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Rapid Lyapunov control of finite-dimensional quantum systems

# Rapid Lyapunov control of finite-dimensional quantum systems

## Abstract

Rapid state control of quantum systems is significant in reducing the influence of relaxation or decoherence caused by the environment and enhancing the capability in dealing with uncertainties in the model and control process. Bang-bang Lyapunov control can speed up the control process, but cannot guarantee convergence to a target state. This paper proposes two classes of new Lyapunov control methods that can achieve rapidly convergent control for quantum states. One class is switching Lyapunov control where the control law is designed by switching between bang-bang Lyapunov control and standard Lyapunov control. The other class is approximate bang-bang Lyapunov control where we propose two special control functions which are continuously differentiable and yet have a bang-bang type property. Related stability results are given and a construction method for the degrees of freedom in the Lyapunov function is presented to guarantee rapid convergence to a target eigenstate being isolated in the invariant set. Several numerical examples demonstrate that the proposed methods can achieve improved performance for rapid state control of quantum systems.

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A1,A2]Sen Kuang, A2]Daoyi Dong, A2]Ian R. Petersen

uantum systems; switching control; approximate bang-bang control; rapid Lyapunov control

## 1 Introduction

Quantum control has the potential to play important roles in the development of quantum information technology and quantum chemistry, and has received wide attention from different fields such as quantum information, chemical physics and quantum optics [1, 2, 3, 4, 5, 6]. Transfer control between quantum states is one of the basic tasks in quantum control. Different control strategies such as optimal control [7, 8, 9, 10], adiabatic control [11, 12], Lyapunov control methods [13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23], and LQG control [24, 25, 26], and sliding mode control [27, 28] have been presented for controller design in quantum systems. Among these control strategies, Lyapunov control methods have been extensively studied for quantum systems due to their simplicity and intuitive nature in the design of control fields [29, 30, 31, 32, 33]. In Lyapunov control, a Lyapunov function is constructed using information on states or operators related to the quantum system [13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 29, 30, 31, 32, 33] and the associated control law is designed based on the Lyapunov function (feedback design). Then the control law can be implemented in an open-loop way. From the viewpoint of control theory, one hopes that any system trajectory converges to a desired target state. Unfortunately, the LaSalle invariance principle used in Lyapunov control methods cannot guarantee convergence of any system trajectory to a target state. Some methods such as using implicit Lyapunov functions or switching control methods have been developed to achieve approximate or asymptotic convergence for some specific quantum control tasks (see, e.g., [14, 16, 19, 21]).

For quantum systems, rapid state control is of importance because a realistic quantum system cannot be perfectly separated from its environment, which will cause a relaxation or decoherence effect. In the context of quantum information processing, rapid control is a basic requirement for performance improvement in quantum computing. In practical applications, robustness has been recognized as a key requirement for the development of quantum technology [34, 35, 36, 37, 38, 39, 40, 41]. Rapid control may make the control law more robust to uncertainties in the model or in the control process. Time optimal control methods have been proposed to achieve rapid control for quantum systems [42, 43, 44]. However, it is very difficult to obtain the optimal control law for general quantum systems [42, 43, 44]. In [22], an optimal Lyapunov design method has been proposed to design a control law for rapid state transfer in quantum systems. Under power-type and strength-type constraints on the control fields, two kinds of Lyapunov control laws were proposed. In particular, the strength-type constraint led to a bang-bang Lyapunov control. In [23], the convergence problem for bang-bang Lyapunov control law is further discussed for two-level quantum systems.

The bang-bang Lyapunov control method in [22, 23] can be used to achieve rapid state control for some practical quantum systems with a high level of fidelity. However, since the control function of bang-bang Lyapunov control is not continuously differentiable, the LaSalle invariance principle cannot be directly used to guarantee convergence. We show that a high-frequency oscillation with an infinitesimal period may occur in bang-bang Lyapunov control, which prevents effective transfer to the target state. Such control fields can also not be realized in the laboratory. In order to achieve rapidly convergent control in state transfer, we propose two classes of new Lyapunov control methods in this paper: switching Lyapunov control and approximate bang-bang Lyapunov control. We first derive a sufficient condition for a two-level system that shows a high-frequency oscillation with an infinitesimal period in bang-bang Lyapunov control. Then we design a switching strategy, i.e., switching between bang-bang Lyapunov control and standard Lyapunov control. For approximate bang-bang Lyapunov control, we design two special control functions that incorporate bang-bang and smoothness properties. These proposed Lyapunov design methods can achieve rapidly convergent control in quantum systems, which is demonstrated by several numerical examples involving a two-level system, a three-level system and a two-qubit superconducting system.

This paper is organized as follows. Section 2 presents the system model, and analyzes the robustness of open-loop quantum control. Section 3 discusses Lyapunov functions with various degrees of freedom, presents several stability results, and develops a construction method for designing the degrees of freedom in the Lyapunov function. A switching strategy between bang-bang and standard Lyapunov control schemes is proposed and the switching condition is investigated in Section 4. In Section 5, we propose two approximate bang-bang Lyapunov control methods. Three numerical examples are presented to demonstrate the performance of the proposed rapid Lyapunov control strategies in Section 6. Conclusions are presented in Section 7.

Notation

• : the imaginary unit, i.e., ;

• : the commutator of the matrices and , i.e., ;

• : the repeated commutator with depth , i.e., ;

• : the induced -norm of the matrix , or the -norm of the vector ;

• : the transpose of the matrix ;

• : the conjugate transpose of the matrix ;

• : the conjugate transpose of the state vector ;

• : the complex conjugate of the complex number ;

• : the modulus of the complex number ;

• : the sets of all real numbers;

• : the sets of all complex numbers;

• : the trace of the matrix ;

• : the spectrum of the matrix , i.e., the set of all eigenvalues of ;

• : the real part of the complex number ;

• : the imaginary part of the complex number .

## 2 System model and robustness of control

### 2.1 Models of finite-dimensional quantum systems

Assume that the quantum system under consideration is an -dimensional and controllable closed system [45], described by the following Liouville-von Neumann equation:

 ˙ρ(t)=−i[H0+m∑k=1Hkuk(t),ρ(t)], (1)

where is a density matrix describing the state of the system; is the internal Hamiltonian, and is the control Hamiltonian that describes the interaction between the external control fields and the system ( and are time-independent Hermitian matrices); and are external real-valued control fields. In the energy representation, has a diagonal form, i.e., . We call the transition frequency between the energy levels and . Denote as the eigenvector of corresponding to the eigenvalue , i.e., where the -th element is 1 and other elements are 0. All form an orthogonal basis of the dimensional complex Hilbert space .

Since the Hamiltonian is a Hermitian matrix, the evolution of the system (1) is unitary. For a given initial state , the quantum state at time , , can always be written as

 ρ(t)=U(t)ρ0U†(t),t≥0 (2)

where is a unitary matrix with . Equation (2) indicates that the system state at any time always has the same spectrum with the initial state .

Using (1) and the property of the unitary operator , we can undertake a differentiation calculation on both sides of (2) to obtain the Schrödinger equation for :

 ˙U(t)=−i(H0+m∑k=1Hkuk(t))U(t)=−iH(t)U(t) (3)

with .

We assume that the control objective is to steer the system to an eigenstate of , . Due to the isospectral evolution property of closed quantum systems and the fact that pure states () and mixed states () have different spectra, we assume that the initial state is a pure state since the target eigenstate is a pure state. Moreover, quantum pure states have wide applications in quantum information processing.

Also, the following conditions are assumed on the system:

 ωaf≠ωbf (a≠b;a,b≠f); (4)
 (Hk′)jf≠0 (j≠f;∃k′∈{1,2,⋯,m}). (5)

Condition (4) means that the transition frequencies between the target eigenstate and other eigenstates are distinguishable, and that is non-degenerate, i.e., its all eigenvalues are mutually different. Condition (5) implies that there exists a direct coupling between the target eigenstate and any other eigenstate.

### 2.2 Robustness of open-loop quantum control

For a closed quantum system, we may use the model (1) to design an open-loop control law and then apply the control law to the practical system. Since any physical system is unavoidably affected by some uncertainties, the robustness of the control law should be considered. Possible perturbations to the quantum system (1) include perturbations of the internal Hamiltonian , and perturbations in the control Hamiltonian , inaccuracy in the control law, and inaccuracy in the initial states. Here, we mainly discuss perturbations in the Hamiltonian. The aim is to show that rapid control can make the control law more robust to uncertainties in the model or in the control process. This is another motivation (besides reducing the relaxation and decoherence effect) to develop rapid Lyapunov control in the following.

We denote the perturbations in the internal and control Hamiltonians as and , respectively, where is a real diagonal matrix and are Hermitian matrices. Thus, the internal and control Hamiltonians with perturbations can be written as and , respectively. We call the model (1) the nominal system and the system with and the perturbed system. Define . Then, the dynamics of the perturbed system can be described as

 ˙~ρ(t)=−i[(H0+m∑k=1Hkuk(t))+ΔH,~ρ(t)], (6)

where the Hermitian matrix is the uncertainty in the Hamiltonian .

