text:
# naginterfaces.library.rand.dist_f¶

naginterfaces.library.rand.dist_f(n, df1, df2, statecomm)[source]

dist_f generates a vector of pseudorandom numbers taken from an (or Fisher’s variance ratio) distribution with and degrees of freedom.

For full information please refer to the NAG Library document for g05sh

https://www.nag.com/numeric/nl/nagdoc_27.1/flhtml/g05/g05shf.html

Parameters
nint

, the number of pseudorandom numbers to be generated.

df1int

, the number of degrees of freedom of the distribution.

df2int

, the number of degrees of freedom of the distribution.

statecommdict, RNG communication object, modified in place

RNG communication structure.

This argument must have been initialized by a prior call to init_repeat() or init_nonrepeat().

Returns
xfloat, ndarray, shape

The pseudorandom numbers from the specified -distribution.

Raises
NagValueError
(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, .

Constraint: .

(errno )

On entry, [‘state’] vector has been corrupted or not initialized.

Notes

The distribution has PDF (probability density function)

dist_f calculates the values

where and are generated by dist_gamma() from gamma distributions with parameters and respectively (i.e., from -distributions with and degrees of freedom).

One of the initialization functions init_repeat() (for a repeatable sequence if computed sequentially) or init_nonrepeat() (for a non-repeatable sequence) must be called prior to the first call to dist_f.

References

Knuth, D E, 1981, The Art of Computer Programming (Volume 2), (2nd Edition), Addison–Wesley

format: plain


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# Use the box plots comparing the number of males and number of females attending the latest superhero

Use the box plots comparing the number of males and number of females attending the latest superhero movie each day for a month to answer the questions. two box plots shown. the top one is labeled males. minimum at 1, q1 at 3, median at 10.5, q3 at 14, maximum at 21, and a point at 30. the bottom box plot is labeled females. minimum at 0, q1 at 15, median at 18, q3 at 21, no maximum shown
part a: estimate the iqr for the males' data. (2 points)

part b: estimate the difference between the median values of each data set. (2 points)

part c: describe the distribution of the data and if the mean or median would be a better measure of center for each. (4 points)

part d: provide a possible reason for the outlier in the data set. (2 points)

## This Post Has 7 Comments

1. dsbdkbv2772 says:

(A) The IQR for the males' data is 11.

(B) The difference between the median values of each data set = 18 - 10.5 = 7.5.

(C) The distribution of the males' data is an asymmetrical distribution or we can say the right-skewed distribution and the median would be a better measure of the center.

The distribution of the females' data is a symmetrical distribution or we can say the normal distribution and the mean would be a better measure of the center.

(D) In the males' data, we see that there is only one outlier at a value of 30.

Step-by-step explanation:

We are given the following information about the two box plots of male and female below;

The top one is labeled Males: Minimum at 1, Q1 at 3, median at 10.5, Q3 at 14, maximum at 21, and a point at 30.

The bottom box plot is labeled as Females: Minimum at 0, Q1 at 15, median at 18, Q3 at 21, no maximum is shown.

(A) The Interquartile range (IQR) is the difference between the third quartile and the first quartile of the data set.

In a box plot representing the males;

$Q_3$ = 14 and $Q_1$ = 3

So, IQR = $Q_3$ - $Q_1$ = 14 - 3 = 11.

Hence, the IQR for the males' data is 11.

(B) The median value of the males' data is given as 10.5 and the median value of the females' data is given as 18.

So, the difference between the median values of each data set = 18 - 10.5 = 7.5.

(C) The distribution of the males' data is an asymmetrical distribution or we can say the right-skewed distribution because the values after the third quartile are very far and also the lower quartile range is much larger than the upper quartile range. Also, there is one outlier in males' data also.

So, for any asymmetrical distribution, the median would be a better measure of the center as it does not take into account the outliers' value and gives the middlemost data value.

The distribution of the females' data is a symmetrical distribution or we can say the normal distribution because the data values are equally spread and the median is at the center of the two quartile values.

So, for any symmetrical distribution, although the mean and median are the same; the mean would be a better measure of the center as it does take into account all the data values and gives the average of the data set.

(D) Yes, there is one outlier in the males' data set. Outliers are those values that fall very far from all other values.

In a box plot data, outliers are represented as single points outside the whiskers.

In the males' data, we see that there is only one outlier at a value of 30. This represents a case in which a male attended the movie theater on the 30th day of the month. This is an invalid case since the maximum of the male plot is at 21, which means that no males attended the movie after the 21st.

2. xMABRYx1991 says:

A) Interquartile range - For males: 11, for femals: 6

B) Difference between the median values: 7.5

C) Males: median, females: mean

D) Outlier:  one male attended the move on the 30th

Step-by-step explanation:

A)

The interquartile range (IQR) of a set of data is the difference between the upper quartile (Q3) and the lower quartile (Q1):

- For the box plot representing males:

Lower quartile is

Upper quartile is

Therefore, interquartile range is

- For the box plot representing femals:

Lower quartile is

Upper quartile is

Therefore, interquartile range is

B)

The median value of a dataset is the "central value" of the dataset, i.e. the value of the dataset for which half of the values in the dataset are lower than the median, and half of the values in the dataset are higher than the median.

The median value of a dataset is indicated as .

Here the median values for the two datasets are given:

- For males,

- For females,

Therefore, the difference between the median values is

C)

The mean of a dataset is the average value of the dataset, calculated as the sum of all data divided by the number of values in the dataset.

On the other hand, the median of a dataset is the central value, i.e. the value for which half of the values in the dataset are lower than the median, and half of the values in the dataset are higher than the median.

For a perfectly symmetrical distribution, mean and median are equal; however, for non-symmetrical distribution, this is not true.

In general, the mean is a good to describe a distribution when the distribution itself is simmetrical; instead, when the distribution is very asymmetrical, the median provides a better indicator for the distribution.

In this case, we can see that the male distribution is very asymmetrical: in fact, the lower quartile range is much larger than the upper quartile range , so the distribution can be better describes using the median. For the female case instead, the distribution is more symmetrical (the median is at the center between and ), therefore the distribution can be better described by using the mean.

4)

An outlier of a dataset is a value of the dataset that falls very far from all other values.

In a box plot, outliers are represented as single points outside the whiskers.

In this case, we see that there is only one outlier, in the male plot, at a value of 30. This represents a single case in which a male attended the movie theater on the 30th. This is an isolated case, since the maximum of the male plot is at 21, which means that no males attended the movie after the 21st.

#LearnwithBrainly

3. iamsecond235p318rq says:

A) Interquartile range - For males: 11, for femals: 6

B) Difference between the median values: 7.5

C) Males: median, females: mean

D) Outlier:  one male attended the move on the 30th

Step-by-step explanation:

A) The interquartile range (IQR) of a set of data is the difference between the upper quartile (Q3) and the lower quartile (Q1):

IQR = Q3 - Q1

- For the box plot representing males:

Lower quartile is Q1=3

Upper quartile is Q3 = 14

Therefore, interquartile range is

IQR = 14 - 3 = 11

- For the box plot representing females:

Lower quartile is Q1 = 15

Upper quartile is Q3 = 21

Therefore, interquartile range is

IQR = 21 - 15 = 6

B) The median value of a dataset is the "central value" of the dataset, i.e. the value of the dataset for which half of the values in the dataset are lower than the median, and half of the values in the dataset are higher than the median.

