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# Math Q&A Game

by Gokul43201
Tags: game, math
 Emeritus Sci Advisor PF Gold P: 11,155 A Q&A game is simple: One person asks a relevant question (it can be research, calculation, a curiosity, something off-the-top-of-the-head, anything ... as long as it's a math question) and other people try to answer. The person who posts the first correct answer (as recognized by s/he who asked the question) gets to ask the next question, and so on. Let me start this off with a simple number theory problem : What is the least number than must be multiplied to 100! (that's a factorial) to make it divisible by $12^{49}$ ? (throw in a brief -couple of lines or so- explanation with the answer)
 P: 617 1/(99!*50)
 Emeritus Sci Advisor PF Gold P: 11,155 Not correct. "Multiplying" by your number gives a "product" of 2 (=100!/50*99!), and 2 is not divisible by 12^{49}.
 P: 617 Math Q&A Game i think i read it backwards...
 Sci Advisor HW Helper P: 9,396 There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97. similiarly there are 33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess 2*3^50 will do
Emeritus
PF Gold
P: 11,155
 Quote by matt grime There are 50 factors of 100 divisibile by 2, 25 by 4, 12 by 8, 6 by 16, 3 by 32 1 by 64 so the power of two in 100! is 97. similiarly there are 33 div by 3, 11 by 9, 3 by 27 and 1 by 81 making 48 times 3 divides, so i guess 2*3^50 will do
I'm not sure I follow the finish...

You've shown that 100! has 2^{97} * 3^{48}. And what happened after that ?

In any case : Next question is yours ...
 Sci Advisor HW Helper P: 9,396 Duh, do i feel stupid from typing too quickly whilst someone is talking to me. should have been 6. Willl think of a question later tonight.
 Sci Advisor HW Helper P: 9,396 Ok, here's one that I hope isn't too tricky. Let A be an nxn matrix over C, the complexes. Suppose that Tr(A^r)=0 for all r. Show A is nilpotent. All the other ones I could think of were too easily looked up.
HW Helper
P: 3,684
 Quote by Gokul43201 I'm not sure I follow the finish... You've shown that 100! has 2^{97} * 3^{48}. And what happened after that ?
$$12^{49}=2^{98}3^{49}$$, so $$100!=2^{97}3^{48}X$$ must be multiplied by $$6=2^{98-97}3^{49-48}$$ before $$12^{49}$$ will divide it.
Emeritus
PF Gold
P: 11,155
Yes, CRG : matt clarified this in his subsequent post. Thanks !

Back to matt's question :

 Quote by matt grime Let A be an nxn matrix over C, the complexes. Suppose that Tr(A^r)=0 for all r. Show A is nilpotent.
 Sci Advisor HW Helper P: 9,396 If there's no progress on this after a couple of days i'll post hints. perhaps people should post what they think they need to do. i like tyhe question cos it uses lots of bits from here there and everywhere.
 P: 127 Well it seems that first you need to show that such a matrix does not have n distinct eigenvectors and then show that not having n distinct eigenvectors implies nilpotents. The first question relates to the fact that if a matrix had a full set of Eigenvectors then diagonalization gives us that $$\sum \lambda_i^n =0$$ for all n>0. For the second half I would gather you need to consider the vectors not in the span of the Eigenvectors and consider where they may be mapped. hmmm... that may not be enough for the first half. You may need to also show that any vector that is a eigenvector is in fact in the kernel. Experimentally it appears like we are dealing with upper or lower triangular matrix if you just assume Tr(A^r) =0 for r=1..n. But that could just be maple not returning all possible answers which it sometimes does. That's what I've been thinking
 Emeritus Sci Advisor PF Gold P: 16,091 Hrm. If a matrix A is diagonalizable, then I claim that there exists an n such that all of the nonzero eigenvalues of An lie in the right half-plane. The requirement that Tr(An) = 0, forces all the eigenvalues to be 0, and thus A is zero... clearly nilpotent! So the trick, then, is when the matrix is not diagonalizable. Then again, this only works for complex valued matrices.
 Sci Advisor HW Helper P: 9,396 I did state the matrix was over C, though this is largely for conveniece. And, snobel, the matrix 0 has a full set of eigen values and is certainly nilpotent an satisfies the criterion.
 P: 127 Oops... of course that is the sole matrix with a full set of 0 eigenvalues and $$\lambda_i=0$$ is the sole set of solutions in C satisfying my condition.
P: 2,751
 Quote by Gokul43201 What is the least number than must be multiplied to 100! (that's a factorial) to make it divisible by $12^{49}$ ?
Sorry but the correct answer is the rational number 12^49 / 100! , it's smaller than the previous answer of 6 by a factor or approx 10^104.

Ok so here's my QA puzzle : Why is it that mathematicians are worse than the general layperson when it comes to not specifying that they require a whole number solution when that is the case. :p
 Sci Advisor HW Helper P: 9,396 If you're going to take that attitude, there is no answer; think negatives.


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# A small block with mass 0.250 kg is attached to a string

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## Solution for problem 96P Chapter 10

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Problem 96P

A small block with mass 0.250 kg is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks? Figure:

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Solution 96P Step 1: Given data: Mass of block m = 0.250 kg Radius of circle r = 0.800 m Tangential speed v = 4.00 m/s The breaking strength of the string T = 30.0 N

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Trudy Instituta Matematiki i Mekhaniki UrO RAN
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 Trudy Inst. Mat. i Mekh. UrO RAN, 2015, Volume 21, Number 2, Pages 168–177 (Mi timm1179)

Damping of a system of linear oscillators using the generalized dry friction

A. I. Ovseevicha, A. K. Fedorovab

a Ishlinsky Institute for Problems in Mechanics of the Russian Academy of Sciences, Moscow
b Bauman Moscow State Technical University

Abstract: The problem of damping a system of linear oscillators is considered. The problem is solved by using a control in the form of dry friction. The motion of the system under the control is governed by a system of differential equations with a discontinuous right-hand side. A uniqueness and continuity theorem is proved for the phase flow of this system. Thus, the control in the form of generalized dry friction defines the motion of the system of oscillators uniquely.

Keywords: optimal control, DiPerna-Lions theory, singular ODE.

Full text: PDF file (176 kB)
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English version:
Proceedings of the Steklov Institute of Mathematics (Supplementary issues), 2016, 293, suppl. 1, 156–165

Bibliographic databases:

UDC: 517.977

Citation: A. I. Ovseevich, A. K. Fedorov, “Damping of a system of linear oscillators using the generalized dry friction”, Trudy Inst. Mat. i Mekh. UrO RAN, 21, no. 2, 2015, 168–177; Proc. Steklov Inst. Math. (Suppl.), 293, suppl. 1 (2016), 156–165

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\Bibitem{OvsFed15} \by A.~I.~Ovseevich, A.~K.~Fedorov \paper Damping of a system of linear oscillators using the generalized dry friction \serial Trudy Inst. Mat. i Mekh. UrO RAN \yr 2015 \vol 21 \issue 2 \pages 168--177 \mathnet{http://mi.mathnet.ru/timm1179} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=3408887} \elib{https://elibrary.ru/item.asp?id=23607929} \transl \jour Proc. Steklov Inst. Math. (Suppl.) \yr 2016 \vol 293 \issue , suppl. 1 \pages 156--165 \crossref{https://doi.org/10.1134/S008154381605014X} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000380005200014} \scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-84978524667}


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A Reaction coordinate mapping

# Environmental dynamics, correlations, and the emergence of noncanonical equilibrium states in open quantum systems

## Abstract

Quantum systems are invariably open, evolving under surrounding influences rather than in isolation. Standard open quantum system methods eliminate all information on the environmental state to yield a tractable description of the system dynamics. By incorporating a collective coordinate of the environment into the system Hamiltonian, we circumvent this limitation. Our theory provides straightforward access to important environmental properties that would otherwise be obscured, allowing us to quantify the evolving system-environment correlations. As a direct result, we show that the generation of robust system-environment correlations that persist into equilibrium (heralded also by the emergence of non-Gaussian environmental states) renders the canonical system steady-state almost always incorrect. The resulting equilibrium states deviate markedly from those predicted by standard perturbative techniques and are instead fully characterised by thermal states of the mapped system-collective coordinate Hamiltonian. We outline how noncanonical system states could be investigated experimentally to study deviations from canonical thermodynamics, with direct relevance to molecular and solid-state nanosystems.

## I Introduction

A quantum system coupled to its macroscopic environment constitutes a challenging theoretical problem in which the large number of environmental degrees of freedom can lead to both conceptual and practical difficulties Leggett et al. (1987); Weiss (2008); Nitzan (2006); May and Kühn (2011); Breuer and Petruccione (2002). The master equation formalism has been developed to offer a simple and intuitive approach for describing such systems; the complex dynamical evolution of the many-body environment is not tracked explicitly, but instead only its effect on the reduced state of the system of interest is considered, eliminating all information on the environmental state. In most cases one must also rely on a series of assumptions in deriving a tractable master equation Note1 (). These customarily neglect the formation of system-bath correlations and lead to the eventual thermalisation of the system with respect solely to its internal Hamiltonian, resulting in canonical equilibrium states.

Here, by incorporating a collective coordinate of the environment into an effective system Hamiltonian, we develop a master equation formalism that can overcome such restrictions. This enables us to straightforwardly determine key environmental properties as well as track the dynamic generation of correlations between the system and bath. Specifically, we characterise the departure of the environment from its initial Gaussian thermal state due to interactions with the quantum system, and show that the resulting correlations—measured in terms of the mutual information—can have a profound effect on the system dynamics too, even persisting into the steady-state. We demonstrate that system-bath correlations are in fact generated on two distinct timescales, with long-lived correlations leading to a departure of the system steady-state from canonical equilibrium, as would otherwise be expected from a perturbative (Born-Markov) treatment, thus heralding the failure of the accepted statistical mechanics view of thermalisation Breuer and Petruccione (2002). Correctly capturing system-environment correlations is hence shown to be crucial in order to properly describe both the system transient and equilibrium behaviour. As a further, and unique, aspect of our approach we illustrate how noncanonical equilibrium states can still be characterised in terms of thermal states, but now with respect to the effective system-collective coordinate Hamiltonian. This also reveals simple experimental signatures by which deviations from canonical thermodynamics can be observed in real physical systems, for example through measurements of system populations.