We assume that the uncertainty satisfies and that the introduction of does not break Conditions (4) and (5), i.e., , where and is the diagonal elements of ; and . Now, we examine the effect of the uncertainty on the quantum system. {thm} We assume . For any initial pure state , the states of the nominal system (1) and the perturbed system (6) satisfy . If at a finite time , then for an arbitrarily given , when , the distance between and the target state satisfies . {pf} Similar to (3), we write the perturbed system (6) in terms of its time evolution operators as follows:

 ˙˜U(t)=−i(H(t)+ΔH)˜U(t),˜U(0)=I. (7)

Since both and are unitary matrices, we let

 ˜U(t)=U(t)Q(t). (8)

Differentiating both sides of (8) with respect to and considering (3) and (7), we have

 i˙Q(t)=(U†(t)ΔHU(t))Q(t),Q(0)=I. (9)

Define . We have . The Dyson series solution of (9) is the following time-ordered integral

 Q(t) =I+∞∑n=1(−i)n∫t0dt1∫t10dt2⋯ ∫tn−10dtnΓ(t1)Γ(t2)⋯Γ(tn)≜I+W(t), (10)

where . Considering , we have

 ∥W(t)∥=∥W†(t)∥≤∞∑n=1εn∫t0dt1∫t10dt2⋯ ∫tn−10dtn=∞∑n=1εntnn!=etε−1. (11)

For any initial state , we have

 ∥Δρ(t)∥ ≜ ∥~ρ(t)−ρ(t)∥ (12) = ∥U(t)Q(t)ρ0Q†(t)U†(t)−U(t)ρ0U†(t)∥ = ∥Q(t)ρ0Q†(t)−ρ0∥ = ∥ρ0W†(t)+W(t)ρ0+W(t)ρ0W†(t)∥ ≤ ∥ρ0W†(t)∥+∥W(t)ρ0∥+∥W(t)ρ0W†(t)∥ ≤ ∥W†(t)∥+∥W(t)∥+∥W(t)∥2 ≤ etε−1+etε−1+(etε−1)2 = e2tε−1.

Considering , we have

 ∥Δρ(t)∥≤min{e2tε−1,2}. (13)

For the perturbed system (6), when ,

 ∥~ρ(T)−ρf∥ ≤∥Δρ(T)∥+∥ρ(T)−ρf∥ ≤e2Tε−1+ξ1. (14)

Hence, when , we have .  Theorem 2.2 can be regarded as a generalization to quantum systems of the continuous dependence on parameters of solutions to differential equations. Theorem 2.2 shows that, for given and , if the nominal system (1) can approach the target state within a shorter time period, the perturbed system (6) can tolerate larger perturbations when guaranteeing given performance. That is to say, a rapidly convergent control for the nominal system (1) may lead to improved robustness. This paper develops rapidly convergent Lyapunov control methods for the system (1).

## 3 Lyapunov quantum control and stability

### 3.1 Lyapunov control design

Consider the following Lyapunov function:

 V=tr(Pρ(t)), (15)

where is a positive semi-definite Hermitian operator that needs to be constructed for completing a given control task.

The time derivative of the Lyapunov function (15) is calculated as

 ˙V = tr(P˙ρ) (16) = tr(−iρ[P,H0])+m∑k=1uk(t)tr(−iρ[P,Hk]).

We design the control laws by guaranteeing in (16). Considering the fact that in (16) is independent of the control field while is an unknown Hermitian matrix to be constructed, we let

 [P,H0]=0. (17)

Since the diagonal matrix is non-degenerate, (17) implies that is also a diagonal matrix. We denote . Using (17), (16) can be written as

 ˙V=m∑k=1uk(t)Tk(t), (18)

where . For notational simplicity, we also denote as hereinafter.

Thus, by guaranteeing in (18), we design a control law with the following general form:

 uk(t)=fk(Tk),(k=1,2,⋯,m) (19)

where the control function satisfies: 1) is continuously differentiable with respect to ; 2) ; and 3) . In particular, we call the following control law the standard Lyapunov control in this paper:

 uk(t)=−KkTk(t),(k=1,2,⋯,m) (20)

where the control gain is used to adjust the amplitude of the control field .

### 3.2 General stability results

The control law (19) means that the whole “closed-loop” system is a nonlinear autonomous system. We use the LaSalle invariance principle to analyze the stability of the system. The LaSalle principle ensures that the system (1) with the control fields (19) necessarily converges to the largest invariant set contained in .

Assume and let . The invariance property guarantees that , which holds when , i.e.

 Tk=tr(−iρ(t)[P,Hk])=0, (k=1,⋯,m). (21)

Substituting the solution of into (21) and using (17), one has

 tr(e−iH0t¯ρeiH0t[P,Hk])=tr(eiH0tHke−iH0t[¯ρ,P]) =tr(∞∑n=0[(iH0t)(n),Hk]n![¯ρ,P]) =∞∑n=0−intnn!tr([H(n)0,Hk][P,¯ρ])=0. (22)

Since the time function sequence is linearly independent, and and are diagonal, we have

 tr([H(n)0,Hk][P,¯ρ])=tr((ωnjl(Hk)jl)((pj−pl)¯ρjl)) =0, (k=1,⋯,m;n=0,1,2,⋯). (23)

Since and are Hermitian matrices, (3.2) reduces to

 Missing or unrecognized delimiter for \Big =0,(k=1,⋯,m;n=0,1,2,⋯). (24)

For even and odd , (3.2) has the following forms, respectively:

 ∑j

We denote , and define

 ξk≜[(Hk)12¯ρ21,⋯,(Hk)1N¯ρN1,(Hk)23¯ρ32,⋯, (Hk)2N¯ρN2,⋯,(Hk)N−1,N¯ρN,N−1]T, (27)
 M ≜ ⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣1⋯11⋯1⋯1ω212⋯ω21Nω223⋯ω22N⋯ω2N−1,Nω412⋯ω41Nω423⋯ω42N⋯ω4N−1,N⋮⋯⋮⋮⋯⋮⋯⋮ωFN12⋯ωFN1NωFN23⋯ωFN2N⋯ωFNN−1,N⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦, (28)
 P≜diag[p2−p1,⋯,pN−p1,p3−p2,⋯, pN−p2,⋯,pN−pN−1], (29)
 Ω≜diag[ω12,⋯,ω1N,ω23,⋯,ω2N,⋯,ωN−1,N]. (30)

 MPI(ξk)=0, (k=1,⋯,m); (31) MΩPR(ξk)=0, (k=1,⋯,m). (32)

Since the system (1) evolves unitarily, the positive limit set of any evolution trajectory has the same spectrum as its initial state. Thus, the invariant set that the system with the control law (19) will converge to can be characterized in the following theorem. {thm} Given an arbitrary initial pure or mixed state , and under the action of the control fields (19), the system (1) converges to the invariant set , where represents the spectrum of “”, and and are defined by (3.2)-(30). From Theorem 3.2, the invariant set that the system converges to is dependent on the transition frequencies of the system, the diagonal values of , the initial state and the connectivity of . When these factors satisfy some particular conditions, it is possible to obtain a simpler form for the invariant set (see [31]).

### 3.3 Construction of Hermitian operator P

In this subsection, we study the construction method of to achieve convergence to the target eigenstate . Thus, we only consider the case of initial pure states. For the system (1), all possible initial pure states can be divided into two classes: initial states satisfying either or . In this subsection, we give a method for constructing to achieve convergence to the target state for these two classes of initial states, respectively.

When the initial state satisfies , we have the following result. {thm} Consider the system (1) satisfying Conditions (4), (5) and with the control fields (19). Assume that the target eigenstate and the initial pure state satisfy . If the diagonal elements of satisfy , then is isolated in the invariant set and the system state starting from necessarily converges to . {pf} Using Conditions (4) and (5), we can simplify the invariant set in Theorem 3.2. For convenience of expression, we assume that the target eigenstate is the -th eigenstate of , i.e., . If , we have and . Thus, (31) and (32) are equivalent to
and , respectively. Using Condition (4), one can obtain . Using Condition (5), we can obtain the relationship . Hence, all states in the invariant set are of the form , where “” represents an arbitrary eigenvalue of the initial state .

Since is a pure state, has one eigenvalue 1 and eigenvalues 0. Hence, the states in the invariant set have the form of or , where and are Hermitian matrices. For , all eigenvalues of are 0, which leads to , i.e., . For , has one eigenvalue 1 with multiplicity 1 and one eigenvalue 0 with multiplicity . In other words, has one eigenvalue 1 and eigenvalues 0. It is clear that the target eigenstate is isolated in .

For any initial pure state which satisfies , one has or . When , the conclusion naturally holds. When , we have . Hence, when and , the system (1) necessarily converges to the target state .  When the initial state satisfies , we have . That is to say, under the construction relation of in Theorem 3.3, the control law (19) cannot enable any state transfer. In this case, there exists a such that . Thus, we may use the following switching control to achieve convergence to the target state:

 uk(t)={fk(sin(ωjft)),t∈[0,t0]fk(Tk),t>t0 (k=1,⋯,m), (33)

where , and is a small time duration.