The median value of a dataset is indicated as Q2.

Here the median values for the two datasets are given:

- For males, Q2 = 10.5

- For females, Q2 = 18

Therefore, the difference between the median values is

Delta Q2 = 18-10.5 = 7.5

C) The mean of a dataset is the average value of the dataset, calculated as the sum of all data divided by the number of values in the dataset.

On the other hand, the median of a dataset is the central value, i.e. the value for which half of the values in the dataset are lower than the median, and half of the values in the dataset are higher than the median.

For a perfectly symmetrical distribution, mean and median are equal; however, for non-symmetrical distribution, this is not true.

In general, the mean is a good to describe a distribution when the distribution itself is symmetrical; instead, when the distribution is very asymmetrical, the median provides a better indicator for the distribution.

In this case, we can see that the male distribution is very asymmetrical: in fact, the lower quartile range (Q2 - Q1) is much larger than the upper quartile range (Q3 - Q2) , so the distribution can be better describes using the median. For the female case instead, the distribution is more symmetrical (the median is at the center between Q_1 and Q_3), therefore the distribution can be better described by using the mean.

4) An outlier of a dataset is a value of the dataset that falls very far from all other values.

In a box plot, outliers are represented as single points outside the whiskers.

In this case, we see that there is only one outlier, in the male plot, at a value of 30. This represents a single case in which a male attended the movie theater on the 30th. This is an isolated case, since the maximum of the male plot is at 21, which means that no males attended the movie after the 21st.

4. eileentennyson says:

Hey There!!

The answer to this is: A) Interquartile range - For males: 11, for femals: 6

B) Difference between the median values: 7.5

C) Males: median, females: mean

D) Outlier:  one male attended the move on the 30th A) The interquartile range (IQR) of a set of data is the difference between the upper quartile (Q3) and the lower quartile (Q1):

IQR = Q3 - Q1

- For the box plot representing males:

Lower quartile is Q1=3

Upper quartile is Q3 = 14

Therefore, interquartile range is

IQR = 14 - 3 = 11

- For the box plot representing females:

Lower quartile is Q1 = 15

Upper quartile is Q3 = 21

Therefore, interquartile range is

IQR = 21 - 15 = 6

B) The median value of a dataset is the "central value" of the dataset, i.e. the value of the dataset for which half of the values in the dataset are lower than the median, and half of the values in the dataset are higher than the median.

The median value of a dataset is indicated as Q2.

Here the median values for the two datasets are given:

- For males, Q2 = 10.5

- For females, Q2 = 18

Therefore, the difference between the median values is

Delta Q2 = 18-10.5 = 7.5

C) The mean of a dataset is the average value of the dataset, calculated as the sum of all data divided by the number of values in the dataset.

On the other hand, the median of a dataset is the central value, i.e. the value for which half of the values in the dataset are lower than the median, and half of the values in the dataset are higher than the median.

For a perfectly symmetrical distribution, mean and median are equal; however, for non-symmetrical distribution, this is not true.

In general, the mean is a good to describe a distribution when the distribution itself is symmetrical; instead, when the distribution is very asymmetrical, the median provides a better indicator for the distribution.

In this case, we can see that the male distribution is very asymmetrical: in fact, the lower quartile range (Q2 - Q1) is much larger than the upper quartile range (Q3 - Q2) , so the distribution can be better describes using the median. For the female case instead, the distribution is more symmetrical (the median is at the center between Q_1 and Q_3), therefore the distribution can be better described by using the mean.

4) An outlier of a dataset is a value of the dataset that falls very far from all other values.

In a box plot, outliers are represented as single points outside the whiskers.

In this case, we see that there is only one outlier, in the male plot, at a value of 30. This represents a single case in which a male attended the movie theater on the 30th. This is an isolated case, since the maximum of the male plot is at 21, which means that no males attended the movie after the 21st.

Hope It Helps!~

$ItsNobody$~

5. clinton1790 says:

c

Step-by-step explanation:

$Use the box plots comparing the number of males and number of females attending the latest superhero$

6. Thomas7785 says:

I good at this and it’s what kind of a girl and you do it is a fake one that I love it so I don’t have a good day tommarow and you do not know how much I am going to do that is

7. PayeA264 says:

yessir

Step-by-step explanation:

i need points


text:
Feedback controlled heat transport in quantum devices: Theory and solid state experimental proposal

# Feedback controlled heat transport in quantum devices: Theory and solid state experimental proposal

Michele Campisi NEST, Scuola Normale Superiore & Istituto Nanoscienze-CNR, I-56126 Pisa, Italy    Jukka Pekola Low Temperature Laboratory, Department of Applied Physics, Aalto University School of Science, 00076 AALTO, Finland    Rosario Fazio ICTP, Strada Costiera 11, Trieste 34151, Italy NEST, Scuola Normale Superiore & Istituto Nanoscienze-CNR, I-56126 Pisa, Italy
###### Abstract

A theory of feedback controlled heat transport in quantum systems is presented. It is based on modelling heat engines as driven multipartite systems subject to projective quantum measurements and measurement-conditioned unitary evolutions. The theory unifies various results presented in the previous literature. Feedback control breaks time reversal invariance. This in turn results in the fluctuation relation not being obeyed. Its restoration occurs by an appropriate accounting of the information gain and information use via measurements and feedback. We further illustrate an experimental proposal for the realisation of a Maxwell demon using superconducting circuits and single photon on-chip calorimetry. A two level qubit acts as a trapdoor which, conditioned on its state is coupled to either a hot resistor or a cold one. The feedback mechanism alters the temperatures felt by the qubit and can result in an effective inversion of temperature gradient, where heat flows from cold to hot thanks to information gain and use.

## 1 Introduction

In a famous thought experiment Maxwell envisioned a method for apparently defying the second law of thermodynamics by means of a feedback control mechanism [1]. Maxwell’s idea is based on a malicious demon, an intelligent being that is able to observe the microscopic dynamics of a system, and acts on it so as to steer it toward defying the second law. In one of Maxwell’s original concepts, the system is a container with two chambers, containing respectively a hot gas and a cold gas. The two chambers are separated by a wall presenting a trap-door which the demon can open and close at will. The demon observes the erratic motion of the gas particle and when sees a particle of the cold chamber approach the trap-door with sufficiently high velocity, she/he swiftly opens the door as to let the particle go through and closes it immediately afterwards. In this way, particle after particle, heat flows from the cold chamber to the hot chamber in contradiction with the second law.

Advance in nanotechnology has made the possibility of bringing Maxwell demons and similar devices from the realm of thought experiments to the realm of real experiments [2, 3, 4, 5]. Both theoretical and experimental studies so far have focused mainly on situations where feedback control is operated as a measurement-conditioned driving on some working substance (classical or quantum) coupled to a single temperature, so as to withdraw energy from the latter in contradiction with the second law as formulated by Kelvin. Interesting realistic proposals have appeared in Refs. [6, 7]. Situations where heat flows between different temperature reservoirs is controlled, however have not been addressed so far, neither theoretically nor experimentally. The main motivation of the present work is that of filling that gap. In the following we shall present the general theory of feedback controlled heat transport in quantum devices, and shall describe a possible experimental realisation thereof.