## Ii Reaction coordinate mapping and master equation

Our method relies on keeping track of a collective environmental degree of freedom, and to do so we make use of the reaction coordinate mapping Garg et al. (1985); Thoss et al. (2001); Cao and Voth (1997); Hughes et al. (2009); Hartmann et al. (2000); Roden et al. (2012); Martinazzo et al. (2011); Woods et al. (2014); Imamoğlu (1994); Garraway (1997); we take a quantum system coupled to a bosonic environment and map to a model in which a collective mode of the environment, known as the reaction coordinate (RC), is incorporated within an effective system Hamiltonian. We then treat the residual environment within a full second-order Born-Markov master equation formalism. By comparing the RC master equation to the numerically exact hierarchical equations of motion (HEOM) (Tanimura, 1990; Tanimura and Kubo, 1989) we demonstrate essentially perfect agreement in the dynamics across all timescales (see below). Thus, all important system-bath, and indeed intra-bath Bera et al. (2014a, b), correlations are incorporated into the system-RC Hamiltonian in the regimes we study.

Though our approach may be applied quite generally, we shall focus in this work on a two-level system (TLS) described by the spin-boson Hamiltonian Leggett et al. (1987); Weiss (2008); Nitzan (2006); Breuer and Petruccione (2002); May and Kühn (2011); Makri and Makarov (1995); Prior et al. (2010); Alvermann and Fehske (2009); Wang and Thoss (2008); Anders et al. (2007); Winter et al. (2009); Silbey and Harris (1984); McCutcheon and Nazir (2011); Jang et al. (2008); Nazir (2009) (with ):

 H=ϵ2σz+Δ2σx+σz∑kfk(c†k+ck)+∑kνkc†kck. (1)

Here, () are creation (annihilation) operators for bosonic modes of frequency , which couple to the TLS with strength , and () are TLS Pauli operators defined such that . In the absence of the bath, the TLS splitting is determined by the bias and tunneling . The effect of the system-bath interaction can be completely characterised by introducing the spectral density Leggett et al. (1987), .

We now apply a normal mode transformation to Eq. (1) to incorporate the most important environmental degrees of freedom into a new effective system Hamiltonian. We carry out this procedure by first defining a collective coordinate of the environment Garg et al. (1985), the RC, which couples directly to the TLS, and is in turn coupled to a residual harmonic environment, as shown schematically in the upper panel of Fig 1. This leads to a mapped Hamiltonian of the form:

 HRC =ϵ2σz+Δ2σx+λσz(a†+a)+Ωa†a+∑kωkb†kbk +(a†+a)∑kgk(b†k+bk)+(a†+a)2∑kg2kωk, (2)

where the collective coordinate is defined such that:

 λ(^a†+^a)=∑kfk(c†k+ck), (3)

with coupling and frequency  Thoss et al. (2001); Cao and Voth (1997); Hughes et al. (2009). The residual bath, denoted by creation (annihilation) operators (), now couples only to the RC and is characterised by an effective spectral density, . To describe the action of the residual bath on the RC mode we need to relate this spectral density to the original spin-boson . To do so, we follow Refs. Garg et al. (1985); Hughes et al. (2009); Leggett (1984) and replace the TLS with a classical coordinate moving in a potential . As outlined in Appendix A, by considering the Fourier transformed equations of motion for both before and after the mapping, we may relate to the original .

To give a concrete example we now consider a Drude-Lorentz form commonly used to describe molecular systems Gilmore and McKenzie (2005): , with cut-off frequency and coupling strength . The mapping we exploit is not limited to this particular spectral density Woods et al. (2014), however it does allow us to benchmark our results against data attained from the numerically exact HEOM. By taking the RC spectral density to be , we find that the equivalent spin-boson form is given by in the limit . Hence, we recover the Drude-Lorentz spectral density by identifying , , and choosing such that  Thoss et al. (2001).

Diagonalising the first line of Eq. (II) we can account for the original spin-boson interaction term in Eq. (1) to all orders, which ensures that our formalism remains non-perturbative in the system-bath coupling strength . We thus derive a Born-Markov master equation for the reduced state of the composite TLS and RC, , which captures their exact internal dynamics, while treating the coupling to the residual bath to second order (see Appendix B). This results in the Schrödinger picture master equation Breuer and Petruccione (2002); Hausinger and Grifoni (2010):

 ∂ρ(t)∂t=−i[H0,ρ(t)]−[^A,[^χ,ρ(t)]]+[^A,{^Ξ,ρ(t)}], (4)

where we have assumed that only the residual environment remains in thermal equilibrium throughout the evolution. Here, ,

 ^χ= γ∞∫0dτ∞∫0dω ωcos(ωτ)coth(βω2)^A(−τ), (5) ^Ξ= γ∞∫0dτ∞∫0dωcos(ωτ)[H0,^A(−τ)], (6)

with and . To solve our master equation we truncate the RC space as necessary for convergence.

## Iii Benchmarking

In order to demonstrate the validity of the reaction coordinate master equation (RCME), we benchmark its predictions for the TLS population dynamics () against converged data generated using the numerically exact HEOM technique, see Appendix C and Refs. Tanimura, 1990; Tanimura and Kubo, 1989; Ishizaki and Fleming, 2009. To give an illustrative example, in Fig. 1 we have taken parameters that are typical for excitonic energy transfer in molecular systems Ishizaki and Fleming (2009); Prior et al. (2010); Thorwart et al. (2009); Pollock et al. (2013), with a representative value of  cm setting the other energy scales (i.e.  ps at ). The TLS is initialised in state , uncorrelated with both the RC and residual bath, which are taken to be in their respective thermal equilibrium states at a temperature ( K for  cm). We observe essentially perfect agreement between the RCME and the HEOM for a slow environment and for spin-boson coupling strengths [(a)  cm, (b)  cm] encompassing the transition from the coherent to the incoherent regime Note2 (). In contrast, a standard weak coupling approach, treating the interaction term in Eq. (1) to second order, fails even qualitatively to capture the correct system behaviour for any parameters shown. It is thus clear that environmental memory and the generation of correlations with the system—both of which are ignored in the weak coupling calculation—are crucial in order to capture the correct dynamical behaviour in this regime. Moreover, their combined impact on the TLS can be accurately described simply through the TLS-RC coupling. This is somewhat remarkable, given that any non-thermal state effects of the original bath have been reduced solely to the action of a single mode on the TLS.

## Iv Environmental dynamics and correlations

The most important aspect of our formalism is that the inclusion of environmental degrees of freedom into the system Hamiltonian allows us to gain additional insight into the dynamics of both the environmental state and system-environment correlations. We do this by calculating two complementary measures; the RC non-Gaussianity Genoni et al. (2008); Genoni and Paris (2010), which probes the environmental evolution:

 δG[ρRC(t)]=S(ϱ)−S(ρRC(t)), (7)

and the TLS-RC quantum mutual information (QMI) Groisman et al. (2005), characterising the correlations:

 I(ρS:ρRC)=S(ρs)+S(ρRC)−S(ρ). (8)

Here, () is the reduced state of the RC (TLS) and is the von-Neumann entropy. The non-Gaussianity determines the distance from to the nearest Gaussian reference state , and is defined such that iff is Gaussian. The QMI quantifies the total classical and quantum correlations shared between the TLS and RC  Groisman et al. (2005). As detailed in Appendix D, both measures act as rigorous lower bounds for the original spin-boson environment, which enables us to explore properties of the multi-mode bath and its correlations with the system simply through the single mode RC. Furthermore, in the limit that the Born approximation holds between the mapped system and residual environment, then the additive nature of the von-Neumann entropy implies that both these measures become exact for the original spin-boson environment.

Fig. 2 shows the dynamics of the non-Gaussianity and the QMI at both strong and weak system-environment coupling . One of the most striking features of these plots is the presence of two distinct timescales, which is most obvious for weak couplings, but is also present at stronger coupling strengths. At short times, the QMI is oscillatory, an indication of the memory effects implied by system-environment correlations, which also push the RC away from its initial Gaussian state. At longer times, we see that system-environment correlations—and consequently non-Gaussian environmental states—are also generated on a second timescale, and in fact persist into the steady state, with values of both the non-Gaussianity and QMI dependent on the coupling strength as demonstrated in Fig. 3. We therefore find a situation in which the Born-Markov approximation breaks down on all timescales. It describes neither the short time transient dynamics, due to the absence of bath memory effects in the Markov approximation, nor the long time behaviour, due to the generation of significant TLS-bath correlations. This leads to non-Gaussian (and hence non-thermal) environmental states upon tracing out the TLS, neglected in the Born approximation. Our line of enquiry also raises an intriguing question: if correlations can so dramatically affect the Gaussian nature of the environment, how do they impact upon the state of the TLS?

## V Noncanonical equilibrium states

A clue can already be taken from Fig. 3, where we find that the steady-state properties of both the RC and TLS-RC correlations may be described by a thermal state with respect to , i.e.

 ρth=e−βH0Z=1Ze−β(ϵ2σz+Δ2σx+λσz(a†+a)+Ωa†a), (9)

where . Likewise, in Fig. 4 we show that the equilibrium behaviour of the TLS (after tracing out the mode) clearly departs from the canonical statistics expected from a perturbative, weak coupling treatment of the environmental influence, which would be given instead by a thermal state with respect only to the TLS Hamiltonian, .

Fig. 4(a) explores the temperature dependence of the TLS equilibrium state in the energy eigenbasis. To emphasise the departure from canonical statistics we have plotted the population ratio on a logarithmic scale. For a canonical distribution we expect a linear dependence on the inverse temperature, shown by the dash-dot line, and given by , where is the TLS splitting. Crucially, however, the RCME steady state shows a clear deviation from this linear behaviour with decreasing temperature. To provide context, taking our estimate again of  cm relevant to molecular systems, we see that deviations from canonical statistics begin to become apparent around  K, and should thus be observable even at room temperature in such systems. Note, from the inset, that a non-zero level of coherence is also apparent around such temperatures (and lower), again pointing to the noncanonical nature of the TLS equilibrium state. These departures demonstrate that we cannot represent the TLS steady state as a Gibbs distribution over the system eigenstates. They also quantify relatively simple experimental signatures of the breakdown of canonical statistics in open quantum systems. For example, by measuring only the TLS populations over a reasonable temperature range, we may infer the emergence of noncanonical equilibrium states simply by observing a nonlinear temperature dependence as shown.

