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Data-driven Voltage Regulation in Radial Power Distribution Systems

# Data-driven Voltage Regulation in Radial Power Distribution Systems

Hanchen Xu,  Alejandro D. Domínguez-García,  Venugopal V. Veeravalli,  and Peter W. Sauer,  The authors are with the Department of Electrical and Computer Engineering at the University of Illinois at Urbana-Champaign, Urbana, IL 61801, USA. Email: {hxu45, aledan, vvv, psauer}@illinois.edu.
###### Abstract

In this paper, we develop a data-driven voltage regulation framework for distributed energy resources (DERs) in a balanced radial power distribution system. The objective is to determine optimal DER power injections that minimize the voltage deviations from a desirable voltage range without knowing a complete power distribution system model a priori. The nonlinear relationship between the voltage magnitudes and the power injections in the power distribution system is approximated by a linear model, the parameters of which—referred to as the voltage sensitivities—can be computed directly using information on the topology and the line parameters. Assuming the knowledge of feasible topology configurations and distribution line resistance-to-reactance ratios, the true topology configuration and corresponding line parameters can be estimated effectively using a few sets of measurements on voltage magnitudes and power injections. Using the estimated voltage sensitivities, the optimal DER power injections can be readily determined by solving a convex optimization problem. The proposed framework is intrinsically adaptive to changes in system conditions such as unknown topology reconfiguration due to its data-driven nature. The effectiveness and efficiency of the proposed framework is validated via numerical simulations on the IEEE 123-bus distribution test feeder.

power distribution system, distributed energy resource, voltage regulation, line parameter estimation, topology estimation, data-driven, sensitivity.

## I Introduction

Voltage regulation, which is of central importance in power distribution systems, is conventionally accomplished by devices such as voltage regulators, load tap changers, and shunt capacitros [1], and more recently, by distributed energy resources (DERs) with fast-responding characteristics [2, 3, 4]. For example, in [2], the authors proposed a two-stage distributed architecture for voltage regulation in power distribution systems, where the required reactive power injections are determined by each local controller in the first stage, and any deficiency is compensated in the second stage by other DERs providing more reactive power so as to uniformly raise the voltage profiles across the network. In [3], the authors formulated the voltage regulation problem as a convex optimization problem leveraging some relaxation technique, by solving which the optimal set-points of DER reactive power injections can be determined. However, most of existing voltage regulation schemes using DERs assume perfect knowledge of the power distribution system model, and thus may fail in the absence of an accurate model, which is typically the case in practice. In addition, the power distribution system model may change from time to time due to operations such as topology reconfiguration for loss minimization or load balancing [5]. As such, voltage regulation schemes which can adapt to changes in system conditions and are robust against model errors are indeed more desirable in power distribution systems.

In situations where an accurate system model is not available, data-driven methods can be applied as an alternative. A key idea in these methods is to approximate the relation between the outputs of interest (e.g., voltage magnitudes) and the controls (e.g., power injections) by a linear model, the parameters of which are referred to as the sensitivities, and estimate the sensitivities from the measurements [6, 7]. For example, this idea has been pursued in estimation of injection shifting factors and power transfer distribution factors [6], and loss factors [8]. The sensitivities have also been utilized in voltage regulation problems [9, 10, 11]. For example, the authors proposed ambient signal based estimation methods for voltage-var sensitivities in transimission systems in [11]. They further developed data-driven sequential voltage control methods based on estimated voltage-var sensitivities and have proven the effectiveness via simulations using realistic data. The sensitivities estimated from measurement enjoy several nice properties, including adaptivity to changes in system conditions such as topology reconfigurations or parameter changes. However, existing approaches require a significant amount of measurements in order to obtain accurate estimates of the sensitivities. This may be feasible in transmission systems equipped with phasor measurement units, yet, it may not be practical for power distribution systems.

In this paper, we develop a data-driven voltage regulation framework for DERs in a balanced radial power distribution system, the objective of which is to determine optimal DER power injections that minimize the voltage deviations from a desirable voltage range without knowing a complete distribution system model a priori. Specifically, we will take advantage of an approximate linear model—the so-called LinDistFlow model (see, e.g., [5])—to simplify the nonlinear relationships between voltage magnitudes and power injections. The coefficients of the LinDistFlow model are essentially the sensitivities of the squared voltage magnitudes with respect to active and reactive power injections—referred to as the voltage sensitivities, and can be computed directly using information on the topology and the line parameters. Assuming the knowledge of feasible topology configurations and distribution line resistance-to-reactance (“-to-”) ratios, which are typically available and do not change during a relatively short time period, the true topology configuration and corresponding line parameters can be estimated effectively using a few sets of measurements on voltage magnitudes and power injections. Using the estimated voltage sensitivities, the optimal DER power injections can be readily determined by solving a convex optimization problem. Theoretical analysis shows that the voltage sensitivities of interest are easily identifiable.