When is small, the state is not in the invariant set . If we take as a new initial state, then Theorem 3.3 guarantees that the control law (19) can achieve convergence to the target state. Thus, we have the following conclusion. {thm} Consider the system (1) satisfying Conditions (4), (5) and with the switching control (33). Assume that the target eigenstate and the initial pure state satisfy . If the switching time satisfies and the diagonal elements of satisfy , then the system state starting from necessarily converges to . For general continuously differentiable control function (19), the construction relation in Theorems 3.3 and 3.3 ensures convergence to the target eigenstate. Based on the construction relation of , we propose two new methods including switching Lyapunov control and approximate bang-bang Lyapunov control to achieve rapidly convergent Lyapunov control.

## 4 Switching between Lyapunov control schemes

To speed up the control process, Ref. [22] proposed two design methods for quantum systems with power-type constraints and strength-type constraints such that in (18) takes the minimum value at each moment. For the case with strength-type constraints, the “optimal” control law is the following bang-bang Lyapunov control:

 uk(t)=⎧⎪⎨⎪⎩−S, (Tk>0)S, (Tk<0)0, (Tk=0)(k=1,⋯,m), (34)

where is the maximum admissible strength of each control field, i.e., .

The bang-bang Lyapunov control in (34) makes in (18) satisfy , and can speed up completing some quantum control tasks. Especially, the state may move rapidly towards the target state at the early stages of the control [22]. However, convergence cannot be guaranteed since the control function is not continuously differentiable. Here, we first show that the bang-bang Lyapunov control may lead to a high-frequency oscillation phenomena [22, 23], which prevents effective state transfer towards the target state. Then, we propose two classes of switching Lyapunov control strategies to achieve rapidly convergent control, i.e., switching between the bang-bang Lyapunov control and the standard Lyapunov control, and switching between bang-bang Lyapunov control schemes with variable control strengths.

### 4.1 High-frequency oscillation in bang-bang Lyapunov control

In this subsection, we present a sufficient condition for two-level quantum systems that high-frequency oscillation phenomena occur in the bang-bang Lyapunov control [22], which can be used to determine switching conditions for the design of switching Lyapunov control. We first give the following definition.

{defn}

The control law (34) is said to have a high-frequency oscillation with an infinitesimal period at time if ,

 inf{τ>0:uk(t+τ)≠uk(t)}=0

for all . Since the control in (34) can take on only one of three constant values (0 and ) at any time, the high-frequency oscillation in Definition 4.1 means that the control always jumps between these values after an arbitrarily small time duration in the interval . Such a control field cannot be realized in practice.

Now consider the system model (1) in the case of two energy levels, and denote its internal Hamiltonian as and its control Hamiltonian as :

 H0=[λ100λ2], H1=[0rr∗0], (35)

where .

We define the first eigenstate as the excited state, as the ground state, and

 ω12≜λ1−λ2>0. (36)

Let the excited state be the target state. According to Theorem 3.3 or Theorem 3.3, can be chosen as . Thus, . From (34), holds at any zero point of the bang-bang Lyapunov control. For convenience of analysis, in this paper we denote such moments as to differentiate them from the initial moment 0. Thus, we have

 T1(~0) =−itr([P,H1]ρ(~0)) =2(p−p1)⋅I(r∗ρ12(~0)) =0. (37)

Equation (4.1) equals that . For the two-level system, we have the following result. {thm} Consider the two-level system

 ˙ρ(t)=−i[H0+H1u1(t),ρ(t)], (38)

with the Hamiltonians (35) and the bang-bang Lyapunov control (34) (where ). Assume that the initial state of the system is an arbitrary pure state. We denote the state at any zero point of the control field (i.e., ) as . Then, a sufficient condition for the bang-bang Lyapunov control (34) to have a high-frequency oscillation with an infinitesimal period is

 |r|(ρ11(~0)−ρ22(~0))|ρ12(~0)|≥ω12S,(ρ12(~0)≠0). (39)
{pf}

Assume that from the state , a constant control acts on the system and lasts to time . Write the state at time as . We have

 T1(t) = −itr([P,H1]ρ(t)) (40) = −itr(ei(H0+H1u)t[P,H1]e−i(H0+H1u)tρ(~0)).

Denote , , and , then in (40) can be calculated as

 ei(H0+H1u)t[P,H1]e−i(H0+H1u)t (41) = [P,H1]+it[H0+H1u,[P,H1]] +(it)22![H0+H1u,[H0+H1u,[P,H1]]] +(it)33![H0+H1u,[H0+H1u,[


text:
# Mr. Johnson gave the sales clerk $60 for 3 sweaters that he purchased. The clerk handed him$7.56 in

Mr. Johnson gave the sales clerk $60 for 3 sweaters that he purchased. The clerk handed him$7.56 in change. Each sweater cost the same amount. What was the cost for each sweater? Record your answer and fill in the bubbles on your answer document. Be sure to use the correct place value. . 00 Olo 9 9 O $17.48 O$17.53 O $22.52 O$26.22

## This Post Has 3 Comments

1. Expert says:

answer:

write 548 as 500 + 40 + 8. multiply each addend by 5. then add the three products to get 2,740.

2. Expert says:

the answer is 0.000652422

but if you ment to ask what is 6131 divided by 4, it is 1532.75.

step-by-step explanation:

3. zarbyp30 says:

$17.48 Work:$60-$7.56=$52.44
$52.44%3=$17.48

format: plain


text:
# Synopsis: Time-Dependent Fields Improve Superlens Resolution

A thin metal slab known as a superlens, which focuses light more sharply than its wavelength, could have an even higher resolution imaging rapidly varying electromagnetic fields.

In principle, a sheet of metal can be a perfect lens, reproducing an image with arbitrarily small features, because the metal surface supports waves with wavelengths much shorter than that of light. But although experiments have confirmed improved resolution, the metal lens is ultimately limited by losses. Now in Physical Review Letters, Alexandre Archambault and co-workers at the Institute of Optics in Palaiseau, France, argue that light whose amplitude varies rapidly in time could overcome this limitation.

Near the frequency of a metal’s internal oscillation, like a plasmon or phonon, the real part of the dielectric constant can pass through the value $-1$. Under these conditions, an object on one side of a metal sheet can create an image on the other side with details finer than the corresponding wavelength of light. But real surface plasmons or phonons decay in picoseconds or less, causing a phase shift that prevents perfect cancellation. Archambault and colleagues suggest that an electromagnetic field with a complex (real plus imaginary) frequency, which decreases rapidly in amplitude as it oscillates, can restore the cancellation.

The researchers tested their argument in numerical simulations in which they abruptly terminated the oscillating field from a point source. As they expected, the blurry image of the source got progressively sharper in the following picoseconds. At the same time, however, the sharpening image became rapidly fainter, which will make this technique hard to exploit. Still, the results show a new way to evade apparent limits on optical manipulation using super lenses. – Don Monroe

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text:
Linear equation in two variables:

When a polynomial equation is expressed in terms of ax + by + c = 0, where a, b, c are real numbers and x, y are variables also, a & b both are nonzero is known to be linear equation in two variables.

Solution of Pair of linear equations:

Each solution (x, y) of a linear equation in two variables, i.e. ax + by + c = 0,      it always corresponds to a sharp point on the line representing the equation, and vice versa.

Pair of Linear Equation in Two Variables

When two linear equations having same variables in both the equation is said to be pair of linear equations in two variables.

The most general form for representation of a pair of linear equations is
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
where a1, a2, b1, b2, c1, c2 are real numbers,
such that a12 + b12 ≠ 0, and a22 + b22 ≠ 0

We can represent and solve pair of linear equation in two variables by these two way:

• Graphical method
• Algebraic method

Graphical Method of solution of linear equation in two variables:

Two lines are used to represent pair linear equation in two variable on the graph:

Type of equation

Intersecting lines Unique Solution Consistent
Coincide Infinite Solution Consistent (dependent)
Parallel No Solution Inconsistent

Algebraic Methods of solution of linear equation in two variables:

We have the following methods for finding the solution(s) of a pair of linear equations algebraically :

• Substitution Method
• Elimination Method
• Cross-multiplication Method

Substitution method: Easy way to solve

Equation:
2x – y = 1
4x – 3y = 2

Step 1. From the two variable in the equation find the value of one variable “y” in terms of the other variable “x” from any one of the two equation whichever is easy to find.
as from 1st  y = 2x – 1

Step 2. Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved easily.
In some cases after substitution you will find that no variable is left if it satisfies the equality it will have infinite solution and if not then it will have no solution and thus inconsistent.
putting the value of y in 2nd equation
4x – 3(2x – 1)= 2
-2x= -1
x = ½

Step 3. Now we will substitute the value of x obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable y accordingly.