Finally, in Fig. 4 (b) we examine the TLS steady state as a function of system-bath coupling strength at constant temperature. We again see significant deviations from the (unchanging) canonical thermal state, such as the development of steady state coherences within the TLS eigenstate basis, which are apparent for any non-vanishing coupling strength. Nevertheless, it is evident from all plots in Fig. 4, by the agreement between Eq. (9) and the HEOM, that the noncanonical steady state is still extremely well described by a thermal state across a wide range of parameters, though now in the mapped TLS-RC representation. This behaviour is a clear result of the TLS-environment correlations; the TLS cannot be considered merely as being in a product state with a (thermal) environment. The non-separability of the steady state thus implies that the equilibrium behaviour of the TLS cannot be described as a thermal distribution over its eigenstates either, but rather, one should also consider states of the environment within the description. This is markedly different to standard master equation techniques and statistical mechanics approaches, where steady states are commonly characterised by the bath temperature and system Hamiltonian alone, an artefact of applying the Born approximation.

## Vi Summary

To summarise, by exploiting a collective coordinate mapping, we have derived a master equation valid in the non-adiabatic regime of the spin-boson model. Notably, besides an accurate description of the system dynamics, our approach also allows us to quantify the accumulation of system-environment correlations with time, as well as probe the dynamic evolution of states of the environment. We have shown that properly accounting for the generation of system-environment correlations is essential for describing both the transient dynamics and equilibrium distributions of open systems in this regime. In particular, we have demonstrated that long-lived correlations lead to the emergence of noncanonical system equilibrium states that can be characterised in a simple and intuitive way within our formalism, as thermal states of the system-collective coordinate Hamiltonian. Our approach can be applied to a number of systems of practical relevance. For example, in molecular and solid-state (e.g. superconducting) devices Leggett et al. (1987); Weiss (2008); Breuer and Petruccione (2002); Le Hur et al. (2007); Le Hur (2012); Würger (1998); Mahan (1990); Goldstein et al. (2013); Nitzan (2006); May and Kühn (2011); Garg et al. (1985); Gilmore and McKenzie (2005); Engel et al. (2007); Lee et al. (2007); Collini and Scholes (2009); Santamore et al. (2013); Franco et al. (2013); Peropadre et al. (2013), system-environment coupling can be strong and memory effects important. Taking parameters relevant for energy transfer processes in molecular dimers, we have shown that deviations from canonical statistics could be observable even at room temperature, which raises the intriguing question of the role of noncanonical equilibrium states in larger molecular aggregates Engel et al. (2007); Lee et al. (2007); Collini and Scholes (2009). Finally, a proper understanding of equilibrium states is a vital component in the growing field exploring the thermodynamics of quantum systems; the full implications of noncanonical steady-states in open quantum systems Lee et al. (2012); Martinez and Paz (2013) thus constitutes a fascinating topic for future exploration.

## Vii Acknowledgements

We would like to thank Alex Chin, Jianshu Cao, Javier Cerrillo, Dara McCutcheon, Tom Stace, and Tommaso Tufarelli for interesting discussions. J. I.-S. is supported by the EPSRC and A. N. by Imperial College and The University of Manchester. N. L. acknowledges the hospitality of the Controlled Quantum Dynamics Group at Imperial College.

## Appendix A Reaction coordinate mapping

As in the main text, we start by considering a two level system (TLS) coupled linearly to a harmonic environment described by the spin-boson (SB) Hamiltonian (with ):

 H=ϵ2σz+Δ2σx+σz∑kfk(c†k+ck)+∑kνkc†kck, (10)

where () are the standard TLS Pauli operators in a basis where , and () are creation (annihilation) operators for bosonic modes of frequency , which couple to the TLS with strength . The system-environment coupling is fully specified by the spin-boson spectral density, .

We now apply a normal mode transformation to Eq. (10), leading to a mapped Hamiltonian of the form Martinazzo et al. (2011)

 HRC =ϵ2σz+Δ2σx+λσz(a†+a)+Ωa†a+∑kωkb†kbk +(a†+a)∑kgk(b†k+bk)+(a†+a)2∑kg2kωk, (11)

where we have defined the collective (reaction) coordinate such that

 λ(a†+a)=∑kfk(c†k+ck). (12)

The TLS-reaction coordinate (RC) coupling strength is given by , such that the RC creation and annihilation operators satisfy the canonical commutation relation . This choice of coupling also fixes the RC frequency to be . The RC is now coupled linearly to a residual harmonic environment characterised by the spectral density . There is also a term quadratic in the system operators in Eq. (A), known as the counter term, which is used to renormalise the mode frequency and avoid divergences due to friction Hartmann et al. (2000).

In order to fully specify the mapping described above, we must relate the RC and SB spectral densities. We do this by following the procedure first outlined by Garg et al. Garg et al. (1985) and derive the spectral density, both before and after the mapping, from the classical equations of motion Leggett (1984). Since the spectral density doesn’t contain any information about the system itself, but rather just the coupling strength between the system and environment, it then follows that in Eq. (10) we can swap the TLS for a continuous classical coordinate moving in a potential . This yields a Hamiltonian of the form

 Hq =P2q2+U(q)+q∑k~fk^xk+q2∑k~f2k2ν2k +12∑k(^p2k+ν2k^x2k), (13)

where, for simplicity, we have written Eq. (A) in the position representation, with coordinate and momentum operators of the environment defined as

 ^xk=√12νk(c†k+ck)%and^pk=i√νk2(c†k−ck), (14)

and the coupling between the TLS and the mode of the environment is given by . From this Hamiltonian, we attain a set of classical equations of motion:

 ¨q(t) =−U′(q)−∑k~fkxk(t)−q(t)∑k~f2kν2k, ¨xk(t) =−~fkq(t)−ν2kxk(t). (15)

We can eliminate the bath variables from the equation of motion for the classical coordinate by making use of the Fourier transform, . This leads to an equation of the form , where the Fourier space operator is defined as

 ~K(z) =−z2⎛⎝1+∑k~f2kν2k(ν2k−z2)⎞⎠ =−z2⎛⎜⎝1+∞∫0dνJSB(ν)ν(ν2−z2)⎞⎟⎠. (16)

We solve the integral using the residue theorem, making use of a single pole at , giving . By writing , it follows that Leggett (1984)

 JSB(ω)=1πlimϵ→0+Im[K(ω−iϵ)]. (17)

We now use the same procedure to write in terms of the RC spectral density. Since the mapping at this stage is exact, then the Fourier transformed operator, , will be identical before and after the normal mode transformation. If we once again swap the TLS for a continuous coordinate, and write the RC Hamiltonian given in Eq. (A) in position space, then we have

 Hq =P2q2+U(q)+κq^x+κ22Ω2q2+12(^p2+Ω2^x2) +^x∑k~gk^Xk+^x2∑k~g2k2ω2k+12∑k(^P2k+ωk^X2k), (18)

where we have scaled the coupling strengths such that and , and the position and momentum operators are defined in the usual way. This Hamiltonian leads to classical equations of motion of the form

 ¨q+κ^x+κ2Ω2q=−U′(q), ¨^x+(Ω2+∑k~g2kω2k)^x+κ2q+∑k~gk^Xk=0, ¨^Xα+~gα^x+ω2α^Xα=0. (19)

By moving to Fourier space and eliminating both the RC and environment from the equation of motion for the classical coordinate, we get the expression for the Fourier space operator

 ~K(z)=−z2−κ2Ω2L(z)Ω2+L(z),

where, by using the definition of the RC spectral density, we have

 L(z) =−z2(1+∑k~g2kω2k(ω2k−z2)) =−z2⎛⎜⎝1+4Ω∞∫0JRC(ω)ω(ω2−z2)dω⎞⎟⎠. (20)

Considering the RC to have an Ohmic spectral density of the form , the integral reduces to , in the limit . By plugging this into Eq. (17), we get the condition

 JSB(ω) =1πlimϵ→0+Im[~K(ω−iϵ)] =1π2πΩγκ2ω(Ω2−ω2)2+(2πΩγ)2ω2. (21)

In order for Eq. (A) to be consistent with the original spin-boson spectral density, we identify the relations

 ωc= Ω2πγ, andα= 2λ2πΩ, (22)

which give us

 JSB(ω)=αωcωω2c+ω4ω2cΩ2+ω2. (23)

Finally, by assuming , we get the spin-boson spectral density in Lorentz-Drude form,

 JSB(ω)=αωcωω2+ω2c. (24)

## Appendix B Reaction coordinate master equation

We now wish to use Eq. (A) to derive a master equation that treats the TLS-RC coupling exactly (up to some number of basis states) while the coupling to the residual environment is treated to second order (within a Born-Markov approximation). To this end, we define the interaction Hamiltonian as

 HI=^A⊗^B+~λ ^A2, (25)

where , and . To derive the RC master equation we first move into the interaction picture using the transformation , where , is the TLS-RC Hamiltonian, and . Tracing out the residual environment, we then find a master equation for the reduced TLS-RC density operator , which in the interaction picture may be written as

 ∂~ρ(t)∂t =−i trB[~HI(t),ρ(0)⊗ρB] −t∫0dτ trB[~HI(t),[~HI(t−τ),~ρ(τ)⊗ρB]], =−i ~λ[^A2(t),ρ(0)] −~λ2t∫0dτ[^A2(t),[^A2(t−τ),~ρ(τ)]] −t∫0dτ([^A(t),[^A(t−τ),~ρ(t)]]Γ+(τ) Missing or unrecognized delimiter for \right (27)

Here, we have made use of the Born approximation, that is, we have assumed that the system and residual environment remain in the product state for all time, where . The correlation functions are defined as . In the continuum limit we may write them as

 Γ+(τ)=∞∫0dωJRC(ω)cothβω2cosωτ (28)

and

 Γ−(τ)=i∞∫0dωJRC(ω)sinωτ. (29)

As outlined in the previous section, the RC spectral density takes the form , in the limit that that the cut-off frequency . We can simplify Eq. (B) further by noticing that

 −i~λ[^A2(t),~ρ(t)] =−i~λ[^A2(t),~ρ(0)] −~λ2t∫0dτ[^A2(t),[^A2(t−τ),~ρ(τ)]]. (30)

By substituting this into the above master equation and assuming a Markov limit, that is, we take the time integrals to infinity, we acquire the RC master equation

 ∂~ρ(t)∂t =−i~λ[^A2(t),~ρ(t)] −∞∫0∞∫0dτdωJRC(ω)cothβω2cosωτ ×[^A(t),[^A(t−τ),~ρ(t)]] −i∞∫0∞∫0dτdωJRC(ω)sinωτ ×[^A(t),{^A(t−τ),~ρ(t)}]. (31)

The definition of assumes an infinite cut-off frequency, hence the first and last terms of Eq. (B) are divergent. However, we can eliminate the divergent contributions by integrating the last term by parts, such that

 ∞∫0dτsinωτ^A(t−τ) =−P(^A(t)ω) +∞∫0dτcosωτω∂^A(t−τ)∂τ. (32)