The theory presented here builds on previous works concerning fluctuation relations in presence of measurements without feedback [8, 9] and with feedback [10], combined with an inclusive approach where quantum heat engines are seen as mechanically driven multipartite systems starting in a multi-temperature initial state [11, 12, 13, 14]. Reference [10] reported on the theory of a one-measurement based feedback control on a quantum working-substance prepared by contact with a single bath. That formalism is here extended to the case of many heat baths, and also repeated measurements, to allow for the study of continuous feedback control of heat flow in a multi reservoir scenario. Previous work concerning repeated measurements appeared in Refs [15] for classical systems in contact with a single bath. Fluctuation relations need to be modified by a mutual information term, which we shall explicitely provide.

Our experimental proposal is based on the fast developing advancements in experimental solid state low temperature techniques: in particular the calorimetric measurement scheme that has been put forward by one of us and co-workers [16, 17]. As proven by some recent theoretical proposals [13, 18] the method opens up a new avenue for the practical management of heat and work on a chip by means of superconducting devices, particularly superconducting qubits. Here we illustrate the possible implementation of very simple feedback controlled heat transport where the trapdoor is realised by a superconducting qubit whose coupling with two resistors at different temperatures is controlled based on the outcomes of continuous calorimetric monitoring of the resistors themselves.

## 2 Theory

Following [14] we model a generic heat transport/heat engine scenario as a driven multi-partite system starting in the factorised state, see Figure 1

 ρ0=⨂le−βlHlZl (1)

where is the Hamiltonian of each partition including a heat bath and possibly a portion of the working substance, and is the corresponding partition function [14]. Let the total Hamiltonian be

 H(t)=∑lHl+V(t) (2)

where is an interaction term that is switched on for the time interval over which the system is monitored. We assume that at times some observable is measured thus causing the wave function describing the compound to collapse onto the subspace spanned by the eigenvectors belonging to the measured eigenvalue . Following [10] we shall assume that there can be a measurement error where the eigenvalue is recorded instead of the actual eigenvalue . This is assumed to happen with probability . The choice of the interaction in the interval is dictated by the sequence of recorded eigenvalues, or more simply the recorded sequence , that is for . The corresponding unitary operator describing the evolution in the time span is where denotes time ordered exponential and . We shall denote the un-conditioned evolution operator from time to the time of the first measurement as . Note that the sequence of recorded labels generally differs from the sequence of labels specifying in which subspace the system state was actually projected at the measurement times . As customary in the context of the fluctuation theorem we shall assume that besides the intermediate measurements of , all ’s are measured at times giving the eigenvalues , respectively.

The quantity of primary interest is the probability that is obtained in the first energy measurement, the sequence is realised, the sequence is recorded and is obtained in the final energy measurement. Here we have introduced the simplified notations , . The explicit expression of is:

 p(m,k,j,n) =TrPmAk,jU0PnU†0A†k,jPmp0n (3) Ak,j =←Πi(Ukiπji√ε[ki|ji]) (4)

where denotes the probability of obtaining the eigenvalue in the first measurement; denotes the corresponding projector; denotes the projector onto the subspace belonging to the eigenvalue of ; the symbol denotes -ordered product, that is, .

Let be the energy change in the partition observed in a single realisation of the feedback driven protocol. Using the cyclic property of the trace and completeness , we obtain the following:

 ⟨e−∑lβlΔEl⟩=γ=∑j,kTrA†k,jρ0Ak,j (5)

The proof is reported in the appendix. This relation extends the result presented in Ref. [10] to the case of multipartite system with initial multi temperature state, and to repeated measurements.111For simplicity we restricted to the case of cyclic . The extension to non-cyclic case is straightforward. The quantity represents the probability that the sequences , , are realised under the backward evolution specified by the adjoint Kraus operators . The total probability does not generally add to one. The reason for that is that the -th evolution occurs before the the -th eigenvalue is realised in the backward map. The feedback loop is evidently not time-reversal symmetric, and such lack of reversibility breaks the fluctuation theorem which in fact is a manifestation of time-reversal symmetry [19]. This is reflected by the fact that the quantum channel specified by the Kraus operators is generally not unital.222We recall that a quantum channel specified by Kraus operators , that is trace preserving , is unital when it maps the identity into itself . The adjoint of a non-unital quantum channel is not trace preserving. In the case of feedback control the quantum channel is generally not unital, as a consequence its adjoint is generally not trace preserving, hence we have generally . Lack of unitality generally reflects lack of time-reversal symmetry. Examples are thermalisation maps, namely maps that have a thermal state (not the identity) as a fixed point. Physically these are realised by means of weak contact of a system with a thermal bath, leading to irreversible dynamics. Likewise feedback control breaks the symmetry. This observation reveals some analogy between feedback control and dissipative dynamics.

Before proceeding let us comment briefly on the origin of lack of unitality in feedback controlled systems, in order to gain insight in the issue. For simplicity let us consider the case of a single measurement . Let us begin by noticing that the quantum channel specified by the is trace preserving. We have , where we have used the cyclic property of the trace, unitarity , idempotence , normalisation , and completeness . Let us now turn to unitality. We have . If the evolution did not dependent on , that is was chosen regardless of the recorded value (e.g, is pre-specified or is completely random), one could perform the sum over using and then use to conclude the map is unital. Feedback, implying explicit dependence on of breaks unitality. Unitality would occur also in the case when does not depend on , meaning the measurement outcome is completely random and has no correlation with the actual state . In sum if the feedback control measurement is off, either because one decides not to use the information gathered in the measurement, or because the measurement gathers no information in the first place, unitality is recovered, and the fluctuation theorem is restored. This result is in agreement with the established fact that projective measurements without feedback control do not alter the validity of the fluctuation theorem [8, 20, 21]. Here we have further learned that noise, i.e. choosing the ’s between the measurements completely randomly, also does not affect the integral fluctuation relation.

Let us now turn to thermodynamics. Using Jensen’s inequality, Eq. (5) implies:

 ∑lβl⟨ΔEl⟩≥−lnγ (6)

In the case when the map is unital it is , and the second law of thermodynamics is recovered [14]. When the condition is not forbidden, and the apparent violation of the second law becomes possible. This occurs with a proper “demonic” design of the feedback control. When instead the second law is more strictly enforced by means of an “angelic” intervention.

As shown in Refs. [10, 22] in the case of a single measurement (in either classical or quantum systems) the fluctuation relation can be restored if an information theoretic term, in the form of a mutual information, is added to the exponent in the exponential average. Ref. [15] reports the extension to the case of repeated measurements in the classical scenario. All these results are for a single-temperature initial state. In the present set-up we find as well an information theoretic correction term (see the appendix for a proof):

 ⟨e−∑lβlΔEl−Jk,j⟩=1 (7)

where is defined by the following set of equations:

 Jk,j =lnp(k,j)p(j:k)p(k) (8) p(k,j) =∑n,mp(m,k,j,n) (9) p(k) (10) p(j:k) =p(k,j)Πiε[ki|ji] (11)

The symbol represents the joint probability that the sequence is realised and the sequence is recorded, while is the probability that is recorded. The symbol stands for the probability that the sequence is realised, conditioned on being the record. More explicitely

 p(j:k) =p(k,j)Πiε[ki|ji]=∑n,mTrPmBk,jU0PnU†0B†k,jPmp0n (12) Bk,j (13)

The operators differ from the operators by the term containing the conditional probability . Note that the Bayes rule does not apply here, i.e. generally it is . The reason is that and are concatenated with each other. An outcome influences the record , which in turn influences the next outcome and so on. The quantity measures the degree of such mutual influence, or correlation between the two sequences and 333Eq. (7) is reminiscent of a similar relation reported by Vedral [23], see Eq. (8) there. The two relations fundamentally differ in various respects. Notably in the meaning of the mutual information term. In our case measuring the correlation between outcomes and their records, in the case of Ref. [23] measuring the correlation between the measurements themselves. In absence of feedback, namely when there is no correlation between the two sequences, is null and the standard relation is recovered. Note that given a feedback rule, generally would grow with the length of the sequences, i.e. the number of measurements. It is accordingly expected that in the large regime.