Part of this work has been published in as a conference paper in [12]. This paper extends our earlier work in [12] in several aspects. First, we have developed an efficient topology estimation algorithm that uses the same set of measurements as the line parameter estimator and incorporated it into the framework. Second, we also present theoretical analysis on the identifiability of the line parameters. Third, we have added results from extensive numerical simulations on a larger distribution test feeder.

The remainder of the paper is organized as follows. Section II presents the preliminaries including the power distribution system model as well as the voltage regulation problem. Section III presents the details of the proposed voltage regulation framework, particularly, a voltage sensitivity estimator and a voltage controller. The identifiability of the voltage sensitivities is analyzed in Section IV. The effectiveness of the proposed framework is validated in Section V through some numerical simulations. Section VI concludes this paper.

## Ii Preliminaries

In this section, we provide the power distribution system model adopted in this work. Subsequently, we describe the voltage regulation problem in power distribution systems.

### Ii-a Power Distribution System Model

Consider a three-phase balanced power distribution system represented by a directed graph , where is the set of buses (nodes), and is set of distribution lines (edges). Let be the index set of distribution lines. Each electrical line is associated with , i.e., is the sending end and is the receiving end of line , with the direction from to defined to be positive. Let and denote the resistance and reactance of line , respectively, and define and . Let denote the “-to-” ratio of line , i.e., , and define , . Throughout the rest of the paper, we make the following assumptions:

1. Bus  corresponds to the substation bus and is a constant;

2. The power distribution system is radial and connected;

3. The power distribution system is lossless; and

4. The “-to-” ratios are known.

Let denote the magnitude of the voltage at bus , and define , . Let denote the index set of DERs. In addition, let and respectively denote the active and reactive power injected by DER , and define and , . Similarly, let and respectively denote the active and reactive power demanded by load , and define , and , . Let and respectively denote the minimum and maximum active power that can be provided by DER , and define and , . Similarly, let and respectively denote the minimum and maximum reactive power that can be provided by DER , and define and , . Let denote the mapping matrix between the DER indices and the buses, of which the entry at the row, column of is if DER is at bus . Define , and .

Let denote the node-to-edge incidence matrix of , with and if line starts from bus and ends at bus , and all other entries equal to zero. Let denote the -dimensional matrix that results from removing the first row in . Under Assumption A2, , and is invertible. Note that the topology of the power distribution system is uniquely determined by ; therefore, we also refer to as a topology configuration. A power distribution system may be operated under various feasible topology configurations. Let denote the set of feasible topology configurations of the power distribution system. Note that each topology configuration is associated with a vector of “-to-” ratios. Let denote the set of “-to-” ratio vectors that correspond to .

Let , and define , , . Under Assumptions A2 and A3, the relation between , , and , can be captured by the so-called LinDisfFlow model as follows [5]:

 ~v=Rp+Xq, (1)

where is the -dimensional all-ones vector, and

 R =2(M−1)⊤diag(r)M−1, (2) X =2(M−1)⊤diag(x)M−1, (3)

where returns a diagonal matrix with the entries of the argument on its diagonals; we refer to the matrices and as the voltage sensitivity matrices, or simply voltage sensitivities.

### Ii-B Voltage Regulation Problem

The objective here is to maintain the voltage magnitude at each bus , , of the power distribution system within a pre-specified interval denoted by . While a number of means, such as load tap changers and capacitor banks can be utilized to achieve the aforementioned objective, it is also possible to utilize the DERs present in the power distribution system. In this paper, we focus solely in this later mechanism for achieving voltage regulation. Then, the problem is to determine the DER active and reactive power injections so that

• the active and reactive power injections from each DER , , do not exceed its corresponding capacity limits, i.e., , ; and

• the voltage magnitude at each bus , , is within the pre-specified interval, i.e., .

In addition, among all feasible values of and , we would like to select the ones that minimize some cost function, which reflects the cost of voltage deviations as well as the cost of active and reactive power provision.

In this paper, we assume no priori information on the voltage sensitivity matrices except and . The voltage regulation problem cannot be solved without knowing the voltage sensitivity matrices. Therefore, we will resort to the data-driven approach to estimate voltage sensitivity matrices from measurements of voltage magnitudes and power injections.