Using the principal value part of this integral we can cancel the counter term, which, after moving back into the Schrödinger picture, gives the master equation

 ∂ρ(t)∂t =−i[H0,ρ(t)] −∞∫0∞∫0dτdωJRC(ω)cosωτcothβω2 ×[^A,[^A(−τ),ρ(t)]] −∞∫0∞∫0dτdωJRC(ω)cosωτω ×[^A,{[^A(−τ),H0],ρ(t)}]. (33)

In order to derive the interaction picture system operators we shall diagonalise the TLS-RC system Hamiltonian, , numerically. Let be an eigenstate of the system Hamiltonian, such that We can now write the position operators in this eigenbasis:

 ^A=∑jkAj,k∣∣φj⟩⟨φk|, (34)

where . In the interaction picture this becomes

 ^A(t)=∑jkAj,keiξjkt∣∣φj⟩⟨φk|, (35)

where is the difference between the and eigenvalues. Using this definition we can include the rates from Eq. (B) into the operators, such that

 ^χ =∞∫0∞∫0dωdτ JRC(ω)cosωτcothβω2^A(−τ) ≈π2∑jkJRC(ξjk)cothβξjk2Ajk∣∣φj⟩⟨φk|, (36) ^Ξ =∞∫0∞∫0dωdτJRC(ω)cosωτω[H0,^A(−τ)] ≈π2∑jkJRC(ξjk)Ajk∣∣φj⟩⟨φk|, (37)

where we have neglected the imaginary Lamb shift terms. We can now write the master equation as

 ∂ρ(t)∂t=−i[H0,ρ(t)]−[^A,[^χ,ρ(t)]]+[^A,{^Ξ,ρ(t)}]. (38)

## Appendix C Heom

The hierarchical equations of motion (HEOM) are a set of time-local equations for the reduced system dynamics, governed by the spin-boson Hamiltonian [Eq. (10)], which capture the bath dynamics and system-bath correlations through a set of auxiliary density matrices. These equations are exact under the assumption of a Lorentz-Drude spectral density, as given in Eq. (24), and an initially separable system-bath state at . Here we assume only a single bath such that the HEOM can be written as

 ˙ρn = −(i[HS,ρn]+K∑m=0nmμm)ρn−iK∑m=0[Q,ρn+m] (39) − iK∑m=0nm(cmQρn−m−c∗mρn−mQ) − Missing or unrecognized delimiter for \right

where is the system-environment coupling operator, , and . The bath correlation functions for the Lorentz-Drude spectral density are

 C=∞∑m=0cmexp(−μmt). (40)

Here , , and the coefficients are

 c0=ωcαH2[cot(βωc/2)−i]/ℏ (41)

and

 cm≥1=2αHωcβμmμ2m−ω2c. (42)

Each density matrix is labelled by an index of positive integers . As here we only have a single bath the integers are defined as . For each “Matsubara” term each index runs from to . The null label defines the system density matrix, and any non-zero label refers to an auxiliary density matrix which encodes the correlations with the bath. The terms in the equation of motion refer to an increase or decrease by of the label index . We take a cut-off in the overall tier and in the Matsubara terms which give convergence in the numerical results.

## Appendix D Quantum mutual information and non-Gaussianity

In this Appendix we show that the mutual information for the TLS and RC acts as a lower bound to the correlations shared between the system and the original multi-mode environment.

Let be the density matrix describing the state of both the system and environment in the original spin-boson representation. We also define the reduced states of the system, , and the environment, . Let be the unitary transformation that maps the spin-boson model to the RC model. Applying this unitary to the reduced state of the system has no effect due to the trace over the environment. However, on the reduced state of the environment this unitary transforms the spin-boson basis to that of the RC and residual bath, that is, .

The quantum mutual information for the system and the spin-boson environment is given by:

 I(ρs:ρE)=S(ρs)+S(ρE)−S(χ),=S(ρs)+S(ρRC+B)−S(~χ), (43)

where is the von-Neumann entropy and is the total state in the RC basis. Here we have used the unitary equivalence of the von-Neumann entropy to write the quantum mutual information in terms of the RC basis.

To proceed we shall make use of the strong subadditivity of the von-Neumann entropy, that is:

 S(~χ)+S(ρRC)≤S(ρS+RC)+S(ρRC+B), (44)

where , with the trace taken over the residual environment. Using this property in conjunction with Eq. (43) gives:

 I(ρs:ρE) ≥S(ρS)+S(ρRC+B)+S(ρRC)−S(ρs+RC) −S(ρRC+B), ≥S(ρs)+S(ρRC)−S(ρS+RC). (45)

Therefore, we have the condition

 I(ρs:ρE)≥I(ρs:ρRC). (46)

Hence, the mutual information between the system and RC acts as a lower bound to the mutual information between the system and spin-boson environment. Furthermore, in the limit that the Born approximation holds between the composite system (TLS and RC) and the residual environment, that is , then the inequality in Eq. (46) becomes an equality due to the additive nature of the von-Neumann entropy.

Similarly, the non-Gaussianity can be shown to be invariant under symplectic transformation (i.e. operators that are quadratic in field operators) and monotonically decreases under partial trace Genoni et al. (2008). This means that the non-Gaussianity of the RC acts as a rigorous lower bound for the non-Gaussianity of the original spin-boson environment, that is:

 δG[ρE]≥δG[ρRC]. (47)

### References

1. A. J. Leggett, S. Chakravarty, A. T. Dorsey, M. Fisher, A. Garg,  and W. Zwerger, Rev. Mod. Phys. 59, 1 (1987).
2. U. Weiss, Quantum Dissipative Systems, 3rd ed. (World Scientific, Singapore, 2008).
3. A. Nitzan, Chemical Dynamics in Condensed Phases: Relaxation, Transfer and Reactions in Condensed Molecular Systems (Oxford University Press, Oxford, 2006).
4. V. May and O. Kühn, Charge and Energy Transfer Dynamics in Molecular Systems, 3rd ed. (Wiley, Berlin, 2011).
5. H. P. Breuer and F. Petruccione, The Theory of Open Quantum Systems (Oxford University Press, Oxford, 2002).
6. The most common being the Born-Markov approximations, wherein the bath is assumed to be static, memoryless, and uncorrelated with the system, remaining in thermal equilibrium throughout the dynamical evolution Breuer and Petruccione (2002).
7. A. Garg, J. N. Onuchic,  and V. Ambegaokar, J. Chem. Phys. 83, 4491 (1985).
8. M. Thoss, H. Wang,  and W. H. Miller, J. Chem. Phys. 115, 2991 (2001).
9. J. Cao and G. A. Voth, J. Chem. Phys. 106, 1769 (1997).
10. K. H. Hughes, C. D. Christ,  and I. Burghardt, J. Chem. Phys. 131, 124108 (2009).
11. L. Hartmann, I. Goychuk,  and P. Hänggi, J. Chem. Phys. 113, 11159 (2000).
12. J. Roden, W. T. Strunz, K. B. Whaley,  and A. Eisfeld, The Journal of Chemical Physics 137, 204110 (2012).
13. R. Martinazzo, B. Vacchini, K. H. Hughes,  and I. Burghardt, J. Chem. Phys. 134, 011101 (2011).
14. M. P. Woods, R. Groux, A. W. Chin, S. F. Huelga,  and M. B. Plenio, Journal of Mathematical Physics 55, 032101 (2014).
15. A. Imamoğlu, Phys. Rev. B 50, 3650 (1994).
16. B. M. Garraway, Phys. Rev. B 55, 2290 (1997).
17. Y. Tanimura, Phys. Rev. A. 41, 6676 (1990).
18. Y. Tanimura and R. Kubo, J. Phys. Soc. Jpn. 58, 101 (1989).
19. S. Bera, S. Florens, H. U. Baranger, N. Roch, A. Nazir,  and A. W. Chin, Phys. Rev. B 89, 121108(R) (2014a).
20. S. Bera, A. Nazir, A. W. Chin, H. U. Baranger,  and S. Florens, Phys. Rev. B 90, 075110 (2014b).
21. N. Makri and D. E. Makarov, J. Chem. Phys. 102, 4600 (1995).
22. J. Prior, A. W. Chin, S. F. Huelga,  and M. B. Plenio, Phys. Rev. Lett. 105, 050404 (2010).
23. A. Alvermann and H. Fehske, Phys. Rev. Lett. 102, 150601 (2009).
24. H. Wang and M. Thoss, New J. Phys. 10, 115005 (2008).
25. F. B. Anders, R. Bulla,  and M. Vojta, Phys. Rev. Lett. 98, 210402 (2007).
26. A. Winter, H. Rieger, M. Vojta,  and R. Bulla, Phys. Rev. Lett. 102, 030601 (2009).
27. R. Silbey and R. A. Harris, J. Chem. Phys. 80, 2615 (1984).
28. D. P. S. McCutcheon and A. Nazir, J. Chem. Phys. 135, 114501 (2011).
29. S. Jang, Y.-C. Cheng, D. R. Reichman,  and J. D. Eaves, J. Chem. Phys. 129, 101104 (2008).
30. A. Nazir, Phys. Rev. Lett. 103, 146404 (2009).
31. A. J. Leggett, Phys. Rev. B 30, 1208 (1984).
32. J. Gilmore and R. H. McKenzie, J. Phys.: Condens. Matter 17, 1735 (2005).
33. J. Hausinger and M. Grifoni, Phys. Rev. A 81, 022117 (2010).
34. A. Ishizaki and G. R. Fleming, J. Chem. Phys. 130, 234111 (2009).
35. M. Thorwart, J. Eckel, J. Reina, P. Nalbach,  and S. Weiss, Chem. Phys. Lett. 478, 234 (2009).
36. F. A. Pollock, D. P. S. McCutcheon, B. W. Lovett, E. M. Gauger,  and A. Nazir, New J. Phys. 15, 075018 (2013).
37. We see similarly excellent agreement for the coherences and for as well.
38. M. G. Genoni, M. G. A. Paris,  and K. Banaszek, Phys. Rev. A 78, 060303 (2008).
39. M. G. Genoni and M. G. A. Paris, Phys. Rev. A 82, 052341 (2010).
40. B. Groisman, S. Popescu,  and A. Winter, Phys. Rev. A 72, 032317 (2005).
41. K. Le Hur, P. Doucet-Beaupre,  and W. Hofstetter, Phys. Rev. Lett. 99, 126801 (2007).
42. K. Le Hur, Phys. Rev. B 85, 140506 (2012).
43. A. Würger, Phys. Rev. B. 57, 347 (1998).
44. G. D. Mahan, Many-Particle Physics (Plenum, New York, 1990).
45. M. Goldstein, M. H. Devoret, M. Houzet,  and L. I. Glazman, Phys. Rev. Lett. 110, 017002 (2013).
46. G. Engel, T. Calhoun, E. Read, T. Ahn, T. Mančal, Y. Cheng, R. Blankenship,  and G. Fleming, Nature 446, 782 (2007).
47. H. Lee, Y.-C. Cheng,  and G. R. Fleming, Science 316, 1462 (2007).
48. E. Collini and G. D. Scholes, Science 323, 369 (2009).
49. D. H. Santamore, N. Lambert,  and F. Nori, Phys. Rev. B 87, 075422 (2013).
50. R. L. Franco, B. Bellomo, S. Maniscalco,  and G. Compagno, International Journal of Modern Physics B 27, 1345053 (2013).
51. B. Peropadre, D. Zueco, D. Porras,  and J. J. García-Ripoll, Phys. Rev. Lett. 111, 243602 (2013).
52. C. K. Lee, J. Cao,  and J. Gong, Phys. Rev. E 86, 021109 (2012).
53. E. A. Martinez and J. P. Paz, Phys. Rev. Lett. 110, 130406 (2013).
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text:
# Self adjoint operators, eigenfunctions & eigenvalues