If a pair of linear equation in two variables is expressed as

a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
The following conditions obtained:

 Comparision of ratios Graphical Representation Algebraic interpretation $$\frac{{a}_{1}}{{a}_{2}}$$ ≠ $$\frac{{b}_{1}}{{b}_{2}}$$ Intersecting Lines Unique solution (Consistent) $$\frac{{a}_{1}}{{a}_{2}}$$ = $$\frac{{b}_{1}}{{b}_{2}}$$ = $$\frac{{c}_{1}}{{c}_{2}}$$ Coincidence Lines Infinite many solutions (Consistent) $$\frac{{a}_{1}}{{a}_{2}}$$ = $$\frac{{b}_{1}}{{b}_{2}}$$ ≠ $$\frac{{c}_{1}}{{c}_{2}}$$ Parallel Lines No Solution (Inconsistent)

Pair of Linear Equation Related Examples:

1. Two lines are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.

Solution:

First let’s take two solutions of each of the equations
x + 2y – 4 = 0
2x + 4y – 12 = 0

 x 0 4 y = $$\frac{4-x}{2}$$ 2 0

 x 0 6 y = $$\frac{12 – 2x}{4}$$ 3 0

Thus we plot the graph accordingly taking the coordinates.
We observe that the lines are parallel i.e., they do not intersect anywhere.

1. Solve the following pair of equations by substitution method:
7x – 15y = 2
x + 2y = 3

Solution:

We take any of equations
x + 2y =3
x = 3 – 2y
Substitute the value of x in 1st Equation
7(3 – 2y) – 15y =2
– 29y = –19
y = $$\frac{19}{29}$$
Substituting this value of y in Equation 2nd equation
x = 3 – 2$$\frac{19}{29}$$ = $$\frac{49}{29}$$

Thus, x = $$\frac{49}{29}$$ and y= $$\frac{19}{29}$$

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text:
In: Physics

# To calculate Energy Radiated by body

A hot black body emits the energy at the rate of 16 J m-2 s-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated

## Solutions

##### Expert Solution

Stefan Boltzmann Law relates the temperature of the blackbody to the amount of the power it emits per unit area. The law states that;

“The total energy emitted/radiated per unit surface area of a blackbody across all wavelengths per unit time is directly proportional to the fourth power of the black body’s thermodynamic temperature. ”

Here, lm becomes half, the Temperature doubles.

T1/T2=1/2 put in eqn

⇒ ε = (σT)^4

e1/e2 = (T1/T2)^4 resiprocal values to get e2

⇒ e2 = (T2/T1)^4 . e1 = (2)^4 . 16

= 16.16 = 256 J m-2 s-1

Wein’s displacement law is, lm.T = b.

T- temperature

i.e. Tμ [1/lm. )

Now from Stefan Boltzmann Law, e = (sT)^4


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## Linear constraints for the frame order models

Linear constraints are implemented for the frame order models using the log-barrier constraint algorithm in minfx, as this does not require the derivation of gradients.

The pivot point and average domain position parameter constraints in Ångstrom are:

These translation parameter restrictions are simply to stop the optimisation in the case of model failures. Converting these to the Ax b matrix notation required for the optimisation constraint algorithm, the constraints become

 ⋅ (12.80)

For the order or motional amplitude parameters of the set , the constraints used are

These reflect the range of validity of these parameters. Converting to the Ax b notation, the constraints are

 ⋅ (12.82)

The pseudo-elliptic cone model constraint θx θy is used to simplify the optimisation space by eliminating symmetry.

The relax user manual (PDF), created 2016-10-28.


text:
# Question

Saylor Enterprises had a capital balance of \$168,000 at the beginning of the period. At the end of the accounting period, the capital balance was \$198,000.
(a) Assuming no additional investment or withdraw als during the period, what is the net income for the period?
(b) Assuming an additional investment of \$13,000 but no withdrawals during the period, what is the net income for the period?

Sales1
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• CreatedJanuary 30, 2014
• Files Included
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text:
Top

Top
 Fractions that have a whole number and a fraction are known as mixed fractions. Any improper fraction can be expressed as mixed numbers and mixed numbers can be expressed as improper fraction.

## Steps for Adding Mixed Fractions

Step 1 :

Change the mixed fractions into improper fractions

Step 2 :

Check the denominators of each fractions.

a) if the denominators are same, Just add the numerators and keep the denominator as same.

b) If the denominators are different, take LCM and make the denominators equal and add the numerator.

Step 3 :

Simplify the result and express the answer as a mixed number.

## Examples for Adding Mixed Fractions Calculator

1. ### Add the fraction as 2 $\frac{3}{4}$ and 5 $\frac{3}{4}$.

Step 1 :

Convert mixed fractions into improper fractions:-

2 $\frac{3}{4}$ = $\frac{8 + 3}{4}$ = $\frac{11}{4}$

and  5 $\frac{3}{4}$ = $\frac{20 + 3}{4}$ = $\frac{23}{4}$

Step 2 :

Denominator are the same, so adding the numerator = 11 + 23 = 34

Step 3 :

Fraction = $\frac{34}{4}$ = $\frac{17}{2}$ .

Mixed fraction = 8 $\frac{1}{2}$. (2 × 8 = 16, 16 + 1 = 17)

8 $\frac{1}{2}$

2. ### Add the fraction as 2 $\frac{3}{4}$ and 5 $\frac{7}{8}$.

Step 1 :

Convert mixed fractions into improper fractions:-

2 $\frac{3}{4}$ = $\frac{8 + 3}{4}$ = $\frac{11}{4}$

and 5 $\frac{7}{8}$ = $\frac{40 + 7}{8}$ = $\frac{47}{8}$

Step 2 :

Denominator are not same,

So, LCM of 8 and 4

8 × 1 = 8.

4 × 2 = 8.

So, LCM of 4 and 8 is 8.

$\frac{11}{4}$  × $\frac{2}{2}$ = $\frac{22}{8}$

$\frac{47}{8}$ * $\frac{1}{1}$ = $\frac{47}{8}$

Adding the numerator = 22 + 47 = 69

Step 3 :

Fraction = $\frac{69}{8}$.

Mixed fraction = 8 $\frac{5}{8}$ (8 × 8 = 64, 64+5 = 69)

8 $\frac{5}{8}$


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# How to find the X coordinate of a point using trigonometry?

1. Sep 20, 2013

### Miroslava89

How to determine the X coordinate of the red point if i know the Y coordinate and the angle between the adjacent side and the hypotenuse in the formed right triangle(see the image)? I don't know the length of the hypotenuse and the adjacent side of the triangle, i know only the angle between the adjacent side and the hypotenuse and the length of the opposite side(the Y coordinate of the red point). Let's say that Y is 40(i.e. the length of the opposite side) and the angle is 35 degrees. How to determine the X coordinate? Or in other words the length of the adjacent side in the triangle. I asked this on stackoverflow.com but the answer i got wasn't helpful at all. It's very simple question, my native language is not english but i think it's pretty clear what i'm asking.

http://img27.imageshack.us/img27/4099/ce4n.png [Broken]

Last edited by a moderator: May 6, 2017
2. Sep 20, 2013

### Office_Shredder

Staff Emeritus
The main trigonometric functions are sin(x), cos(x) and the one you are going to want to use, tan(x). sin(x) is literally the ratio between the opposite side and hypotenuse of a right triangle with an angle of x in it, cos(x) is the ratio between the adjacent side and hypotenuse of a right triangle with an angle of x in it, and tan(x) is the ratio between the opposite and adjacent sides of a right triangle with an angle of x in it.

As a side point, it should be clear that tan(x) = sin(x)/cos(x).

Your example right triangle has an adjacent leg of length X to be determined, the opposite leg has a length of 40, and your angle is 35 degrees. So we have

tan(35) = 40/X
X = 40/tan(35).

You can calculate tan(35) on your calculator (possibly needing the sin(x)/cos(x) thing depending on the type) or by using any of a number of websites. You do have to be careful because there are two ways of measuring angles - radians and degrees - and a lot of places might assume you are inputting your angle in radians. You should try getting tan(35) from a computing source yourself, and the number you should get is very close to .7

As an exercise to practice, if that point has an X coordinate of 12 and an angle of 50 degrees, what is the length of the hypotenuse?

3. Sep 20, 2013

### Miroslava89

So, if the adjacent side is 12 and if by "an angle of 50 degrees" you mean the angle formed by the hypotenuse and the adjacent side(like in my drawing), then cos(50) = 12 / hypotenuse => 0.642 = 12 / hypotenuse => 12 / 0.642 = 18.67

Thanks. Actually the answer on the website i mentioned was helpful but i didn't paid much attention to it initially. But your answer is more uuh...descriptive and generally better, so thank you again.

Last edited: Sep 20, 2013
4. Sep 20, 2013

### HallsofIvy

Staff Emeritus
You don't mention tangent which is what Office Shredder said was the function you need. tangent is "opposite side over near side" so, in this xy- coordinate system [itex]tan(\theta)= y/x[/tex]. Given that $$\theta= 35$$ degrees and y= 40, $$x= y/tan(\theta)= 40/tan(35)= 40/.7002$$.

5. Sep 21, 2013

### Office_Shredder

Staff Emeritus
Miro, that looks good to me.