With Jensen’s inequality Eq. (7) implies

 ∑lβl⟨ΔEl⟩≥−⟨Jk,j⟩ (14)

We thus have found two bounds to .

By looking directly at the as in Ref. [14] we have found a third bound whose interpretation is most direct and straightforward. Let

 ρτ =∑n,j,k,mPmAk,jU0Pnρ0PnU†0A†k,jPm=∑j,kAk,jU0ρ0U†0A†k,j (15)

be the system density matrix at time . In the second equality we have used completeness and the fact that the initial state has no coherences in the energy eigenbasis . Simple manipulations, similar to those employed in Ref. [14] lead to the following salient result

 ∑lβl⟨ΔEl⟩=∑iD[ρlτ||ρl0]+I[ρτ]+ΔH (16)

where

 D[ρlτ||ρl0] =Trρlτlnρlτ−Trρlτlnρl0 (17) I[ρτ] =−∑lTrρlτlnρlτ+Trρτlnρτ (18) ΔH =−Trρτlnρτ+Trρ0lnρ0 (19)

denote the Kullback Leibler divergence between the final state and the initial state , Eq. (17); the total amount of correlations (mutual information) that builds up among the partitions as a consequence of their interaction during the time span , Eq. (18); and the total change in von-Neumann entropy of the whole compound, Eq. (19). Here is the reduced state of partition at time ( denotes trace over all partitions but the -th). The mutual information among the partitions of the system (measuring all correlations, quantal and classical), which develops generally due to their interaction (and can also occur in absence of measurements and feedback [14]), should not be confused with the classical mutual information between the realisation sequence and the record sequence caused by the feedback mechanism.

Both the Kullback Leibler divergence and the mutual information are non negative quantities. We thus arrive at the central inequality:

 ∑lβl⟨ΔEl⟩≥ΔH (20)

In the standard no measurement case, is linked to via a unitary map, hence and one recovers the result of Ref. [14], namely , and the second law in its standard form. Note that when there are measurements, but no feedback, the is linked to via a unital map, implying , , and hence , meaning that, as is already known [8, 20, 21] the second law is not altered by the mere application of projective measurements that interrupt an otherwise unitary dynamics. However Eq. (20) clearly indicates that there is a dissipation term associated with quantum-mechanical measurements, which is not present in the classical case. In sum through Eq. (20) we see that there is a thermodynamic cost associated to quantum measurements.

Combining Eqs. (6,14,20) the second law of thermodynamics, in presence of feedback control takes the form

 ∑lβl⟨ΔEl⟩≥max[−lnγ,−⟨Jk,j⟩,ΔH] (21)

## 3 Illustrative example

To exemplify the theory above we consider a prototypical model of quantum heat engine whose working substance is made of two qubits [13, 14, 24]. Their Hamiltonian reads

 H=H1+H2=ℏω2σ1z+ℏω2σ2z (22)

where denote Pauli operators. We assume the two qubits have same level spacing and are initially in the state:

 ρ0=e−β1H1Z1⊗e−β2H2Z2 (23)

with their partition functions. At the ’s are measured collapsing the two qubits in the state , with . We assume classical error in the measurement of each qubit , for some . Accordingly the eigenvalues are recorded with probability . If the states , are recorded we do nothing: ; else, i.e., if we apply a swap operation, , that maps into . The system is now in a joint eigenstate of the two qubits Hamiltonian , hence the final measurement of the is irrelevant. At the end of the process each qubit is allowed to relax to thermal equilibrium with their respective thermal baths of inverse temperatures so as to re-establish the initial state . Accordingly the average energies acquired by each qubit during the process equals the average heats that they release in the baths in the thermal relaxation step. Due to the feedback mechanism energy may be withdrawn from the cold bath and released in the hot one. Note that, due to the fact that the two qubits have same level spacing the SWAP operation does not alter their total energy. Namely there is no energy injection by the Demon: to steer the energy flow he only uses information. The set-up is illustrated in Fig. 2 panel a).

The relevant probability chain is a bit simpler than in the general case because the first energy measurement is itself here also the first feedback measurement. It reads with . For we have . The final state is . The probability that the outcome is realised conditioned on being recorded is simply the marginal probability that is realised because the record comes chronologically after the realisation of and hence cannot have any influence on it. The quantity boils down then to the logarithm of the ratio [10] hence its expectation is the non-negative mutual information between and : .

Panels b,c) of Fig. 2 show for two choices of and same , as a function of the error probability . In accordance with Eq. (21) we see that is bounded from below by and . Independent of all other parameters the refrigerator cannot work in the region where and are anti-correlated, while it may only work if . This is captured by being positive in the region and negative for . At outcome and recording are fully uncorrelated, which restores unitality as discussed above and implies . Regarding , while it tends to be closer to in the operation region (), it greatly departs from it in the non-operation region, where it can even get negative values. Notably in both panels there is a value of for which the bound is saturated by . Regarding we note it is everywhere non-positive as expected. Furthermore it is symmetric with respect to . This reflects the fact that the mutual information does not distinguish between correlation and anti-correlation. The maximum is attained at where are uncorrelated, and the standard fluctuation relation is recovered (i.e., ). In both panels we see that . Whether this a generic bound is yet to be understood. We note that while at both and are null, is non-negative, reflecting the fact that in absence of feedback there is nonetheless an entropic cost associated to measurements, as discussed above. Such cost can be counterbalanced in presence of feedback (note that may be negative for ). Confronting now the two panels, we see that the higher the thermal gradient , the larger is the point where the engine starts operating, i.e. where turns from positive into negative: As intuition suggests the more the gradient the better must your measurement be. This feature is captured also by but not by . Also the smaller the gradient the more the shape of the function resembles that of , with the shift between the two being approximately the value of at : that is .

## 4 Experimental proposal

The general theory developed above allows for a joint information theoretic and thermodynamic analysis of feedback controlled dynamics in the broad scenario where a demon can influence not only the amount of work being provided by the outside as in previous works [2, 3, 4], but also the heat flow between the various parts of a compound system, e.g. the heat flow between various heat baths.

The progress of solid state technology on the other hand allows to realise such feedback controlled heat transport mechanisms in real devices. The example illustrated above can be experimentally realised by introducing a feedback mechanism in the two-superconducting qubits scheme illustrated in in Ref. [13]. Below we illustrate a design that is of more immediate realisation. It is a based on a single qubit and it does not involve any qubit-operation, but only manipulations of qubit-bath couplings. The proposal that we put forward here is based on two ingredients that enable unique capabilities allowing for the implementation of a Maxwell demon based on a most simple concept. The two ingredients are a two-level-system acting as quantum trap door and the calorimetric measurement scheme developed in Refs. [16, 17].