## Homework Statement

Consider the space ##P_n = \text{Span}\{ e^{ik\theta};k=0,\pm 1, \dots , \pm n\}##,
with the hermitian ##L^2##-inner product
##\langle f,g\rangle = \int_{-\pi}^\pi f(\theta) \overline{g(\theta)}d\theta##.
Define operators ##A,B,C,D## as
##A = \frac{d}{d\theta}, \; \; B= i\frac{d}{d\theta}, \; \; C= \frac{d^2}{d\theta^2}, \; \; D: f\to D f(\theta) = f(\theta) + f(-\theta)##.
Which of the operators are self-adjoint? Find the eigenvalues and eigenfunctions for each operator.

## Homework Equations

The operator ##T## is self adjoint iff
##\langle Tf,g \rangle = \langle f,Tg\rangle## for all ##f,g## in the ##P_n##.

## The Attempt at a Solution

Mostly looking for some quick input if I'm getting this roughly right or not since there's no answer and I'm a bit unsure.
I first write ##f(\theta) = \sum_{-n}^n c_k e^{ik\theta}##. We're allowed to differentiate since ##f(\theta)## is in ##C^\infty##. Differentiating we have ##f'(\theta) = \sum_{-n}^n ikc_k e^{ik\theta}## and ##f''(\theta ) = \sum_{-n}^n -k^2c_k e^{ik\theta}##. every part of ##f(\theta )## is ##2\pi##-periodic so the only time we get a contribution to the integral is when the the exponential terms cancel each other exactly.
A) In this case when the index matches so we have##\langle Af, g\rangle = \int_{-\pi}^\pi \sum_{-n}^n \left( ika_kb_k \right) d\theta = 2\pi \sum_{-n}^n ika_k\overline{b_k}##.
and ##\langle f, Ag\rangle = \int_{-\pi}^\pi \left( \sum_{-n}^n -ika_k\overline{b_k} \right) d\theta = -2\pi \sum_{-n}^n ika_k\overline{b_k}## so not self-adjoint.
Similary we see that both B and C are self-adjoint.

The eigenvalues would be the functions ##e^{ikx}## for ##k = \pm 1, \pm 2,\dots , \pm n## and the eigenvalues for for ##A,B## and ##C## are ##ik, -k## and ##-k^2## respectively.

The case ##D## is slightly harder we have
##\langle Df,g \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(-\theta)\overline{g(\theta)}d\theta## and
##\langle f,Dg \rangle = \langle f,g \rangle + \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##
So we need to prove that ##\int_{-\pi}^\pi f(-\theta)\overline{g(\theta)} = \int_{-\pi}^\pi f(\theta)\overline{g(-\theta)}d\theta##. Again matching the coefficients we have for the left side
##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}} \right) d\theta##
and the right side
##\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{n-k}}\right) d\theta## which matches so self adjoint. The eigenvalues should be ##\frac{a_k+a_{n-k}}{a_k}## and the eigenfunctions the same as before. Did I understand this right?

## Answers and Replies

andrewkirk
Science Advisor
Homework Helper
Gold Member
Your work looks mostly sound.
You have omitted the case ##k=0##, which gives an additional eigenvalue of 0 for A, B, C, with the eigenvector being any constant.

For D, I also conclude it is Hermitian but I get the integrals being like this

$$\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{-k}} \right) d\theta$$

The eigenvalues won't be of the form you gave because the ##a_k## coefficients are properties of the function ##f##, not the operator D.

Given the revised integral, can you work out what eigenvalues and eigenfunctions of D will be?

Incand
Your work looks mostly sound.
For D, I also conclude it is Hermitian but I get the integrals being like this

$$\int_{-\pi}^\pi \left( \sum_{-n}^n a_k\overline{b_{-k}} \right) d\theta$$

The eigenvalues won't be of the form you gave because the ##a_k## coefficients are properties of the function ##f##, not the operator D.

Given the revised integral, can you work out what eigenvalues and eigenfunctions of D will be?
Thanks for taking the time replying! You're absolutely right about the index in that sum and thanks for clarifying that I can't have those constants from the function as an eigenvalue.

If I write out ##Df(\theta)## I have ##\sum_{-N}^N \left(c_k e^{ik\theta} + c_ke^{-ik\theta} \right) = \sum_{-N}^N (c_k + c_{-k})e^{ik\theta}##
I'm not sure I get all eigenfunctions here but I was thinking if I take eigenfunctions of the form ##f\in P_n## with the added constraint that ##c_k+c_{-k}=0## I have the eigenvalue zero?

I also possibly see other eigenvalues For example
if I choose the constrant that ##c_{-k} = ac_{k}## with ##a## being a complex number I have the eigenvalue ##(1+a)##. But not sure if I'm allowed too do this? The eigenvalue doesn't vary like in the last case(which was obviously wrong) but I do use properties from the function I guess.

andrewkirk
Science Advisor
Homework Helper
Gold Member
I was thinking if I take eigenfunctions of the form ##f\in P_n## with the added constraint that ##c_k+c_{-k}=0## I have the eigenvalue zero?
That sounds right. What would be a neat basis for the eigenspace with eigenvalue zero?

I also possibly see other eigenvalues For example
if I choose the constrant that ##c_{-k} = ac_{k}## with ##a## being a complex number I have the eigenvalue ##(1+a)##.
To see if this works, apply D to ##f(\theta)=e^{-ik\theta}+ae^{ik\theta}## and see what you get. Is it a complex scalar multiple of ##f(\theta)##?

What about other nonzero eigenvalues?

Incand
That sounds right. What would be a neat basis for the eigenspace with eigenvalue zero?
Perhaps ##\{e^{ik\theta}-e^{-ik\theta} \}_0^n ## that is the basis ##\{ 1, e^{i\theta}-e^{-i\theta}, \dots , e^{in\theta}-e^{-in\theta} \}##.
The basis vector aint orthogonal nor normalized but if one wants we could get there with Gram-Schmidts method.

To see if this works, apply D to ##f(\theta)=e^{-ik\theta}+ae^{ik\theta}## and see what you get. Is it a complex scalar multiple of ##f(\theta)##?

What about other nonzero eigenvalues?
##D\left( e^{-ik\theta}+ae^{ik\theta} \right) = e^{-ik\theta} + ae^{ik\theta} + e^{ik\theta} + e^{-ik\theta} = (1+a) \left(e^{-ik\theta}+e^{ik\theta} \right)##. Right this doesn't work since I should have a ##a## in there as well.

So instead let's look at the case when ##c_{-k} = c_{k}## we have
##D(e^{ik\theta}+ e^{-ik\theta}) =e^{ik\theta}+ e^{-ik\theta} + e^{-ik\theta}+ e^{ik\theta} = 2(e^{ik\theta}+ e^{-ik\theta})##. So we have the eigenvalue two for the eigenfunctions as a linear combination of ##\{e^{ik\theta}+e^{-ik\theta} \}_0^n##. I don't think there's actually any more eigenvalues.


text:
# On Zaremba's Conjecture on Continued Fractions

## On Zaremba's Conjecture on Continued Fractions - Jean Bourgain

Jean Bourgain
Institute for Advanced Study
February 14, 2012
Zaremba's 1971 conjecture predicts that every integer appears as the denominator of a finite continued fraction whose partial quotients are bounded by an absolute constant. We confirm this conjecture for a set of density one.


text:
From: Robert Stam <robert@tallcomponents.com>
Date: Wed, 27 Jan 2010 09:44:16 -0600
Message-ID: <31DBA4F3A35EEE408177B6D1E50A4C5002307E1C@34093-C9-EVS1.exchange.rackspace.com>

Hi all,

We have the following questions about the Microsoft testset
(available at: http://samples.msdn.microsoft.com/ietestcenter/css.htm)

---------
css\chapter_16\Rules\text-align-white-space-008.htm
css\chapter_16\Rules text-align-white-space-003.htm

We consider both test as incorrect tests: only pre and pre-line (not
pre-wrap) cause text-align value 'justify' to be replaced by its initial
value.

---------
css\chapter_16\white-space-control-characters-001.htm

#22 is the Unicode character for 'Synchronous Idle', but no font was
found that displays it as mentioned in the test description.

---------
css\chapter_16\word-spacing-remove-space-002.htm

Looks like IE8 specific behavior.

---------
css\chapter_17\rules\empty-cells-exceptions-004.htm

Whitespace is preserved in the table, so whitespace between <table> and
<tr> is considered non-empty content and is wrapped in an anonymous
table-row and an anonymous table-cell.

---------
css\chapter_17\rules\table-height-algorithm-005.htm

CSS 2.1 specification does not specify how extra table height space
should be distributed over the rows. Since a cell-height percentage
computes to auto, all rows will get an equal part of the remaining
space. A more sophisticated algorithm could inspect cell and row
percentage heights and try to satisfy them.

---------
css\chapter_18\properties\outline-width\outline-width-095.htm

We consider this test as invalid test: In div1, the computed value of
outline-width is 0, because outline-style is none (the default value).
The 'outline-width: inherit' inherits this computed value, so the width
of the outline is zero.

---------
css\chapter_6\Rules\HTML-attribute-019.htm

The DOCTYPE of this test is XHTML 1.0 Transitional in which there's no
baseline space reserved below an image. This causes the images to
overlap slightly. Changing the DOCTYPE to strict will results in the
correct output..