Halls, he didn't use tan because he was answering the other practice question I gave him


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## arXiv Analytics

### arXiv:1708.03316 [math.QA]AbstractReferencesReviewsResources

#### Noncommutative Catalan numbers

Published 2017-08-10Version 1

The goal of this paper is to introduce and study noncommutative Catalan numbers $C_n$ which belong to the free Laurent polynomial algebra in $n$ generators. Our noncommutative numbers admit interesting (commutative and noncommutative) specializations, one of them related to Garsia-Haiman $(q,t)$-versions, another -- to solving noncommutative quadratic equations. We also establish total positivity of the corresponding (noncommutative) Hankel matrices $H_n$ and introduce accompanying noncommutative binomial coefficients.


text:
Browse Questions

# A vessel A with a capacity $V_1=2lt$ contains gas at a pressure $P_1=2atm$ and a vessel B with a capacity of $V_2=4lt$ contains the same gas at a pressure of 1atm.The temperature is $0^{\large\circ}C$ in A & $27^{\large\circ}C$ in B.What will be the pressure in the vessels if they are connected by a tube?

$\begin{array}{1 1}(A)\;1.43atm\\(B)\;14.3atm\\(C)\;0.72atm\\(D)\;7.2atm\end{array}$

$PV=P_1V_1+P_2V_2$
$P(V_1+V_2)=P_1V_1+P_2V_2$
$\Rightarrow P=1.43atm$
Hence (A) is the correct answer.


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# GeometricBoundingVolume¶

class GeometricBoundingVolume

Bases: BoundingVolume

This is another abstract class, for a general class of bounding volumes that actually enclose points in 3-d space, such as BSP’s and bounding spheres.

Inheritance diagram

int contains(GeometricBoundingVolume const *vol) const
int contains(LPoint3 const &point) const
int contains(LPoint3 const &a, LPoint3 const &b) const

Returns the appropriate set of IntersectionFlags to indicate the amount of intersection with the indicated volume.

Returns the appropriate set of IntersectionFlags to indicate the amount of intersection with the indicated point.

Returns the appropriate set of IntersectionFlags to indicate the amount of intersection with the indicated line segment.

bool extend_by(GeometricBoundingVolume const *vol)
bool extend_by(LPoint3 const &point)

Increases the size of the volume to include the given volume.

Increases the size of the volume to include the given point.

virtual LPoint3 get_approx_center(void) const = 0
static TypeHandle get_class_type(void)
virtual void xform(LMatrix4 const &mat) = 0


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## Why this ugly looking formula to calculate ‘target’ from ‘nBits’ in block header: target = coefficient * 256**(exponent-3)

Why this ugly looking formula to calculate ‘target’ from ‘nBits’ in block header: target = coefficient * 256**(exponent-3).

Why not? : target = coefficient * 256**(exponent). What is the need of subtracting 3. All we need is to be able to generate 256 bit long number and enough precision (which we already have with 3 bytes dedicated for coefficient)

Even better, why not? : target = coefficient * 2**(exponent)

## co.combinatorics – Random gaussian binomial coefficient

Pick a lattice path which starts at $$(0,0)$$ and finishes at $$(X,Y)$$ in
$$(0,X) times (0,Y) capmathbb{Z}^2$$,
and
consider the area $$A$$ enclosed beneath. If $$X,Y$$ are independent Poisson random
variables, what is the p.g.f. of the area $$A$$?

To explain further, consider the $$q$$-analogue of the binomial coefficient, known as the Gaussian binomial coefficient:

$${X + Y choose Y}_{q} = 1 +q + dots + q^{XY}$$

but where we let $$X,Y$$ be independent Poisson random variables with mean $$lambda_{X}$$ and $$lambda_{Y}$$:

$$X sim text{Poisson}(lambda_{X}) \ Y sim text{Poisson}(lambda_{Y})$$

We could write

$$f(lambda_X,lambda_Y,q)=sum_{x=0}^{infty} sum_{y=0}^{infty} left(frac{e^{-lambda_X}lambda_X^x}{x!}frac{e^{-lambda_Y}lambda_Y^y}{y!}left(frac{1}{{x+ychoose y}}{x + y choose y}_{q}right)right) \ ,,,= e^{-(lambda_X +lambda_Y )}sum_{x=0}^{infty} sum_{y=0}^{infty} frac{lambda_X^xlambda_Y^y}{(x+y)!}{x + y choose y}_{q}$$

This gives the p.g.f. of the area under a lattice path in $$(0,X) times (0,Y) cap mathbb{Z}^2$$ from $$(0,0)$$ to $$(X,Y)$$ i.e. a lattice path in a random rectangle, from $$(0,0)$$ to $$(X,Y)$$, with Poisson dimensions $$X times Y$$.

What can be said about this random area? It is not just a random lattice path and the p.g.f. of its enclosed area, but now, the lattice it traverses is random.

## vba – User-defined function for estimating coefficient of performance

I am using hourly weather data to estimate performance of commercial refrigeration systems under various parameters. The formula for estimating co-efficient of performance (COP) of these systems is repeated for each hour.

I defined a formula to do this calculation:

``````Public Function estimatedCOP(dischargeTemp As Double, suctionTemp As Double)

Dim a As Double, b As Double, c As Double, d As Double, e As Double, f As Double

a = 9.12808037
b = 0.15059952
c = 0.00043975
d = -0.09029313
e = 0.00024061
f = -0.00099278

estimatedCOP = a + b * suctionTemp + c * suctionTemp * suctionTemp + _
d * dischargeTemp + e * dischargeTemp * dischargeTemp + _
f * suctionTemp * dischargeTemp

End Function
``````

This gives the expected result, but takes a long time to calculate and update (as it’s repeated ~10,000 times). How can I improve the performance of this function?

## Bayesian Statistics – Posterior distribution of the logistic regression coefficient

I have a binary logistic regression with the following characteristics

Consider the logistic regression for binary data Yi ∈ {0,1} and the covariate vector xi = (xi, 1, xi, 2, …, xi, p). Assuming logistic regression, the sample distribution is Yi (1). We assume a normal prior for βj for j = 1, …, n as in (2) and βjs are independent of each other. (μj, σj2) are previous parameters and must be specified.

Prob (Yi = 1 | β) = exp (x⊤i β) / 1 + exp (x⊤i β)
βj ∼N (μj, σj2) (2)

I need to find the back distribution for β, i.e. H. P (β | y, μ1, …, μp, σ12, …, σp2).

## at.algebraic topology – when the local coefficient system in the Leray-Serre spectral sequence is simple

To let $$F to E to B$$ a vibration and $${E_ {r} ^ { ast, ast}, d_ {r} }$$ the Leray-Serre spectral sequence converges to $$H ^ { ast} (E; R),$$ so that
$$E_ {2} ^ {p, q} = H ^ {p} (B; mathcal {H} ^ {q} (F; R))$$
the cohomology of $$B$$ with local coefficients in the fiber cohomology $$F.$$ We know that when the action of $$pi_ {1} (B)$$ induced by vibration on the cohomology of $$F$$ is trivial, then the system of local coefficients is simple. My question is: there is something to say (provided that $$R = mathbb {Z} _ {2}$$ and $$pi_ {1} (B) = mathbb {Z} _ {2}$$) If the action is trivial except for ONE element in cohomology? More specifically, there is only one element $$c in H ^ {1} (F; R)$$ so that $$mu (g, c) = gc neq c,$$ to the $$g in mathbb {Z} _ {2} = langle g rangle$$when $$mu: pi_ {1} (B) times H ^ { ast} (F) to H ^ { ast} (F)$$ denotes the induced effect.

## Google Sheets – Calculate the performance trend line coefficient for the array over a period of time

I have a database with 3 columns (code, week, value), where code is project code, week is the date of the beginning of the week and value is the aggregate amount for the respective week.

For each project, I have to determine a performance trend line coefficient for a certain period (e.g. the last 4 weeks since the beginning of 2020).

``````INDEX(LINEST(LN(Values),LN({1,..,N})),1)
``````

where N is the number of weeks from the start (e.g. 4 for the last 4 weeks)

My problems started with trying to do this for a number of projects.

I only found a solution for one project (example for the last 4 weeks, others in the file), but there is a way to do it for all projects.

``````=IFERROR(INDEX(
LINEST(
LN(
VLOOKUP(
SORT(
LARGE(UNIQUE(FILTER(DB!\$B:\$B,DB!\$A:\$A = A2)),
SEQUENCE(4,1,2,1)),1,true),
QUERY({DB!\$A:\$C},
"Select Col2, SUM(Col3)
Where Col1 ='"&A2&"'
Group by Col2
Order by Col2
Label SUM(Col3) ''",0),
2,0)
),
LN(
SEQUENCE(4,1,2,1)
)),
1),0)
``````

Here is Test_Doc,
Thank you in advance.

## Classical analysis and odes – Limitation of a Fourier coefficient of a non-negative periodic function with respect to its \$ L ^ 2 \$ norm

This question is motivated by the earlier MO question: Show that \$ ( sum_ {k = 1} ^ {n} x_ {k} cos {k}) ^ 2 + ( sum_ {k = 1} ^ {n}) x_ {k} sin {k}) ^ 2 le (2+ frac {n} {4}) sum_ {k = 1} ^ {n} x ^ 2_ {k} \$.
It is an adjusted asymptotic version of this question.