The one qubit set-up is illustrated in Fig. 3. The two-level system is embodied by a superconducting qubit of level spacing . The two chambers are embodied by two resistors being kept at different temperatures. Qubit and resistors con exchange energy (i.e. heat) in the form of photons of energy associated to the TLS absorbing/emitting one photon from/to one of the two baths. The resistors are embedded into an RLC loop of tunable resonance frequency. This results into a tuneable TLS/resistor coupling. When an RLC circuit is far detuned from , the qubit is effectively decoupled from the resistor, while maximal coupling occurs when it is in tune with the qubit. The resonance frequency can be tuned by using a SQUID as a non-linear and tuneable inductor, its inductance being governed by a controllable threading magnetic flux.

When a photon enters/exits one of the two resistors, its electronic temperature undergoes a positive/negative jump followed by a fast decay. Two calorimeters [16, 17] continuously monitor the two resistors, and count how many photons enter/exit them. This allows for a directional full counting statistics of heat. Most remarkably it also allows to infer the state of the TLS at each time. If an absorption (in either resistor) is observed, it means the TLS jumped down, hence it was up before the absorption was detected, and is down afterwards. This allows to experimentally access the quantum state trajectory of the TLS.

The feedback concept is extremely simple: as soon as a jump-down is observed, turn on the interaction with the cold resistor and turn off the interaction with the hot resistor. Vice-versa for the observation of a jump up. This results in a net flow of heat from the cold resistor to the hot one. Based on the above general analysis the apparent violation of the second law is understood in terms of lack of time-reversal symmetry of feedback control, leading to an overall non-unital dynamics of resistors plus TLS. In a practical realisation one is realistically not able to fully turn off the interactions. Furthermore there will be some delay time between measurement being performed and feedback being realised, giving rise effectively to possible error between measured state and actual state of the qubit.

## 5 Modelling

In the following we model the dynamics of the proposed experiment. We model the evolution of the two level system via a standard Lindblad master equation

 ˙ρ=−i[HS,ρ]+LLρ+LRρ (24)

where is the two level system Hamiltonian expressed in terms of the Pauli matrices , and are Lindblad operators

 Llρ =Γ↓lD[σ]ρ+Γ↑lD[σ†]ρ (25)

expressed in terms of and the super-operator and the rising and lowering spin operators of the Hamiltonian , defined via , where is the ground (excited) state of . Here denote either the left or the right reservoir. The rates for jump down/up in the ’th resistor are given by

 Γ↓↑l =E20M2lℏ2Φ0Δ2(q2+Δ2)SI,l(±ω) (26)

where is the current noise spectrum expressed in terms of the voltage noise spectrum , is the quality factor and the resonance frequency of resonator , expressed in terms of its resistance, inductance and capacitance . By increasing the rates , can be quenched, namely the interaction between the TLS and the -th resistor can be turned off. The symbol stands for the mutual inductance between the qubit and the -th resistor and is the flux quantum. Note that the rates are detailed balanced:

 Γ↓l=eβlℏωΓ↑l (27)

The study of heat and work fluctuations requires the study of the dynamics to be performed at the level of single quantum-jump trajectories [13, 25], resulting from the unravelling of the master equation. This is here achieved by means of the Monte Carlo wave function (MCWF) method [26, 27]. In the specific case under study of a two level system subject to dissipation terms leading to full wave function collapse in either state or , this results in a classical dichotomous Poisson process with rates [13].

The basis of our numerical experiment is the generation of such dichotomous Poisson random trajectories. We chose the right reservoir as the cold one and the left as the hot one. The TLS is assumed to be initially in equilibrium with the left bath. We produce a large sample of trajectories and build the normalised historgram of the number of photons entering the right reservoir. Since the heat entering the right reservoir is given as , the statistics is the heat statistics. In absence of feedback it satisfies the fluctuation relation

 h(NR)h(−NR)=e−ΔβℏωNR,no feedback (28)

The feedback is introduced as follows. At each moment in time we distinguish between the actual state of the system and the knowledge we have about it. The latter does not necessarily coincide with the former because we allow for some delay-time between a jump occurring in the TLS and our knowledge of the state of the qubit being updated accordingly. The delay time thus effectively introduces an error probability between the actual state and the knowledge about the state, at each time. At each time, conditioned on the knowledge of the state we use either one set of rates favouring the interaction with either the cold or hot bath. More explicitly, let be the rate for jump down (up) in -th bath conditioned on TLS being measured to be in state . In accordance with Eq. (26) we use the following rates

 Γ↓|+L=A1−e−βLℏωΓ↑|+L=Γ↓|+Le−βLℏω (29) Γ↓|+R=B1−e−βRℏωΓ↑|+R=Γ↓|+Re−βRℏω (30) Γ↓|−L=B1−e−βLℏωΓ↑|−L=Γ↓|−Le−βLℏω (31) Γ↓|−R=A1−e−βRℏωΓ↑|−R=Γ↓|−Re−βRℏω (32)

where are determined by the circuitry parameters, and can be tuned via external fluxes . With , this means that energy exchange with the right (cold) bath is larger when the TLS is believed to be down, so that it becomes more likely that energy flows out of the cold reservoir. Similarly energy exchange with the left (hot) bath is larger when the TLS is believed to be up, so that it becomes more likely that energy flows in the hot reservoir. Overall this results in an effect that contrasts the natural flow from hot to cold. The largest effect can be achieved when turning off the unwanted interaction completely, namely when . Having in mind a realistic set-up here we keep the ratio finite, meaning partial turning-off is considered.

Because of the feedback the fluctuation relation (28) is not obeyed. However it can be proved (see appendix) that, due to the feedback mechanism, the TLS feels the effective temperature gradient

 Δβeff=βeffL−βeffR=Δβ+2ℏωlnε[+|+]A+ε[−|+]Bε[+|−]A+ε[−|−]B (33)

we thus see that by tuning the ratio the effective temperature gradient can be manipulated and if the errors associated to the measurement is not too big, it can even be inverted as compared to the original thermal gradient . So the overall effect of the demon is to change the “temperatures felt” by the TLS. Accordingly the following fluctuation relation

 h(NR)h(−NR)=e−ΔβeffℏωNR (34)

is obeyed by the histogram . This immediately allows to interpret the quantity

 Jexp=−2QRℏωlnε[+|+]A+ε[−|+]Bε[+|−]A+ε[−|−]B (35)

via Eq. (7) as the mutual information encoded in a trajectory along which a heat is exchanged with the bath. Note that when , the feedback has no effect and accordingly . Likewise if (hence ) meaning no correlation between state and knowledge thereof, feedback control does not work and again . Most importantly the experimental mutual information is proportional to the heat exchanged. This allows for accessing a fluctuating information theoretic quantity by means of a thermodynamic measurements in a realistic experimental scenario.

Figure 4 shows typical histograms for realistic parameters. We also plotted the quantity finding a good agreement with the theoretical prediction . The effective conditional probabilities were obtained by recording for each trajectory the total time when state was and knowledge was , and averaging their value over the whole ensemble of trajectories. The observed deviation is a consequence of the fact that error here is not introduced in the form of an outcome being missed (as assumed in deriving Eq. (34)), but rather being reported with some delay. With the histogram we computed , , , for the chosen parameters. The computed values are in agreement with the prediction of Eq. (21). The proposed experiment does not allow to measure , which would require accessing the full system+baths density matrix.

