$\begin{aligned} C_{1} \frac{d v_{C_{1}}}{d t} &=G\left(v_{C_{2}}-v_{C_{1}}\right)-g\left(v_{C_{1}}\right) \\ C_{2} \frac{d v_{C_{2}}}{d t} &=G\left(v_{C_{1}}-v_{C_{2}}\right)+i_{L} \\ L \frac{d i_{L}}{d t} &=-v_{C_{2}} \end{aligned}$$
Where $G=1/R$.
The circuit diagram is given at the bottom, where the quantity represented by each letter is indicated clearly.
When Chua invented his circuit, he deliberately designed it in such a way that the three equilibrium points were unstable. But I don't know how he did this, although I can analyze the stability using eigenvalues.
For small $x$, $f(x)=-mx$ for some positive $m$. So the equations above are linear for samll $x$. In order to analyze the stability of the zero solution $v_{C_1}=v_{C_2}=i_L=0$, I write $\mathbf x=[v_{C_1},v_{C_2},i_L]^T$ and $$ M=\begin{bmatrix}-a & b & 0\\ c & -c & d \\ 0 & -f & 0 \end{bmatrix} $$ where all letters are positive constants, and $a=1/RC_1,b=1/RC_1+m/C_1$ and so on. So the differential equations above can be written as $\dot{\mathbf x}=M\mathbf x$.
The characteristics equation $\det(M-\lambda I)=0$ is $\lambda^3+(a+c)\lambda^2+(ac-bc+fd)\lambda+fda=0$.
If the trivial solution $\mathbf x=0$ is a stable solution, then
according to a theorem(see here), the real parts of all eigenvalues of $M$ have to be negative. Thus it is not hard to deduce that the coefficients of the characteristic equation must be all positive. Therefore $ac-bc+fd>0$. Substituting the expressions for $a,b,\ldots,f$ in terms of $R,C_1,...$ and rearranging gives $m<\frac{RC_1}{L}$. Questions: So does it mean that when $m\geq\frac{RC_1}{L}$ the zero solution is always unstable? Does this result help us to choose the values of parameters? Intuitively, for the trivial solution to be unstable, we want the Chua's diode to develop power at a faster rate than the resistor consuming power. If we ignore the capacitors for a moment, then this means that $m<1/R$. However, the $R$ in $m\geq\frac{RC_1}{L}$ is on the numerator rather than the denominator. why are the two inequalities ($m\geq\frac{RC_1}{L}$ and $m<1/R$) so different?this seems quite counter-intuitive. Also, intuitively, for the trivial solution to be unstable, we want the Chua's diode to develop power at a faster rate than the resistor consuming power. If we ignore the capacitors for a moment, then this means that $m<1/R$. However, the $R$ in $m\geq\frac{RC_1}{L}$ is on the numerator rather than the denominator. why are the two inequalities ($m\geq\frac{RC_1}{L}$ and $m<1/R$) so different?this seems quite counter-intuitive. Also, can I improve my results? This is just a sufficient condition for the equilibrium to be unstable. Can I somehow make it a necessary condition as well? Or are there any better methods I can use instead of evaluating eigenvalues?
EDIT: It appears that the condition $m\geq\frac{RC_1}{L}$ should be also sufficient if all eigenvalues are real. |
LaTeX supports many worldwide languages by means of some special packages. In this article is explained how to import and use those packages to create documents in
Italian.
Contents
Italian language has some accentuated words. For this reason the preamble of your document must be modified accordingly to support these characters and some other features.
documentclass{article} \usepackage[utf8]{inputenc} \usepackage[italian]{babel} \usepackage[T1]{fontenc} \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Questo è un breve riassunto dei contenuti del documento scritto in italiano. \end{abstract} \section{Sezione introduttiva} Questa è la prima sezione, possiamo aggiungere alcuni elementi aggiuntivi e tutto digitato correttamente. Inoltre, se una parola è troppo lunga e deve essere troncato babel cercherà per troncare correttamente a seconda della lingua. \section{Teoremi Sezione} Questa sezione è quello di vedere cosa succede con i comandi testo definendo \[ \lim x = \sin{\theta} + \max \{3.52, 4.22\} \] \end{document}
There are two packages in this document related to the encoding and the special characters. These packages will be explained in the next sections.
If your are looking for instructions on how to use more than one language in a sinlge document, for instance English and Italian, see the International language support article.
Modern computer systems allow you to input letters of national alphabets directly from the keyboard. In order to handle a variety of input encodings used for different groups of languages and/or on different computer platforms LaTeX employs the
inputenc package to set up input encoding. In this case the package properly displays characters in the Italian alphabet. To use this package add the next line to the preamble of your document: \usepackage[utf8]{inputenc} The recommended input encoding is utf-8. You can use other encodings depending on your operating system.
To proper LaTeX document generation you must also choose a font encoding which has to support specific characters for Italian language, this is accomplished by the
package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in Italian, using this specific encoding will avoid glitches with some specific characters. The default LaTeX encoding is
OT1.
To extended the default LaTeX capabilities, for proper hyphenation and translating the names of the document elements, import the
babel package for the Italian language. \usepackage[italian]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the Italian words "Sommario" and "Indice" are used.
Sometimes for formatting reasons some words have to be broken up in syllables separated by a
- (
hyphen) to continue the word in a new line. For example, matematica could become mate-matica. The package babel, whose usage was described in the previous section, usually does a good job breaking up the words correctly, but if this is not the case you can use a couple of commands in your preamble.
\usepackage{hyphenat} \hyphenation{mate-mati-ca recu-perare}
The first command will import the package
hyphenat and the second line is a list of space-separated words with defined hyphenation rules. On the other side, if you want a word not to be broken automatically, use the
{\nobreak word} command within your document.
For more information see |
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