To let $$f$$ be a non-negative function, periodically with period $$1$$and square can be integrated $${ Bbb R} / { Bbb Z}$$. Is it true that
$$| { widehat f} (1) | ^ 2 = Big | int_0 ^ 1 f (x) e ^ {- 2 pi ix} dx Big | ^ 2 le frac 14 int_0 ^ 1 f (x) ^ 2 dx ?$$
For example, equality is achieved when $$f (x) = max (0, cos (2 pi x))$$.

Note that $$| widehat f (1) | = | widehat f (-1) |$$ and since $$f$$ is not negative $$| widehat f (1) | le widehat f (0)$$. Therefore
$$int_0 ^ 1 f (x) ^ 2 dx = sum_n | widehat f (n) | ^ 2 ge 3 | widehat f (1) | ^ 2,$$
so that the estimate applies $$1/3$$ instead of $$1/4$$. There is a lot of scope to improve this argument, and with a more careful application of Bessel's inequality I could get the constant $$1/4 + 1/4 pi$$. But the claimed inequality looks very clean, and I wonder if (i) it is true! (Ii) is known in a certain context and (iii) (hopefully) has elegant evidence?

## Geometry – problem with equating the coefficient of the elliptic curve

To let, $$E: = y ^ 2 = x ^ 3 + Ax + B$$ an elliptic curve, 2 points $$P, -Q$$ on $$E$$ so that $$2P = -Q$$, we can write

$$y ^ 2-x ^ 3 + Ax + B = (x – e_1) (x – e_2) ^ 2 = 0$$
Where, $$e_1 = x (Q), e_2 = x (P)$$ a double root. So,
$$y = m (x-x (Q)) + y (Q)$$, Here $$y$$ is the line that goes thoroughly $$P, Q$$, and $$m$$ is the slope.

$$(x – e_1) (x ^ 2 – 2xe_2 + e_2 ^ 2)$$
$$= (x ^ 3 – 2x ^ 2e_2 + xe_2 ^ 2) – (x ^ 2e_1 – 2xe_1e_2 + e_1e_2 ^ 2)$$
$$= x ^ 3 – 2x ^ 2e_2 + xe_2 ^ 2-x ^ 2e_1 + 2xe_1e_2-e_1e_2 ^ 2$$
$$= x ^ 3 + x ^ 2 (-1) (2e_2 + e_1) + x (e_2 ^ 2 + 2e_1e_2) + (- 1) e_1e_2 ^ 2$$

Of $$y ^ 2-x ^ 3 + Ax + B = 0$$ we get coefficients of $$x ^ 2$$ is $$(- 1) m ^ 2$$, We find the equality coefficient, $$m ^ 2 = (2e_2 + e_1)$$,

The problem is, if I simplify both sides and then equate the coefficient, I get the correct coefficient of $$x ^ 2$$ but not for $$x$$ (if plug-in value of $$P, Q$$, the coefficients of $$x ^ 2$$ just doesn't satisfy $$x$$).

What's the problem?

## Number theory – Question about the coefficient of the Dirichlet series in relation to \$ frac { zeta (s + 2)} { zeta (s)} \$

This question concerns the evaluation of $$a (n)$$ defined in (1) below, which refers to the Riemann zeta function $$zeta (s)$$ as shown in (2) below.

(1) $$quad a (n) = sum limit_ {d | n} frac { mu left ( frac {n} {d} right)} {d ^ 2} = frac {A046970 (n)} {n ^ 2}$$

(2) $$quad frac { zeta (s + 2)} { zeta (s)} = sum limit_ {n = 1} ^ infty a (n) n ^ {- s} ,, quad Re (s)> frac {1} {2} quad text {(assuming the Riemann hypothesis)}$$

A046970 in formula (1) above is the Dirichlet inverse of the Jordan function $$J_2 (n)$$ (A007434) where $$J_k (n)$$ is a generalization of the Euler totient function $$phi (n)$$ both of which are defined below.

(3) $$quad phi (n) = n prod limit_ {p | n} left (1- frac {1} {p} right)$$

(4) $$quad J_k (n) = n ^ k prod limit_ {p | n} left (1- frac {1} {p ^ k} right)$$

The following figure shows $$a (n)$$ always seems to rate that $$-1 except for $$n = 1$$ Where $$a (1) = 1$$and there is also a noticeable streaking in the values of $$a (n)$$,

Illustration 1): Discrete act of $$a (n)$$

Question 1): Can this be proven? $$-1 for all $$n> 1$$? I checked $$-1 to the $$1 ,

Question 2): Is there an explanation for the perceptible streaking in the values of $$a (n)$$ in Figure (1), and this stripe pattern continues as $$n to infty$$?

## Approximation theory – are there good references to the decay rate of the Legendre coefficient?

To let $$P_n: (- 1,1) rightarrow mathbb {R}$$ be the $$n$$Legendre polynomial. and let
$$a_n: = int _ {- 1} ^ 1 f (t) P_n (t) , dt$$
for some $$f: (- 1,1) rightarrow mathbb {R}$$,

There is good evidence of the decay rate of $$vert a_n vert$$? I don't know about this type of problem, but I think there have to be many methods.

From the following similar mathoverflow question: reference for the exponential decay of Legendre coefficients, I found a work. I also found a book by Atkinson on "Spherical Harmonics and Approaches to the Unity Sphere". After reading these references, it appears that the smoothness of $$f$$ is a method. But the application in my head is the case when $$f = arccos ^ 2 (t)$$what is not smooth enough.

So I wonder if there are other references that explain different methods of calculating the decay rate of $$vert a_n vert$$, Especially if there are some techniques that can be used for slippery $$f$$, I really want to know.


text:
I’ve started working on a HLS project. Initially, I have heard a lot of good things about HLS, but now I’ve also started to realize some of its problems. This is an unfinished post, and it’s based on my personal experience as someone who is not an expert on FPGA programming.

Pros

• Provide a really good environment for co-development between firmware and software (including test bench).
• Provide the firmware emulation with bitwise accurate results. The emulation can be used outside of Vivado for a lot of studies.
• Allow non-FPGA engineers to collaborate on the algorithm development. This helps physicists to take over some work load, which is useful for projects where the num of physicsts >> num of engineers.
• Facilitate interfacing with C++ and Python. Xilinx has started providing first-class HLS libraries (e.g. Vitis Libraries) to interact with its FPGAs. This helps lowering the barrier to adopt FPGAs.

Cons

• The output Verilog code generated by the Vivado HLS tool is not human readable.
• Due to the above, it’s very hard to figure out how to optimize the algorithm. Maybe an experienced FPGA engineer could do that, but it’s really difficult for me. I have no idea how should I structure a particular for loop, or which pragma directive should I use, or whether a function should be inlined, etc in order to achieve the results that I want, since I couldn’t understand the translation process.
• The Vivado HLS tool is way too slow. It takes several hours (sometimes >3 hours) to compile my codes after a single change.
• A lot of configurations need to be passed as compile-time arguments (e.g. loop bounds), rendering the C++ codes also kind of unreadable. In C++11 (the standard supported by the Vivado HLS tool), the use of constexpr is very restrictive.


text:
# How do you factor 64v^2-81?

$64 {v}^{2} - 81$
$= {\left(8 v\right)}^{2} - {\left(9\right)}^{2}$
$= \left(8 v + 9\right) \left(8 v - 9\right)$ by the difference of squares formula


text:
## Intermediate Algebra (12th Edition)

$\frac{0}{-8}=0\div-8=0$ In general, $\frac{0}{a}=0\div a=0$, where $a$ is a nonzero number. That is, 0 divided by a nonzero number will have a quotient of 0.


text:
# maximum value at given interval

• February 22nd 2010, 10:16 AM
bigwave
maximum value at given interval
$
f(x)=\sqrt{1+x^3}$

find the maximum value of |f''(x)| on the given Interval of $[0,2]$

calculated $f''(x)$ to be

$
f''(x) = \frac{3x(x^3+4)}{4(x^3+1)^{3/2}}$

thot setting this to zero would give where the maximum value is

but the answer is $f''|(\sqrt[3]{-10+\sqrt{108}})|=1.47$

graphing this can see that the top of the graph is 1.47 but don't see how this was derived.
• February 22nd 2010, 11:13 AM
Keithfert488
To find the max of f''(x) you have to set it's derivative (f'''(x)) to zero
• February 22nd 2010, 12:06 PM
bigwave
i did try this but did not get the answer as expected.
f'''(x) is pretty complicated.


text:
# How do you find the roots, real and imaginary, of y=-5(x-2)^2+3x^2-15  using the quadratic formula?

Jun 9, 2018

color(violet)(x = 5 + sqrt 15, 5 - sqrt15

#### Explanation:

$y = - 5 {\left(x - 2\right)}^{2} + 3 {x}^{2} - 15$

$y = - 5 {x}^{2} + 20 x - 20 + 3 {x}^{2} - 15$

$y = - 2 {x}^{2} + 20 x - 35$

$x = \frac{- 20 \pm \sqrt{{20}^{2} - 280}}{-} 4$

$x = \frac{- 20 \pm \sqrt{120}}{-} 4$

color(violet)(x = 5 + sqrt 15, 5 - sqrt15


text:
$$\renewcommand\AA{\unicode{x212B}}$$

# Framework Changes¶

• The ALF IDF has been updated following a detector array alteration.