### 5.1 Energy spent by the Demon

What is the energy cost incurred by the demon to open/close the trap-door? To roughly estimate that we model the LCR circuit as a classical harmonic oscillator (LC circuit) in contact with a heat bath (the resistor) at temperature . To open/close the door towards one of the two reservoirs, the demon switches the LC frequency from to another frequency so as to put it in/off resonance with the qubit. If the operation is carried in a quasi static manner, the work done is equal to the free energy change: . The operation would in this case be reversible, and the work lost when opening the door will be retrieved when opening it. The overall cost of a open/close cycle would be null in this limiting case. The other limiting case is when the switch is infinitely fast. The overall cost of a single open/close cycle in this case would be non-negative in accordance with the second law of thermodynamics, and amounts to . The overall work incurred in a repeated feedback operation is proportional to the number of open/close cycles, which in turn is proportional to the net number of energy quanta being transported, namely the total heat transported. Interestingly we note that the faster the open/close operation, the more effective is the feedback mechanism, the more energy needs to be invested.

## 6 Conclusions

We have developed a general quantum theory of repeated feedback control in a multiple heat reservoir scenario. The main effect of feedback control is that it induces a generally non-unital dynamics of the full reservoirs+system compound. As a consequence the standard bound set by the second law od thermodynamics on the dissipation quantifier is shifted and may become negative. We have illustrated an experimental proposal where a single superconducting qubit plays the role of a trap-door that is subject to feedback control. The envisaged method for simultaneously measuring the qubit state and the heat exchanged by each reservoir is single photon calorimetry.

## Acknowledgements

This research was supported by a Marie Curie Intra European Fellowship within the 7th European Community Framework Programme through the project NeQuFlux grant n. 623085 (M.C.), by Unicredit Bank (M.C.), by the Academy of Finland contract no. 272218 (J.P.), and by the COST action MP1209 “Thermodynamics in the quantum regime”.

## Appendix A Derivation of Eq. (5)

 ⟨e−∑lβlΔEl⟩ =∑n,j,k,mp(m,k,j,n)e−∑βlElme∑βlEln =∑n,j,k,mTrPmAk,jU0PnU†0A†k,jPme−∑βlElnZe−∑βlElme∑βlEln =∑j,k,mTrPmAk,jA†k,jPme−∑βlElmZ =∑j,kTrA†k,jρ0Ak,j

Eq. (3) and have been used to obtain the second line. Completeness and unitarity led to the third line. Fourth line follows from the cyclical property of the trace, idempotence and .

## Appendix B Derivation of Eq. (7)

Using Eq. (11), the exponentiated fluctuating mutual information can be conveniently expressed as

 e−Jk,j=p(j:k)p(k)p(k,j)=p(k)Πiε[ki|ji] (36)

hence

 ⟨e−∑lβlΔEl−Jk,j⟩ =∑n,j,k,mp(m,k,j,n)e−∑βlElme∑βlElnp(k)Πiε[ki|ji] (37) =∑n,j,k,mTrPmBk,jU0PnU†0B†k,jPme−∑βlElnΠlZle−∑βlElme∑βlElnp(k) =∑j,k,mTrPmBk,jB†k,jPme−∑βlElmΠlZlp(k) =∑k,mTrPme−∑βlElmΠlZlp(k) =Trρ0∑kp(k) (38) =1 (39)

Eq. (3), and Eq. (13) have been used to obtain the second line. Completeness and unitarity led to the third line. The fourth line follows from which follows by expanding the -ordered products, apply idempotence , completeness , and unitarity . Cyclical property of the trace, idempotence and lead to the fifth line. The final result is a consequence of normalisation of and of .

## Appendix C Derivation of Eq. (33)

Under the operation of the demon the TLS experiences effective temperatures of the baths that differ from their actual value. To fix ideas, let us for the moment, assume no delay time and no error in the measurement. The qubit is effectively subject to the following effective rates . Accordingly, the detailed balance temperatures are shifted:

 eβeffLℏω =Γ↓,effL/Γ↑,effL=(A/B)eβLℏω (40) eβeffRℏω =Γ↓,effR/Γ↑,effR=(B/A)eβRℏω (41)

where we used the explicit expressions Eq. (26). This implies the effective temperatures

 βeffL =1ℏωln(A/B)+βL (42) βeffR =1ℏωln(B/A)+βR (43)

Let us now introduce the errors related to the measurement. The stochastic process describing the dynamics of the TLS is still Poissonian with one rate occurring in case of right measurement and one rate occurring in the other case. The idea is that monitoring is continuous, or better, occurring with a sampling time interval , which we assume short compared to all rates . Let us imagine the system is in state . There is a probability the observation is and a probability the observation is . Thus the probability to undergo a jump down in the reservoir in the interval is

 p(dt) =ε[+|+]e−Γ↓|+sdt+ε[−|+]e−Γ↓|−sdt (44) ≃ε[+|+](1−Γ↓|+sdt)+ε[−|+](1−Γ↓|−sdt) (45) =1−(ε[+|+]Γ↓|+s+ε[−|+]Γ↓|−s)dt (46) =e−(ε[+|+]Γ↓|+s+ε[−|+]Γ↓|−s)dt (47)

Similarly for the jump up. Overall the TLS experience the new rates

 Γ↓s,eff =ε[+|+]Γ↓|+s+ε[−|+]Γ↓|−s (48) Γ↑s,eff =ε[+|−]Γ↑|+s+ε[−|−]Γ↑|−s (49)

Accordingly

 eβeffsℏω =Γ↓s,effΓ↑s,eff=ε[+|+]Γ↓|+s+ε[−|+]Γ↓|−sε[+|−]Γ↑|+s+ε[−|−]Γ↑|−s (50) βeffs =1ℏωlnε[+|+]Γ↓|


text:
# Estimating Resistance

1. Nov 13, 2012

### Peter G.

"Estimating" Resistance

Hi,

So, suppose I have a bulb labelled 10 V, 12 W. This is connected to a 4 V e.m.f battery. Can I calculate the Resistance from 10 V, 12 W (Using P = VI and R = V/I) and assume that it is the resistance when it connected to a 4V battery?

Thanks!

2. Nov 13, 2012

### liblenovo

Re: "Estimating" Resistance

Generally speaking, the resistance of a bulb vary with temperature. If ignore this feature, I think your answer is right.

3. Nov 13, 2012

### SammyS

Staff Emeritus
Re: "Estimating" Resistance

If the bulb is rated 12W at 10V, that likely means that it produces a reasonable of light at 10 V. The filament of the bulb, under such conditions, will be near its maximum operating temperature.

At 4 V, the same bulb would be operating at a considerably lower temperature. The resistivity of tungsten is very temperature dependent, so the resistance of the bulb would be much less than when it operates at 10 V .

4. Nov 13, 2012

### Peter G.

Re: "Estimating" Resistance

Without a graph, however, is there any more precise method of estimation?


text:
1. ## \int{\frac{x^2}{\sqrt{4x-x^2}}}dx

Having trouble with this one.