---------
css\chapter_8\properties\border-width\border-width-005.htm
css\chapter_8\properties\border-width\border-width-005.htm

Incorrect test: the computed value of property border-width at element
'body' is 0, because border-style is 'none'. The value 'inherit'
inherits the computed value, so at element div.test the border-width is
also 0 and no rectangle is shown. (Chrome, FireFox and Opera render this
correctly, by displaying only one square).

---------
---------

Best Regards,
Robert Stam

TallComponents


text:
# Using Systemd as a Socket Server for a Shell Command

In the past I’ve run socat as a daemon to run a shell command and send the output to a socket. It works well, but I wanted to try the systemd way of doing what “xinetd” has done in the past: Run a command and send the output to a socket.

A quick word on my use case: I’m using headscale, but at the moment we are in a transition phase and I will be having much of the use by users going to our old firewalls that previously were VPN concentrators. Previously we had different VPNs, and hence different networks, for Operations and Development. With headscale I’m using ACLs to control access, but I need a way to tell the firewalls what IPs belong to what groups of users.

I wrote a script that converts headscale ACL groups and node IPs into “ipset” commands. So the firewall can check “is the source headscale-ops” and pass it to the existing “operations allowed” rules. But to get this script to the firewalls, I decided to set up a port that the firewalls can connect to and get the rules.

It took me quite a while to get the exact right set of configs. In this example I called my service “headscale-ipset”.

You need a “/etc/systemd/system/headscale-ipset.socket” file:

[Unit]
Description=Socket server to display ipset for headscale groups

[Socket]
ListenStream=8080
Accept=yes

[Install]
WantedBy=sockets.target

And a “/etc/systemd/system/headscale-ipset@.service” file:

[Unit]
Description=Headscale IPSet Socket Server
After=network.target

[Service]
Type=simple
Restart=always
RestartSec=1
ExecStart=/usr/bin/cat /usr/local/lib/10-headscale-rules
StandardOutput=socket

[Install]
WantedBy=multi-user.target

After adding these files, run a “systemctl daemon-reload” and “systemctl start headscale-ipset.socket” and you should then be able to telnet to port 8080 and have the contents of “/usr/local/lib/10-headscale-rules” spit out.


text:
# $||f||_1 =(\int_a^b [|f|^2+|f'|^2]dx)^{1/2}$. Is this normed space complete?

Define $C_1^1[a,b]$ to be the space of continuously differentiable functions on $[a,b]$, with norm

$$||f||_1 =\left(\int_a^b \left(|f|^2+|f'|^2\right) dx \right)^{1/2}$$

Is this normed space complete?

So far I have shown that this is a normed space by satisfying the four axioms of a normed space. Now I need to show that it is complete. That is, show that every Cauchy sequence in this space is convergent. I'm stuck on how to do this part. Any hints or solutions are greatly appreciated.

• Try showing that it is not complete. Since the space is not complete under that norm, that has higher chances of success. – Daniel Fischer Jun 16 '15 at 21:08
• hmmm ok, so a counter-example – 1233dfv Jun 16 '15 at 21:21
• And it would be sporting to note that this is the $H^1$ ($L^2$) Sobolev norm, so the completion is denoted/called the Sobolev space $H^1[a,b]$. Not to recommend googling rather than finding your own counter-example, but for follow-up later. – paul garrett Jun 16 '15 at 21:23
• I can't figure out a counter-example. Any recommendations or links? – 1233dfv Jun 16 '15 at 21:57
• Hint: concentrate on the derivative. You want that to have a discontinuous $L^2$-limit. – Daniel Fischer Jun 16 '15 at 23:40

Hint

Pointwise limit of differentiable functions needn't be differentiable.

• I think that my answer is a hint to find a counterexample, and this is precisely my intention. A pointwisely convergent sequence of functions of $C_1^1[a,b]$ will be a Cauchy sequence with this norm, but if the limit function is not differentable, the sequence will not converge in $C_1^1[a,b]$. – ajotatxe Jun 16 '15 at 23:59
• Pointwise convergence does not imply $L^2$-convergence, much less $L^2$-convergence of the derivatives. – Daniel Fischer Jun 17 '15 at 8:57

I was working on a similar problem. Here is a way of showing that $C^1[a,b]$ is not a Banach space. We will assume $a=0$ and $b=1$; the method behind the proof below will motivate the general idea. Let $\{f_j\}_{j\geqslant 1}\subset C^1[0,1]$ be a sequence, defined as follows.

$$f_j(t)=\left\{ \begin{array}{ll} \frac{1}{2(j+1)}(2t)^{(j+1)}, ~~0\leqslant t<\frac{1}{2}\\ (t-\frac{1}{2})+\frac{1}{2(j+1)}, ~~\frac{1}{2}\leqslant t\leqslant 1 \end{array} \right.$$

It's easy to check that $f_j$ is differentiable on $(0,1)$ and continuous on $[0,1]$. Now we claim that $\{f_j\}_{j\geqslant 1}$ is Cauchy.

Proof. Check that for any $m,n\in\mathbb{N}$ (assuming $m>n$),

\begin{eqnarray} d(f_m,f_n) & = & \Vert f_m-f_n\Vert \\ & = & \left(\int_0^1(f_m-f_n)^2+[(f_m-f_n)']^2dt\right)^{\frac{1}{2}} \\ & = & \left(\int_0^{1/2}(f_m-f_n)^2+[(f_m-f_n)']^2dt+\int_{1/2}^1(f_m-f_n)^2+[(f_m-f_n)']^2dt\right)^{\frac{1}{2}}. \end{eqnarray}

The two integrals in the squareroot are easily bounded above by $\left[\frac{1}{2(n+1)}\right]^2+\left(\frac{1}{2}\right)^{2n+1}$, which approaches zero. Hence, for any $\varepsilon>0$ picked, there exists $N\in\mathbb{N}$ such that for all $m,n\geqslant N$ the expression above is strictly bounded above by $\varepsilon$. $\blacksquare$

We claim that $\{f_j\}_{j\geqslant 1}$ does not converge to a member of $C^1[0,1]$. What is $f=\lim_{j\rightarrow\infty}f_j$? It is given by

$$f(t)=\left\{ \begin{array}{ll} \lim_{j\to\infty}\frac{1}{2(j+1)}(2t)^{(j+1)}, ~~0\leqslant t<\frac{1}{2}\\ \lim_{j\to\infty}(t-\frac{1}{2})+\frac{1}{2(j+1)}, ~~\frac{1}{2}\leqslant t\leqslant 1 \end{array} \right.$$

In other words, it's the function $f(t)=0$ when $0\leqslant t<\frac{1}{2}$ and $f(t)=t-\frac{1}{2}$ when $\frac{1}{2}\leqslant t\leqslant 1$. It's straightforward to check that $f$ isn't differentiable on $(0,1)$.

• Your argument is true, but you use the pointwise limit to calculate the limit function $f$. Is that correct? Shouldn't it be the convergence in norm, that is, a function $f$ will be the limit of the sequence considered when $\Vert f_j-f\Vert\rightarrow 0$ and not when $f=\lim f_j$. – Senna Mar 7 at 12:24


text:
# [−][src]Struct plexus::graph::FaceView

pub struct FaceView<M, G> where    M: Reborrow,    M::Target: AsStorage<Face<G>>,    G: GraphGeometry,  { /* fields omitted */ }

View of a face in a graph.

Provides traversals, queries, and mutations related to faces in a graph. See the module documentation for more information about topological views.

Faces are notated by the path of their associated ring. A triangular face with a perimeter formed by vertices $A$, $B$, and $C$ is notated $\overrightarrow{\{A,B,C\}}$. While the precise ordering of vertices is determined by a face's leading arc, the same face may be notated using rotations of this set, such as $\overrightarrow{\{B,C,A\}}$.

## Methods

### impl<M, G> FaceView<M, G> where    M: Reborrow,    M::Target: AsStorage<Face<G>>,    G: GraphGeometry, [src]

#### pub fn key(&self) -> FaceKey[src]

Gets the key for the face.

### impl<'a, M, G> FaceView<&'a mut M, G> where    M: 'a + AsStorageMut<Face<G>>,    G: 'a + GraphGeometry, [src]

#### pub fn into_orphan(self) -> FaceOrphan<'a, G>[src]

Converts a mutable view into an orphan view.

#### pub fn into_ref(self) -> FaceView<&'a M, G>[src]

Converts a mutable view into an immutable view.

This is useful when mutations are not (or no longer) needed and mutual access is desired.

# Examples

use decorum::N64;
use nalgebra::Point3;
use plexus::graph::MeshGraph;
use plexus::prelude::*;
use plexus::primitive::cube::Cube;
use plexus::primitive::generate::Position;

let mut graph = Cube::new()
.polygons::<Position<Point3<N64>>>()
.collect::<MeshGraph<Point3<f64>>>();
let key = graph.faces().nth(0).unwrap().key();
let face = graph
.face_mut(key)
.unwrap()
.extrude(1.0)
.unwrap()
.into_ref();

// This would not be possible without conversion into an immutable view.
let _ = face.into_arc();
let _ = face.into_arc().into_next_arc();

### impl<M, G> FaceView<M, G> where    M: Reborrow,    M::Target: AsStorage<Arc<G>> + AsStorage<Face<G>> + Consistent,    G: GraphGeometry, [src]

#### pub fn into_ring(self) -> RingView<M, G>[src]

Converts the face into its ring.

#### pub fn into_arc(self) -> ArcView<M, G>[src]

Converts the face into its leading arc.

#### pub fn ring(&self) -> RingView<&M::Target, G>[src]

Gets the ring of the face.

#### pub fn arc(&self) -> ArcView<&M::Target, G>[src]

Gets the leading arc of the face.

#### pub fn traverse_by_breadth(    &self) -> impl Clone + Iterator<Item = FaceView<&M::Target, G>>[src]

Gets an iterator that traverses the faces of the graph in breadth-first order beginning with the face on which this function is called.

The traversal moves from the face to its neighboring faces and so on. If there are disjoint faces in the graph, then a traversal will not reach all faces in the graph.

#### pub fn traverse_by_depth(    &self) -> impl Clone + Iterator<Item = FaceView<&M::Target, G>>[src]

Gets an iterator that traverses the faces of the graph in depth-first order beginning with the face on which this function is called.

The traversal moves from the face to its neighboring faces and so on. If there are disjoint faces in the graph, then a traversal will not reach all faces in the graph.

#### pub fn interior_arcs(    &self) -> impl Clone + Iterator<Item = ArcView<&M::Target, G>>[src]

Gets an iterator of views over the arcs in the face's ring.