• The LARMOR IDF has been updated following the addition of a new detector to the instrument.

• Instrument definition for IN4, 5, and 6 at the ILL have been updated

• REF_L and CNCS has update instrument files and EQSANS, REF_M, ARCS and CNCS are now configured for loading live data.

## Stability¶

• We have introduced a Project Recovery mechanism for Mantidplot in order to be able to recover the lost state of the interface in the event of a crash or unexpected shutdown. There are more details in the UI section of the release notes.

• The error reporter can now catches hard crashes to desktop, allowing us to get more information on causes of hangs or crashes in Mantid. Since the last release error reports sent to us led directly to the identification of and fixes for 3 separate bugs in Mantid.

• Mantid now handles poor network stability better when reading live data from the ISIS DAE. Mantid will now timeout after a couple of minutes of loss of network connectivity and remains responsive during this time. You can alter the duration of this timeout by adding a line to the mantid.user.properties file like:  ISISDAE.Timeout = 100 #seconds

## Algorithms¶

### New feature¶

• Sometimes the algorithm you are looking at is close to what you want, but not quite, to help you find the right one a list of related algorithms has been added to each algorithm, and is displayed in the documentation page of each algorithm as part of it’s summary.

### New Algorithms¶

• LoadSampleShape loads a shape into the sample in a workspace from an ASCII STL file, which contains a list of triangles or an ASCII OFF file, which has a list of vertices and triangles.

• CalculateCarpenterSampleCorrection outputs a group workspace with the separate absorption and multiple scattering corrections for flexibility to the User to apply them to the sample workspace

• CarpenterSampleCorrection replaces MultipleScatteringCylinderAbsorption and uses CalculateCarpenterSampleCorrection for calculating its corrections.

• SaveBankScatteringAngles was added to save theta and phi values for diffraction banks to a MAUD-readable grouping.new file format

• ExportSampleLogsToHDF5 saves a workspace’s samples logs to an HDF5 file

• SaveGDA saves a focused diffraction workspace to MAUD-readable .gda format

• SaveGEMMAUDParamFile, which acts as a partner to SaveGDA, saves a MAUD calibration file to convert the output of SaveGDA back to d-spacing

• Algorithm FitPeaks is implemented as a generalized multiple-spectrum multiple-peak fitting algorithm.

### Bugfixes¶

• In LoadMcStas internally reduce number of event workspaces created. If n mcstas event components now create n*(n-1) fewer.

• The documentation of the algorithm CreateSampleWorkspace v1 did not match its implementation. The axis in beam direction will now be correctly described as Z instead of X.

• The ExtractMask algorithm now returns a non-empty list of detector ID’s when given a MaskWorkspace.

• Fixed a crash when the input workspace for GroupDetectors contained any other units than spectrum numbers.

• ConvertToMD can now be used with workspaces that aren’t in the ADS.

• Fixed SumSpectra to avoid a crash when validation of inputs was called with a WorkspaceGroup.

• Fixed a bug in TableWorkspaces where vector column data was set to 0 when the table was viewed

• The output workspace of LineProfile now has correct sample logs, instrument and history.

• TimeSeriesProperty::splitByTimeVector’s behavior on a boundary condition is changed. In the set of splitters toward a same target split workspace, if a splitter’s beginning time is after the last entry of the TimeSeriesProperty to be split, then this last entry shall be included in its output TimeSeriesProperty.

• Fixed a bug in MergeRuns which could cause the runs to be merged in a different sequence than indicated in the InputWorkspaces property.

• Fixed a bug where the values entered for basis vector properties in BinMD were not being remembered.

• Fixed a bug which prevented Load and LoadAndMerge from parsing advanced run ranges such as 1-3+5-7+10+15-20.

## Python¶

### New¶

• Added a new MDFrameValidator which can check that a MD workspace passed to a python algorithm has the expected MD frame (e.g. HKL, QLab, QSample etc.).

### Improvements¶

• Python fit functions that use from IPeakFunction as a base no longer require a functionDeriveLocal method to compute an analytical derivative. If the method is absent then a numerical derivative is calculate.

### Bugfixes¶

• Checks on the structure of Python fit function classes have been improved to avoid scenarios, such as writing function1d rather than function1D, which would previously have resulted in a hard crash.

• Fit functions defined in a python script can be used with the new fit function API right after subscription.

• Child algorithms now respect their parent algorithm’s EnableLogging setting when invoked using the function-style calling. Previously, some messages could appear in the log even though EnableLogging was set to False.

• Fixed a bug in detectorSignedTwoTheta method in MatrixWorkspace where the sign of the angle depended on the axis pointing up, not on the actual theta-sing axis defined in the IDF.

• Fixed a bug where calling setCell method on a PeaksWorkspace did not set the cell value.

Release 3.13.0


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1. Help?

If you walked 10 km at an average speed of 4km/h how many hours did you walk for?

2. rate is 4km/h

you need to walk 10 km.

note that..

rate = distance/time meaning time = distance/rate


text:
Proposition 66.10.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X$ is reasonable, then there exists a dense open subspace of $X$ which is a scheme.

Proof. By Properties of Spaces, Lemma 64.13.1 the question is local on $X$. Hence we may assume there exists an affine scheme $U$ and a surjective étale morphism $U \to X$ (Properties of Spaces, Lemma 64.6.1). Let $n$ be an integer bounding the degrees of the fibres of $U \to X$ which exists as $X$ is reasonable, see Definition 66.6.1. We will argue by induction on $n$ that whenever

1. $U \to X$ is a surjective étale morphism whose fibres have degree $\leq n$, and

2. $U$ is isomorphic to a locally closed subscheme of an affine scheme

then the schematic locus is dense in $X$.

Let $X_ n \subset X$ be the open subspace which is the complement of the closed subspace $Z_{n - 1} \subset X$ constructed in Lemma 66.8.1 using the morphism $U \to X$. Let $U_ n \subset U$ be the inverse image of $X_ n$. Then $U_ n \to X_ n$ is finite locally free of degree $n$. Hence $X_ n$ is a scheme by Properties of Spaces, Proposition 64.14.1 (and the fact that any finite set of points of $U_ n$ is contained in an affine open of $U_ n$, see Properties, Lemma 28.29.5).

Let $X' \subset X$ be the open subspace such that $|X'|$ is the interior of $|Z_{n - 1}|$ in $|X|$ (see Topology, Definition 5.21.1). Let $U' \subset U$ be the inverse image. Then $U' \to X'$ is surjective étale and has degrees of fibres bounded by $n - 1$. By induction we see that the schematic locus of $X'$ is an open dense $X'' \subset X'$. By elementary topology we see that $X'' \cup X_ n \subset X$ is open and dense and we win. $\square$

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text:
Home > INT2 > Chapter 1 > Lesson 1.3.3 > Problem1-103

1-103.

Throughout this book, key problems have been selected as “checkpoints”. Each checkpoint problem is marked with an icon like the one at right. These checkpoint problems are provided so you can check to be sure you are building skills at the expected level. When you have trouble with checkpoint problems, refer to the review materials and practice problems that are available in the “Checkpoint Materials” section at the back of your book.

This problem is a checkpoint for solving problems with linear and exponential relationships. It will be referred to as Checkpoint 1.

Write and solve an equation or equations for each of the following problems.

1. In the last election, candidate C received $15,000$ fewer votes than candidate B. If a total of $109,000$ votes were cast, how many votes did candidate B receive?

2. The price of gourmet coffee beans at the corner market has increased $10 \%$ per year over the past several years. If the current cost is $14$ per pound, what did a pound of gourmet coffee beans cost two years ago?

3. Gage bought a new car for $29,000$ to use while he is away at college. The car depreciates (decreases in value) by $11\%$ annually. If he plans to sell the car once he graduates in four years, what will the value of the car be?

4. Ms. B and Ms. D are writing problems for an algebra book. Ms. B started with ten problems already written, and she can write six problems an hour. Ms. D had no problems written, but she writes ten problems an hour. When will Ms. D have the same number of problems written as Ms. B? How many problems will they each have written at that time?

Check your answers by referring to the Checkpoint 1 materials located at the back of your book.

Ideally, at this point you are comfortable working with these types of problems and can solve them correctly. If you feel that you need more confidence when solving these types of problems, then review the Checkpoint 1 materials and try the practice problems provided. From this point on, you will be expected to do problems like these correctly and with confidence.

Answers and extra practice are located in the back of your printed textbook or in the Reference Tab of your eBook. If you have an eBook for INT2, login and then click the following link: Checkpoint 1: Solving Problems with Linear and Exponential Relationships


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plot,TrinarizationResult-method {Binarize} R Documentation

## Visualization of trinarization results.

### Description

Visualizes a trinarization as a ray or a two-dimensional plot.