$\int{\frac{x^2}{\sqrt{4x-x^2}}}dx$

2. Originally Posted by th%$&873 Having trouble with this one. $\int{\frac{x^2}{\sqrt{4x-x^2}}}dx$ Note that $4x - x^2 = 4 - (x + 2)^2$. Make the substitution $x + 2 = 2 \sin t$. You might want to check the details (as I rushed this) - I get something like $8 \int (\sin t - 1)^2 \, dt$. Expand, integrate term-by-term and finally substitute back that $x + 2 = 2 \sin t$ (tedious, I know, but not too hard). Ask for more help if you get stuck at any point. Edit: There is a small mistake - you should spot it easily. A small detail that doesn't change the general game plan ..... 3. Originally Posted by th%$&873
Having trouble with this one.

$\int{\frac{x^2}{\sqrt{4x-x^2}}}dx$
$\int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx$

Let $x-2=2\sin u$ then,

$\int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2 u+2\sin u+1\:du$, now you integrate each term

now you should get, $4\int \sin^2 u+2\sin u+1\:du=6u-2\sin u\cos u-8\cos u$

$\int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6\arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)\sqrt{4x-x^2}}{2}+C$

4. Originally Posted by polymerase
$\int{\frac{x^2}{\sqrt{4x-x^2}}}\:dx = \int \frac{x^2}{\sqrt{4-(x-2)^2}}\:dx$

Let $x-2=2\sin\:u$ then,

$\int \frac{x^2}{\sqrt{4-(x-2)^2}}dx=4\int \sin^2\:u+2\sin\:u+1\:du$, now you integrate each term

now you should get, $4\int \sin^2\:u+2\sin\:u+1\:du=6u-2\sin u\cos u-8\cos\:u$

$\int{\frac{x^2}{\sqrt{4x-x^2}}}\;dx=6arcsin\left(\frac{x-2}{2}\right)-4\sqrt{4x-x^2}-\frac{(x-2)(\sqrt{4x-x^2})}{2}+C$
5. I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get $6\theta-8\cos\:\theta-\sin 2\theta$.
It appears that polymerase used the double-angle formula to get $-2\sin u\cos u$ instead of $-\sin 2 u$? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around.
I don't really understand either of your methods (you combined two steps?), but I used u-substitution then trig-substitution to get $6\theta-8\cos\:\theta-\sin 2\theta$.
It appears that polymerase used the double-angle formula to get $-2\sin u\cos u$ instead of $-\sin 2 u$? I just wasn't sure what to do with that last term. I also made a small mistake with the coefficient of the first term, but I caught that the second time around.
You're right, I did use the double angle formula, the reason is because I needed it when I'm re-substituting and expressing everything in terms of $x$. Thats all.