#### pub fn neighboring_faces(    &self) -> impl Clone + Iterator<Item = FaceView<&M::Target, G>>[src]

Gets an iterator of views over neighboring faces.

### impl<M, G> FaceView<M, G> where    M: Reborrow,    M::Target: AsStorage<Arc<G>> + AsStorage<Face<G>> + AsStorage<Vertex<G>> + Consistent,    G: GraphGeometry, [src]

#### pub fn distance(    &self,     source: Selector<VertexKey>,     destination: Selector<VertexKey>) -> Result<usize, GraphError>[src]

Gets the distance (number of arcs) between two vertices within the face.

#### pub fn vertices(    &self) -> impl Clone + Iterator<Item = VertexView<&M::Target, G>>[src]

Gets an iterator of views over the vertices that form the face.

### impl<M, G> FaceView<M, G> where    M: ReborrowMut,    M::Target: AsStorageMut<Arc<G>> + AsStorage<Face<G>> + Consistent,    G: GraphGeometry, [src]

#### pub fn interior_arc_orphans(&mut self) -> impl Iterator<Item = ArcOrphan<G>>[src]

Gets an iterator of orphan views over the arcs in the face's ring.

### impl<M, G> FaceView<M, G> where    M: ReborrowMut,    M::Target: AsStorage<Arc<G>> + AsStorageMut<Face<G>> + Consistent,    G: GraphGeometry, [src]

#### pub fn neighboring_face_orphans(    &mut self) -> impl Iterator<Item = FaceOrphan<G>>[src]

Gets an iterator of orphan views over neighboring faces.

### impl<M, G> FaceView<M, G> where    M: ReborrowMut,    M::Target: AsStorage<Arc<G>> + AsStorage<Face<G>> + AsStorageMut<Vertex<G>> + Consistent,    G: GraphGeometry, [src]

#### pub fn vertex_orphans(&mut self) -> impl Iterator<Item = VertexOrphan<G>>[src]

Gets an iterator of orphan views over the vertices that form the face.

#### pub fn flatten(&mut self) -> Result<(), GraphError> where    G: FacePlane,    G::Vertex: AsPosition,    VertexPosition<G>: EuclideanSpace + FiniteDimensional<N = U3>, [src]

Flattens the face by translating the positions of all vertices into a best-fit plane.

Returns an error if a best-fit plane could not be computed or positions could not be translated into the plane.

### impl<'a, M, G> FaceView<&'a mut M, G> where    M: AsStorage<Arc<G>> + AsStorage<Edge<G>> + AsStorage<Face<G>> + AsStorage<Vertex<G>> + Default + Mutable<G>,    G: 'a + GraphGeometry, [src]

#### pub fn split(    self,     source: Selector<VertexKey>,     destination: Selector<VertexKey>) -> Result<ArcView<&'a mut M, G>, GraphError>[src]

Splits the face by bisecting it with a composite edge inserted between two non-neighboring vertices within the face's perimeter.

The vertices can be chosen by key or index, where index selects the $n^\text{th}$ vertex within the face's ring.

This can be thought of as the opposite of merge.

Returns the inserted arc that spans from the source vertex to the destination vertex if successful. If a face $\overrightarrow{\{A,B, C,D\}}$ is split from $A$ to $C$, then it will be decomposed into faces in the rings $\overrightarrow{\{A,B,C\}}$ and $\overrightarrow{\{C,D,A\}}$ and the arc $\overrightarrow{AC}$ will be returned.

# Errors

Returns an error if either of the given vertices cannot be found, are not within the face's perimeter, or the distance between the vertices along the ring is less than two.

# Examples

use nalgebra::Point2;
use plexus::graph::MeshGraph;
use plexus::prelude::*;
use plexus::primitive::Tetragon;

let mut graph = MeshGraph::<Point2<f64>>::from_raw_buffers(
vec![Tetragon::new(0usize, 1, 2, 3)],
vec![(0.0, 0.0), (1.0, 0.0), (1.0, 1.0), (0.0, 1.0)],
)
.unwrap();
let key = graph.faces().nth(0).unwrap().key();
let arc = graph
.face_mut(key)
.unwrap()
.split(ByIndex(0), ByIndex(2))
.unwrap()
.into_ref();

#### pub fn merge(self, destination: Selector<FaceKey>) -> Result<Self, GraphError>[src]

Merges the face into a neighboring face over their shared composite edge.

The neighboring face can be chosen by key or index, where index selects the $n^\text{th}$ neighbor of the face.

This can be thought of as the opposite of split.

Returns the merged face if successful.

# Errors

Returns an error if the destination face cannot be found or is not a neighbor of the initiating face.

# Examples

use nalgebra::Point2;
use plexus::graph::MeshGraph;
use plexus::prelude::*;
use plexus::primitive::Tetragon;

let mut graph = MeshGraph::<Point2<f64>>::from_raw_buffers(
vec![Tetragon::new(0usize, 1, 2, 3), Tetragon::new(0, 3, 4, 5)],
vec![
(0.0, 0.0),  // 0
(1.0, 0.0),  // 1
(1.0, 1.0),  // 2
(0.0, 1.0),  // 3
(-1.0, 1.0), // 4
(-1.0, 0.0), // 5
],
)
.unwrap();

let key = graph.faces().nth(0).unwrap().key();
let face = graph
.face_mut(key)
.unwrap()
.merge(ByIndex(0))
.unwrap()
.into_ref();

#### pub fn bridge(self, destination: FaceKey) -> Result<(), GraphError>[src]

Connects faces with equal arity with faces inserted along their perimeters.

The inserted faces are always quadrilateral. Both the initiating face and destination face are removed.

# Errors

Returns an error if the destination face cannot be found or the arity of the face and its destination are not the same.

#### pub fn triangulate(self) -> Self[src]

Decomposes the face into triangles. Does nothing if the face is triangular.

Returns the terminating face of the decomposition.

#### pub fn poke_with<F>(self, f: F) -> VertexView<&'a mut M, G> where    F: FnOnce() -> G::Vertex, [src]

Subdivides the face about a vertex. A triangle fan is formed from each arc in the face's perimeter and the vertex.

Poking inserts a new vertex with geometry provided by the given function.

Returns the inserted vertex.

# Examples

Forming a pyramid from a triangular face:

use nalgebra::Point3;
use plexus::graph::MeshGraph;
use plexus::prelude::*;
use plexus::primitive::Trigon;
use plexus::AsPosition;

let mut graph = MeshGraph::<Point3<f64>>::from_raw_buffers(
vec![Trigon::new(0usize, 1, 2)],
vec![(-1.0, 0.0, 0.0), (1.0, 0.0, 0.0), (0.0, 2.0, 0.0)],
)
.unwrap();
let key = graph.faces().nth(0).unwrap().key();
let mut face = graph.face_mut(key).unwrap();

// See poke_with_offset, which provides this functionality.
let mut geometry = face.centroid();
let position = geometry.as_position().clone() + face.normal().unwrap();
face.poke_with(move || {
*geometry.as_position_mut() = position;
geometry
});

#### pub fn poke_at_centroid(self) -> VertexView<&'a mut M, G> where    G: FaceCentroid,    G::Vertex: AsPosition, [src]

Subdivides the face about its centroid. A triangle fan is formed from each arc in the face's perimeter and a vertex inserted at the centroid.

Returns the inserted vertex.

#### pub fn poke_with_offset<T>(    self,     offset: T) -> Result<VertexView<&'a mut M, G>, GraphError> where    T: Into<Scalar<VertexPosition<G>>>,    G: FaceCentroid + FaceNormal,    G::Vertex: AsPosition,    VertexPosition<G>: EuclideanSpace, [src]

Subdivides the face about its centroid. A triangle fan is formed from each arc in the face's perimeter and a vertex inserted at the centroid. The inserted vertex is then translated along the initiating face's normal by the given offset.

Returns the inserted vertex if successful.

# Errors

Returns an error if the geometry could not be computed.

# Examples

Constructing a "spikey" sphere:

use decorum::N64;
use nalgebra::Point3;
use plexus::graph::MeshGraph;
use plexus::prelude::*;
use plexus::primitive::generate::Position;
use plexus::primitive::sphere::UvSphere;

let mut graph = UvSphere::new(16, 8)
.polygons::<Position<Point3<N64>>>()
.collect::<MeshGraph<Point3<f64>>>();
let keys = graph.faces().map(|face| face.key()).collect::<Vec<_>>();
for key in keys {
graph.face_mut(key).unwrap().poke_with_offset(0.5).unwrap();
}

#### pub fn extrude<T>(self, offset: T) -> Result<FaceView<&'a mut M, G>, GraphError> where    T: Into<Scalar<VertexPosition<G>>>,    G: FaceNormal,    G::Vertex: AsPosition,    VertexPosition<G>: EuclideanSpace, [src]

Extrudes the face.

Returns the extruded face if successful.

# Errors

Returns an error if the geometry could not be computed.

#### pub fn remove(self) -> Option<RingView<&'a mut M, G>>[src]

Removes the face.

Returns the remaining ring of the face if it is not entirely disjoint.

## Trait Implementations

### impl<M, G> Deref for FaceView<M, G> where    M: Reborrow,    M::Target: AsStorage<Face<G>>,    G: GraphGeometry, [src]

#### type Target = Face<G>

The resulting type after dereferencing.

### impl<M, G> DynamicArity for FaceView<M, G> where    M: Reborrow,    M::Target: AsStorage<Arc<G>> + AsStorage<Face<G>> + Consistent,    G: GraphGeometry, [src]

#### fn arity(&self) -> Self::Dynamic[src]

Gets the arity of the face. This is the number of arcs that form the face's ring.

## Blanket Implementations

### impl<T> Same<T> for T

#### type Output = T

Should always be Self

### impl<T> ToOwned for T where    T: Clone, [src]

#### type Owned = T

The resulting type after obtaining ownership.

### impl<T, U> TryFrom<U> for T where    U: Into<T>, [src]

#### type Error = Infallible

The type returned in the event of a conversion error.

### impl<T, U> TryInto<U> for T where    U: TryFrom<T>, [src]

#### type Error = <U as TryFrom<T>>::Error

The type returned in the event of a conversion error.


text:
A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

I need a clarification on how to use binomial distribution formula in this problem. What i got was:

$$x = 55, \quad \mu = 50, \quad SD = 5$$ $$z = (55-50)/5 \implies z = 0.84 \implies 1- z = 0.16$$ $$p(x) = 0.16$$

The exact answer is obtained from the Binomial distribution. Various pieces of software, including Wolfram Alpha, will do the computation for you. This approach is painful to carry out by hand.