### Usage

## S4 method for signature 'TrinarizationResult,ANY'
plot(x,
twoDimensional=FALSE,
showLegend=TRUE,
showThreshold=TRUE,
...)
## S4 method for signature 'numeric,TrinarizationResult'
plot(x,
y,
showLegend=TRUE,
showThreshold=TRUE,
...)

### Arguments

 x If y is supplied, this is a vector of x coordinates for the trinarization values iny, which are plotted on the y axis. If y is not supplied, this is object of class TrinarizationResult containing the trinarized values to visualize. y If x is a vector of x coordinates, this is object of class TrinarizationResult containing the trinarized values to visualize. twoDimensional Specifies whether the trinarization is depicted as a ray or as a two-dimensional curve (see details). showLegend If set to true, a legend is included in the plot. showThreshold If set to true, the trinarization thresholds are depicted as a horizontal or vertical lines (depending on twoDimensional). ... Further graphical parameters to be passed to plot. The parameters col and pch can be supplied in different ways: If supplied as vectors of size 3, the first value corresponds to a 0 in the trinarization, the second value corresponds to a 1, and the third value corresponds to a 2. col can also have length 4, in which case the fourth entry is the color of the threshold line. If col or pch have the size of the input vector, the corresponding colors and symbols are assigned to the data points.

### Details

The function comprises two different plots: If twoDimensional = TRUE, the positions in the input vector are aligned with the x axis, and the y axis corresponds to the values. The trinarization thresholds are shown as a horizontal lines, and the trinarization is indicated by three different symbols.

If twoDimensional = FALSE, the trinarized values are aligned with a one-dimensional ray, and the separating thresholds are depicted as a vertical lines.

plot, TrinarizationResult

### Examples

# plot a binarization in one and two dimensions
res <- TASC(iris[,"Petal.Length"])
plot(res)
plot(res, twoDimensional = TRUE)
plot(res, twoDimensional = TRUE,
pch = c("x", "+"),
col = c("red", "black", "royalblue", "green"),
lty = 4, lwd = 2)

[Package Binarize version 1.3 Index]


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# Replacing parts of a neural net during training using TrainingProgressFunction

NetReplacePart is interesting because it allows to change, for instance, the weights of a layer (in a net), however it does return a new net and not mutate the existing net.

TrainingProgressFunction on the other hand allows us to call arbitrary functions during training.

Is is possible to combine NetReplacePart and TrainingProgressFunction so that we can change the net we are currently training?

An useful application could a soft tying of weights in an auto encoder net. Usually tying of weights means that the decoder weight matrix is just the transposed encoder weight matrix. As far as I am aware there is currently no method to do so directly, so I though one could call a function that takes both encoder and transposed decoder matrix and replaces both by the average of the two every few batches. My naive approach of just using Set on the #Net parameter supplied to TrainingProgressFunction prints out the usual warnings of misusing Set.


text:
# intersectionLattice -- generates the intersection lattice of a hyperplane arrangement

## Synopsis

• Usage:
P = intersectionLattice(L, R)
• Inputs:
• L, a list, which gives the equations defining the hyperplane arrangement
• R, a ring, which the hyperplanes are defined over
• Outputs:
• P, an instance of the type Poset,

## Description

The intersection lattice of a hyperplane arrangement is the lattice of intersections in the arrangement partially ordered by containment.

 i1 : R = QQ[x,y,z]; i2 : intersectionLattice({x+y, x+z, y+z}, R) o2 = Relation Matrix: | 1 0 1 0 1 0 1 0 | | 0 1 1 0 0 1 1 0 | | 0 0 1 0 0 0 1 0 | | 0 0 0 1 1 1 1 0 | | 0 0 0 0 1 0 1 0 | | 0 0 0 0 0 1 1 0 | | 0 0 0 0 0 0 1 0 | | 1 1 1 1 1 1 1 1 | o2 : Poset


text:
# Math Help - Arcs

1. ## Arcs

Please give me full solution of this question with image also:
If X and Y are the middle points of arcs AB and CD of circle and sigment joining by X and Y cuts the AB and CD at point P and Q then prove that AP=AQ

2. Hello, soni!

There must be a typo . . . The problem makes no sense.

$X\text{ and }Y$ are midpoints of arcs $AB\text{ and }CD$ of a circle
and segment $XY$ cuts chords $AB\text{ and }CD$ at $P\text{ and }Q$, resp.

Prove: . $AP=AQ$ . ??
Code:
                  C
A   * * o
o       |   *
* |       |     *
*  |       |      *
|       |
*   |P      |Q      *
X o - + - - - + - - - o Y
*   |       |       *
|       |
*  |       |      *
* |       |     *
o       |   *
B   * * o
D

Draw segment $AQ.$

We can see that $AP$ is never equal to $AQ.$


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# In 2013 The Andrews International BID Patrol Arrested Homeless People at More than 57 Times the Rate that the LAPD Did and were Responsible for 1 in 14 Homeless Arrests in Entire City of Los Angeles

(I apologize in advance for this necessarily data-heavy post, but it’s essential information).

In 20131 the BID Patrol arrested homeless people at more than 57 times the rate that the LAPD did. Furthermore, they were responsible for more than 1% of all arrests made in the entire City of Los Angeles that year even while working only 0.13% of the hours that the LAPD did. Approximately one in fourteen arrests of homeless people in the entire city of Los Angeles that year was made by the BID Patrol.

Here’s how I calculated these figures: That year the LAPD made 14,838 arrests of homeless people2 whereas the Andrews International BID Patrol made 1,096 arrests.3 Reading through A/I’s 2013 arrest reports and examining A/I’s 2013 arrest photos I see no reason to believe that the BID Patrol arrested non-homeless people in 2013 in any significant number.4

I have been unable to find an accurate account of the number of hours the LAPD worked in 2013, but I estimate it to be 19,200,000 (details in the footnote).5 Thus in 2013 the LAPD arrested homeless people at a rate of

$\frac{19,200,000}{14,838}\frac{hours}{arrest}=1,294.0\frac{hours}{arrest}$.

According to the 2013 Planning Report of the Hollywood Property Owners Alliance, the BID Patrol managed 416 patrol hours per week and about 80 hours for other employees. I’m using the total figure of 496 hours per week.6 Thus in 2013 the BID Patrol arrested homeless people at a rate of

$\frac{496 \times 52}{1,096}\frac{hours}{arrest}=22.6\frac{hours}{arrest}$.

Note that the figure for the LAPD is $\frac{1294.0}{22.6}=57.3$ times bigger than the BID Patrol. In other words, in 2013, it took the LAPD 57.3 times more hours to arrest each homeless person than it took the BID Patrol.

This figure is even more astonishing when you consider that there are about 23,000 homeless people in the City of Los Angeles7 and, according to Kerry Morrison, during 2013 there were never more than 175 homeless people at a time in the two BIDs covered by the BID Patrol. That is, the LAPD made

$\frac{14,838}{23,000}\frac{arrests}{homeless}=0.65\frac{arrests}{homeless}$

$\frac{1,096}{175}\frac{arrests}{homeless}=6.3\frac{arrests}{homeless}$

Another way to look at the shocking import of these calculations is that in 2013 the LAPD made a total of 104,255 arrests.8 In 2013, as we’ve seen, the BID Patrol made 1,096 arrests. Thus more than 1% of all arrests made in the entire City of Los Angeles were made by the BID Patrol.9 This is amazing, considering that, as we’ve seen, the BID Patrol worked about 24,752 hours that year whereas the LAPD worked about 19,200,000. In other words, the BID Patrol worked about 0.13% as much as the LAPD.

I hope to write a follow-up post to this quite soon examining the effect that this off-the-chain behavior on the part of the BID could have on the homeless policies of the City of Los Angeles.

1. This is a year for which I have extraordinarily accurate statistics available. I have no reason to think there’s anything exceptional about it.
2. See Homelessness and the City of Los Angeles by Miguel Santana, City Administrative Officer. April 16, 2015. P.18.
3. See 2013-YTD-TOTALS-WE-12-31-13.xls, which I obtained from the Hollywood Property Owners Alliance pursuant to the California Public Records Act.
4. According to HPOA Executive Director Kerry Morrison the BID does not track this data, so it won’t be possible to find the actual figure.
5. According to the California Policy Center, who analyzed a number of public records, LAPD officers work very little overtime. I’ll assume they work none, so 40 hours per week, 48 weeks per year, per officer. Note that I’m assuming 48 weeks per year per officer because that makes the calculations more favorable to A/I. In 2013 the LAPD had very close to 10,000 sworn officers. This is 40 hours per week x 48 weeks per year per officer x 10,000 officers = 19,200,000 hours per year.
6. Making this number larger yields an estimate that’s more favorable to Andrews International. The information appears on page 5 of the linked report.
7. Again according to Santana’s report
8. Santana’s report again.
9. It’s not possible to determine from the data I have whether the BID Patrol’s arrests are included in the LAPD’s figure. It would make sense if they were, because the LAPD has to accept custody of every person that the BID Patrol arrests. It would also make sense if they were not, as in some sense private persons’ arrests are not made by the LAPD. In any case, the numbers change very little between the case where they’re included and the case where they’re not.


text:
#### Problem 9E

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. g(s)=sx(tt2)8dt