text:
# [NTG-context] calling up chapter title

Wolfgang Schuster wolfgang.schuster.lists at gmail.com
Fri Apr 26 13:03:08 CEST 2019

```
Hans van der Meer schrieb am 26.04.2019 um 12:21:
> This does not what I have in mind:
>
> Instead of [\pagenumber chapter] I thought [\pagenumber \chapter]
> would give me in the header both the pagenumber and the recorded
> chapter title.
> However \chapter turns out to be undefined, neither did \chaptertitle.
> What is the macro to be called?

I doubt \chapter is undefined otherwise you can add a chapter title to

What you're looking for is the \getmarking marking which is used to
access section
titles and number in the header and footer lines, e.g.


text:
# USB Wi-Fi adapter does not wakeup on Windows Vista

I'm making use of the Linksys WUSB300N USB Wi-Fi adapter under Windows Vista Home Premium x64. The trouble is it does not wakeup when computer system heads out of standby. It's actually aggravating.

I have 1.0.3.2 variation of vehicle drivers dated 2007-09-11, and also I assume they are the latest ones.

In the network adapters/Wireless-N USB Network Adapter/Properties/Power Management I have "Allow the computer system to switch off this tool to conserve power" uncontrolled and also "Allow this tool to wake the computer system" is non-active (impaired).

I have a number of various other USB tools and also there is no worry with them after the computer system awakens.

0
2019-05-04 02:36:27
Source Share

First point to examine is that the computer system is not set to transform it off to conserve power :

• Open tool supervisor
• Click the Power Management tab.
• The Allow the computer system to switch off this tool to conserve power check box is presented, see to it it is uncontrolled.
0
2019-12-04 06:24:47
Source

It seems like Vista SP2 could be of usage to you

"Microsoft claims that the Service Pack will certainly additionally bring ... better Wi - Fi efficiency when returning to from Sleep setting

0
2019-12-04 06:24:24
Source

Also examine the Advanced Power Options. Panorama can be readied to switch off the USB ports to conserve power.

Given that your WiFi tool is USB - based, that might be the trouble.

0
2019-12-04 06:24:16
Source

On Windows Vista, I had some problems with the Wi - Fi start-up till I add the adhering to manuscript at start-up.

net stop Wlansvc
net start Wlansvc

0
2019-12-04 06:19:41
Source

The vehicle drivers are the latest ones, and also it is a concern that will possibly never ever be dealt with by Linksys. The only remedy I have actually located can be located in A USB device may no longer work correctly after Windows Vista resumes from sleep or from hibernation .

Adhere to the guidelines beginning with the Post - installment guidelines area.

0
2019-12-04 06:19:13
Source

I have actually seen this concern with Wi - Fi cards and also several USB tools. The most up to date vehicle drivers and also having fun with your power conserving setups are the means to go. Quiting and also reactivating the tool has actually offered me some success, yet it is trial and error.

0
2019-12-04 06:19:06
Source

Have you searched in your BIOS setups to see if there are any kind of alternatives for USB setups? Additionally, have you attempted another thing like a USB disk drive to see if that rotates back up after waking the computer system from rest? That might aid see if the trouble is with the adapter or the port.

Last but not least, I would certainly relocate the adapter to an additional equipment to see if it does the very same point (preferably one making use of XP or a various OS than you're presently making use of).

0
2019-05-09 02:14:23
Source

I've had a comparable trouble with intel network cards. My job about was to run a set documents when you wake the COMPUTER up.

If you intend to do it in this manner, the energy you require is DevCon. You after that utilize this in a.bat documents making use of the reactivate argument and afterwards name of the tool which you can locate making use of the findall argument.

0
2019-05-09 01:53:41
Source

In the Windows Device Manager , I would certainly examine to see if the USB Wi-Fi adapter has any kind of power-related setups in its equipment arrangement. I recognize that I've treated numerous wakeup concerns with Wi-Fi cards and also Windows Vista by readjusting their setups within the Device Manager (yet never ever a USB adapter, so this is a little a shot in the dark ).

0
2019-05-07 16:44:44
Source


text:
# Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?

I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).

Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?

I would greatly appreciate it if people would please take the time to explain this.

• What about hyperconjugation comparing primary, secondary, tertiary radicals; or for the row of phenyl, phenoxyl, benzyl radicals? Mar 14 '20 at 21:49
• @Buttonwood What are you referring to? I'm not sure how this relates to my question. Mar 14 '20 at 22:06
• hyperconjugation is one of the forms stabilizing cations and radicals if a $\sigma$ orbital may interact with an adjacent empty or partially filled orbital, e.g., $\pi*$. (en.wikipedia.org/wiki/Hyperconjugation). One methyl group may not offer much stablization (two or three are increasingly better) which eventually lowers the enthalpy of formation for example of tertiary radicals over the one of primary ones. Mar 14 '20 at 22:17
• @Buttonwood I have thought about your comments and posted a related question here chemistry.stackexchange.com/q/129076/75460 Mar 15 '20 at 14:39

Here is a rather simple structure explanation presented on this site. To quote, where 'pieces' refer to an organic molecular with a non-symmetric branching structure:

You’ll notice something – the simpler the pieces are, the easier they are to stack together, which provides a tighter fit with fewer spaces. Here, by putting a kink in the block, we make them harder to stack.

What’s this got to do with chemistry?

When compounds freeze, the process is a lot like stacking bricks. The more symmetrical the molecules are, the easier it will be and the fewer spaces there will be between the molecules. Fewer spaces = better stacking. Hence, when you compare hexane to its structural isomer, 2-methylpentane, hexane has a much higher melting point due to the regular arrangement of its structure.

Better stacking, higher melting point. Case closed. Right? Not quite.

It’s also about surface area [...]

It looks like as we increase branching, we’re increasing melting point and decreasing boiling point. What’s going on? [...]

Going from “branched” to “highly branched” makes a molecule more compact and sphere-like. As the surface area of the molecule decreases (remember that spheres have the lowest surface area/volume ratio of any shape) they will become more compact and thus easier to pack. This explains the melting point phenomenon [...]

Boiling point is related to the forces between molecules, which in the case of hydrocarbons is Van Der Waals interactions. If you’ve ever seen microscope images of a gecko’s feet – which allow it to climb walls – you’ll see that there is no adhesive but the pads contain a tremendous amount of surface area. It’s all about the Van der Waals interactions.

As we decrease surface area, we are going to decrease intermolecular Van Der Waals interaction and therefore decrease boiling point.

The summary relationships are presented as follows:

linear versus branched —> higher melting/boiling points due to better stacking and surface area contact.

highly branched vs. branched —> more sphere-like –> better stacking –> higher melting point

highly branched vs. branched —>more sphere-like – -> lower surface area —> lower boiling point.

The above connects boiling point to branching structure, and this paper "Hypothetical Thermodynamic Properties: The Boiling and Critical Temperatures of Polyethylene and Polytetrafluoroethylene" found here links BP to thermodynamics. Per the opening reference statement: "A lower heat of formation is equivalent to a higher thermodynamic stability".

This completes the argument as to why the heat of formation decreases for organic radicals with decreased size and degrees of branching.


text:
# Let the width of a cell be set by a picture

I wish to have a table with two rows - first one with a picture, and the second is some text (I know I can probably do it within the figure environment, but I want to do it on my own). I want the text to be split whenever necessary, so I use \begin{tabular}{p{width}}. However, I decided that I want to firstly set the height of the picture, and then let the width of the scaled picture determine the width of the text cell. How can I do it?

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“I want the text to be split whenever necessary…” Could you elaborate: Split to where? – morbusg Feb 3 '12 at 11:04

First set the picture into a box, then you can get the width of the box to use in your p argument using \settowidth (or just the \wd primitive) then use \usebox to put the picture into your table.

\documentclass{article}

\newsavebox\foo

\begin{document}

\savebox\foo{%
%\includegraphics...
a picture here}

\begin{tabular}{|p{\wd\foo}|}
\usebox\foo\\
some text here some text here
\end{tabular}

\end{document}

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Perhaps you should elaborate on how the OP can accomplish this, perhaps an example would make things clearer. – Roelof Spijker Feb 3 '12 at 11:23
yes well if the OP had provided a MWE I'd have probably edited it with an explicit fix, but OK I'll edit my answer in a bit. – David Carlisle Feb 3 '12 at 11:27
Supplementary question on personal preferences: I would have put the \savebox in the preamble (to keep it out of the way). Would you care to comment on that? – Brent.Longborough Feb 3 '12 at 12:10
The savebox contents are logically part of the document flow and so belong after begin{document}. If it's just a graphic it probably works here (but would mean you'd need one box per table, which is expensive) but if it contains text (as in my cut down example) it's only really reliable to do it after begin{document} as that is when all the font setup is finalised. – David Carlisle Feb 3 '12 at 12:32
Thank you. It's always rewarding to learn from a Master... – Brent.Longborough Feb 3 '12 at 14:32


text:
# REBOUNDx

REBOUNDx: A Library for Adding Conservative and Dissipative Forces to Otherwise Symplectic N-body Integrations. Symplectic methods, in particular the Wisdom-Holman map, have revolutionized our ability to model the long-term, conservative dynamics of planetary systems. However, many astrophysically important effects are dissipative. The consequences of incorporating such forces into otherwise symplectic schemes is not always clear. We show that moving to a general framework of non-commutative operators (dissipative or not) clarifies many of these questions, and that several important properties of symplectic schemes carry over to the general case. In particular, we show that explicit splitting schemes generically exploit symmetries in the applied external forces which often strongly suppress integration errors. Furthermore, we demonstrate that so-called symplectic correctors’ (which reduce energy errors by orders of magnitude at fixed computational cost) apply equally well to weakly dissipative systems and can thus be more generally thought of as weak splitting correctors.’ Finally, we show that previously advocated approaches of incorporating additional forces into symplectic methods work well for dissipative forces, but give qualitatively wrong answers for conservative but velocity-dependent forces like post-Newtonian corrections. We release REBOUNDx, an open-source C library for incorporating additional effects into REBOUND N-body integrations, together with a convenient Python wrapper. All effects are machine-independent and we provide a binary format that interfaces with the SimulationArchive class in REBOUND to enable the sharing and reproducibility of results. Users can add effects from a list of pre-implemented astrophysical forces, or contribute new ones.

## Keywords for this software

Anything in here will be replaced on browsers that support the canvas element

## References in zbMATH (referenced in 2 articles )

Showing results 1 to 2 of 2.
Sorted by year (citations)


text:
# Surjctive,Bijective,Injective Examples

Is $f : Z_9 \rightarrow Z_9$ injective, surjective, bijective och/eller inverterbar $d^{\circ}$ for

1. $f(x) = x^3 + 3*x + 7$
2. $f(x) = x^3 + 4*x + 5$
3. $f(x) = x^3 + 5*x + 4$

How should I prove it? I think like a table but shall these values point to the same or?

As you say, the easiest way to do it is to draw up a table of the values that the function $f$ takes in each case.

If all of the values 0 to 8 appear in your table, then $f$ is surjective.

If no value is repeated, then $f$ is injective.

If both, then $f$ is bijective.

Of course, the "table" method only works for (small!) finite sets, and for these sets $f$ will either be bijective, or not injective and not surjective.

• Which x-values should be same? – user3704516 Nov 5 '16 at 21:50
• Ah, no $x$-values will be the same, it's the $f(x)$-values that might be the same! – Josh Hunt Nov 5 '16 at 21:51
• So it can be different x-values but same f(x)? – user3704516 Nov 5 '16 at 21:52
• I'm sorry, I'm not sure I understand your question. A function is said to be injective, if for every $x$ you have a different value of $f(x)$. That is, $f(x) = f(y)$ implies $x = y$ – Josh Hunt Nov 5 '16 at 21:57
• I am test for all this x-values for b example – user3704516 Nov 5 '16 at 23:13

Calculate the tables for $x=0,1,\cdots,8$.

The function is injective if and only if all values $0,1,\cdots ,8$ appear.

Furthermore, the function is bijective if and only if it is injective if and only if it is surjective.