Perhaps you are expected to use the normal approximation to the binomial. If $X$ is the number of heads, then $X$ has approximately normal distribution, mean $50$, standard deviation $\sqrt{100(1/2)(1/2)}=5$.

We want to find the probability that $X\ge 56$. We find, approximately, the probability of the complement, the probability that $X\le 55$. This, approximately, is the probability that a normal with mean $50$ and standard deviation $5$ is $\le 55$.

But perhaps you are expected to use the continuity correction. Then $$\Pr(X\le 55)\approx\Pr\left(Z\le \frac{55.5-50}{5}\right)\tag{1}$$ where $Z$ is standard normal. If you are not expected to use the continuity correction, use $55$ instead of $55.5$.

In either case, the answer to the original question is, approximately, $1$ minus the number obtained in (1).

• so what i got for Less than and equal to 55 is 0.84, for 56 above would be 1-0.84= 0.16? to obtain more than 55 heads? – Kit lai Aug 16 '14 at 4:15
• Yes, if you do not use the continuity correction, it is $\approx 1-0.8413$. If you use the correction, it is $\approx 1-0.8643$. Note how close this gets us to the exact answer. – André Nicolas Aug 16 '14 at 4:18
• thanks , i Kinda got what you were you trying to say – Kit lai Aug 16 '14 at 4:27
• You are welcome. – André Nicolas Aug 16 '14 at 4:27
• @AndréNicolas I just want make sure. We can use Chernoff Bound instead and we get $1/e^{1/6}$ – YOUSEFY May 25 '17 at 11:26

As lab bhattacharjee mentioned, the probability of exactly $k$ heads is

$\binom{n}{k} p^k (1-p)^{n-k}$

Let us write $X$ to denote the number of heads, and $Q$ to denote the desired probability, $P(X > 55)$. Note that by symmetry we have $P(X > 55) = P(X < 45)$. Also,

$P(X > 55) + P(X < 45) + P(45 \leq X \leq 55) = 1$

so $2Q + P(45 \leq X \leq 55) = 1$. Therefore the problem reduces to evaluating $Z = P(45 \leq X \leq 55)$.

We can use another symmetry argument to evaluate $Z$ by observing that $P(45 \leq X < 50) = P(50 < X \leq 55)$. This means that we will only need to calculate $\binom{n}{k} p^k (1-p)^{n-k}$ for $50 \leq k \leq 55$, substantially reducing the workload.

• so p is the probability for obtaining H or T which is 1/2? n = 100 trials, k = 55 heads ? – Kit lai Aug 16 '14 at 4:33
• Yes, that's right, p = 1/2, n = 100. k = number of heads. – Bungo Aug 16 '14 at 4:44

HINT:

From this,

the probability of getting exactly $k$ successes in $n$ trials is $$\binom nk p^k(1-p)^{n-k}$$

Here $\displaystyle p=\dfrac12\implies p^k(1-p)^{n-k}=\left(\frac12\right)^n, n=100$

and $\displaystyle\sum_{k>55}^{100}\binom nk=\frac{(1+1)^{100}}2-\sum_{k=50}^{55}\binom nk$

• my answer was 0.0048 – Kit lai Aug 16 '14 at 3:46
• ^That's a bit off. See this – JimmyK4542 Aug 16 '14 at 3:48
• @JimmyK4542 i dont really get why was (Sum[Binomial[100, k]/2^100, {k, 56, 100}]) 2^100 where did the 2 come from or from which formula? – Kit lai Aug 16 '14 at 3:58
• @Kitlai, $$\sum_{k=0}^n\binom nk=(1+1)^n$$ and $$\binom nk=\binom n{n-k}\implies\sum_{k=0}^{49}\binom nk=\sum_{k=51}^{100}\binom nk$$ – lab bhattacharjee Aug 16 '14 at 4:01


text:
# Qanat! {Time limit: 2 seconds}

Computer Science Level pending

A qanat is an irrigation system widely used to deliver water in hot, arid climates. The technology was originally developed by Persians over 2000 years ago. In Morocco, qanats are known as khettara and are still used today in the southern part of the country.

The basic feature of a qanat is an essentially horizontal channel that brings water from an underground water source to an outlet near a civilization. There is also a shaft known as a mother well that rises vertically from the underground water source to the surface of a mountain or hill. Creating such a system is extremely expensive, and was especially so in ancient times, since all of the materials excavated from the channel and mother well must be carried above ground, either through the channel outlet or the top of the mother well. To aid in the construction, there are often one or more additional vertical shafts placed at strategic locations above the underground channel. Although these shafts must also be excavated, they provide a means for lifting additional dirt from the horizontal channel as illustrated in Figure H.1.

For this problem, model the cross-section of a qanat as shown in Figure H.2, with the channel outlet at (0; 0), the water source at $$(\omega; 0)$$, and the top of the mother well at $$(\omega; h)$$ with $$\omega> h$$. The surface of the mountain extends along a straight line from $$(\omega; h)$$ to $$(0;0)$$.

Every qanat must have a vertical mother well from the water source to the mountain surface above, along with n additional vertical shafts. The channel and all shafts are modeled as line segments. Your goal is to determine the placement for those additional shafts so as to minimize the overall excavation cost. This cost is equal to the sum of the distances that each piece of excavated dirt must be transported to reach the surface (using any combination of horizontal and vertical movement). For example, the cost of excavating a continuous section of dirt starting from the surface and going along a path of length ` (possibly including turns) is $$∫_0^l xdx=l^2 /2$$.

Input The input consists of a single line containing three integers $$\omega (1  ,\omega,  10 000), h (1,  h < w)$$, and n (1  n  1 000). The value w is the horizontal distance from the water source to the qanat outlet. The value h is the vertical distance from the water source to the mountain surface. The value n is the number of vertical shafts that must be used in addition to the mother well.

Output First, display the minimum overall excavation cost. Next, display the x-coordinates, in increasing order, for n optimally placed vertical shafts. If n > 10, display only the first 10 x-coordinates. Answers within an absolute or relative error of 10􀀀4 will be accepted. You may assume that there is a unique solution. No test case will result in a shaft within 0:001 units from the outlet of the qanat channel or from another shaft.

if

What is $$⌊(A-B+C-D+E)*10^4⌋$$?

this problem is a last ICPC problem.

×


text:
Volume 105 - The XXVIII International Symposium on Lattice Field Theory (Lattice 2010) - CHIRAL SYMMETRY
Analysis of the $N_f=2+1$ lattice QCD results on the lowest-lying baryon masses using covariant ChPT
J. Martin Camalich*, L. Geng and M.J. Vicente Vacas
Full text: pdf
Published on: June 06, 2011
DOI: https://doi.org/10.22323/1.105.0089
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access
Copyright owned by the author(s) under the term of the Creative Commons Attribution-NonCommercial-ShareAlike.


text:
# Which of the following terms means suturing of the rectum? a. proctorrhaphy b. rectostenosis c....

## Question:

Which of the following terms means suturing of the rectum?

a. proctorrhaphy

b. rectostenosis

c. rectoplasty

d. sigmoid surgery

## The Rectum

The rectum is a portion of the gastrointestinal tract that occurs between the sigmoid colon and the anus. The rectum generally measure up to 15 cm in length in the human body.


text:
# What are critical differences between conventional lasers and matter lasers?

In what ways would a coherent stream of matter waves/De Broglie wavelength interacting with the environment differ from a conventional coherent stream of photons? What have been observed and what have been proposed and/or theorized?

Here is a link to what I mean, as it seems it is not completely clear what I am talking about: https://en.wikipedia.org/wiki/Atom_laser

• Do you mean the free electron laser en.wikipedia.org/wiki/Free-electron_laser ? That system emits electromagnetic radiation and the differences are seen in the link. Jun 12 at 4:59
• Just updated the question to a link to what I am talking about. Jun 12 at 7:57
• I think the wiki site emphasizes the similarities and differences all through. Jun 12 at 8:53

Physical: There is a huge difference in particle flux. A BEC will typically contain around $$10^6$$ particles. This can be limited by the number of atoms loaded into the initial magneto-optical trap and the efficiency of the evaporative cooling. The number of photons in a laser depends on the power. For example, in our lab we use a 1W laser with a wavelength of 780nm, this gives a photon flux of $$\approx 4\times10^{18}$$s$$^{-1}$$. The propagation speed is also a difference, obviously photons in a laser travel at the speed of light. In contrast, the speed record for a BEC (from what I can find) is 28mm s$$^{-1}$$. Another difference could be the achievable coherence times.
Applications: There are a number of applications for both lasers and BECs, leaving out the obvious differences (BECs can't burn holes through solid objects) and focusing on an application that they have in similar, interferometry. There is much research and development in implementing interferometers that use BECs or matterwaves. This is due to the massive increase in sensitivity that can be achieved when using atoms instead of lasers with optical frequencies. This increase in sensitivity comes from comparing the energy of a photon at a frequency $$f$$ and the rest mass energy of an atom of mass $$m$$. For example, this ratio for a photon at an optical frequency and an atom of Rubidium comes out to be $$\approx10^{10}$$. However, the possible increase in sensitivity also depends on the particle flux, so this ratio is significantly reduced in practice.


text:
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text:
## 2008年12月6日 星期六

### A Solution to the "Ten-Minute" Problem

It should be easy to observe that z must be odd. We are looking for positive integer solution to

$4024x+4178y=3757057-4609z$

By manually doing division algorithm once, you can show that $gcd(4024,4178)=2$. Simultaneously, you can find a particular integer solution to $4024x'+4178y'=2$, which is

$x'=-624$
$y'=601$

Now, set $2a=3757057-4609z$ Then we have

$x =-624a-2089K$
$y =601a+2012K$

as general solution. To keep x and y both positive, we may have a "sense" that when $-624a(\mod 2089)$ is small, we can adjust K to keep x small but positive, and hence "allow room" for y to be positive.

Take $z=1$, we have $a=1876224$. However, $-624a\equiv 1651(\mod 2089)$. The remainder 1651 is too large to be x.

Now observe that when z is increased by 2, a drops by 4609. Notice $-624\times 4609\equiv 537(\mod 2089)$. That means when we increase z by 2, $-624a(\mod 2089)$ drops by 537.

Increase z from 1 to 7, $-624a(\mod 2089)$ drops from 1651 to $1651-3\times 537=40$. Setting $x=40$, we get $y=853$. A possible solution is given by

$x = 40$
$y = 853$
$z = 7$


text:
## Algebra 1

First, you must turn the problem into an equation. The problem is asking for a number times 0.75 (or 75%) that will equal 45. turning that into an actual equation would look like this: $x\times0.75=45$ then solve the equation and divide both sides by 0.75 $x=60$