## Iii Voltage Regulation /media/arxiv_projects/500772/Framework

In this section, we propose an adaptive data-driven framework for voltage regulation using DERs. We first give an overview on the framework and then present the details of the fundamental building blocks of the framework.

### Iii-a /media/arxiv_projects/500772/Framework Overview

The proposed voltage regulation framework consists of two components, a voltage sensitivity estimator and a voltage controller. The interaction between the different components is illustrated via the block diagram in Fig. 1. The estimator component contains a topology estimator that estimates the topology of the power distribution system (essentially, ), and a parameter estimator that estimates the line parameters ( and ), using measurements of power injections and voltage magnitudes. The estimated voltage sensitivity matrices, and , are computed using , , and . After that, the estimated and , denoted respectively by and , are sent to the voltage controller. The voltage controller then computes the set-points for the DER active and reactive power injections that minimize some cost function subject to constraints C1 and C2. The DERs will be instructed to inject active and reactive power by the amount determined by the voltage controller. A new set of measurements will be available once the DERs have modified their power injections. These measurements will be used by the estimator to update and so as to reflect any changes in them. The detailed formulations for the voltage sensitivity problem and the voltage regulation problem are presented next.

### Iii-B Voltage Sensitivity Estimator

Assume at time instant , we have measurements , , , , , where the index indicates the corresponding measurement is obtained at time instant . To reduce the computational burden, we select a subset of measurements, denoted by . The voltage sensitivities can be estimated based on the LinDistFlow model in (1). The objective of the voltage sensitivity estimator at time instant is to estimate from the measurements in the values of and , which can be computed using , , and .

We propose a voltage sensitivity estimator that consists of two components, a parameter estimator and a topology estimator. The former aims to estimate the line parameters, given the topology information, i.e., , while the later aims to determine from , based on the results from the parameter estimator, the details of which are presented next.

#### Iii-B1 Parameter estimator

Given the topology information , to estimate and is essentially to estimate and . Let and denote the estimate of and , respectively. We can then formulate the parameter estimation problem by using the relation in (1) as

 ^r,^x=argminr,x∑k′∈Kγk−k′∥Rp[k′]+Xq[k′]−~v[k′]∥2,

subject to

 R =2(M−1)⊤diag(r)M−1, (4a) X =2(M−1)⊤diag(x)M−1, (4b)

where denotes the -norm, is a discount factor. Essentially, the objective of the parameter estimator is to find the line parameters that can fit the LinDistFlow model best, for the given topology configuration.

We next show that (4) has a closed-form solution. First note that the matrix can be decomposed as follows:

 diag(x)=L∑ℓ=1xℓeℓe⊤ℓ, (5)

where is the basis vector in , i.e., all entries in are except the entry, which equals to . Using (5), we obtain that

 Xq[k′] =2(M−1)⊤diag(x)M−1q[k′] =2(M−1)⊤L∑ℓ=1xℓeℓe⊤ℓM−1q[k′] =L∑ℓ=1Ξℓq[k′]xℓ, (6)

where . Similarly,

 Rp[k′]=L∑ℓ=1Ξℓp[k′]rℓ=L∑ℓ=1Ξℓzℓp[k′]xℓ. (7)

Let , , and define

 Ψ[k]=⎡⎢ ⎢⎣Ξ1ρ1[k−m]⋯ΞLρL[k−m]⋮⋮⋮Ξ1ρ1[k]⋯ΞLρL[k]⎤⎥ ⎥⎦, (8)

and

 ψ[k]=[γm2~v[k−m]⊤,⋯,γ02~v[k]⊤]⊤. (9)

Note that and are dependent on . Then (4) can be equivalently formulated in the classical form of a linear regression problem as follows:

 minimizex∥Ψ[k]x−ψ[k]∥2, (10)

the solution to which is given by

 ^x=Ψ[k]†ψ[k] (11)

where denotes the pseudo-inverse of , obtained via singular value decomposition. Note that needs to have full rank, i.e., , in order to estimate . The resistance vector can be estimated using

 ^r=diag(z)^x. (12)

Define a residual vector, denoted by , as follows:

 ε=^Rp+^Xq−~v, (13)

where

 ^R =2(M−1)⊤diag(^r)M−1, (14) ^X =2(M−1)⊤diag(^x)M−1. (15)

Given a set of measurements, we can compute a residual vector for each deterministically.

#### Iii-B2 Topology estimator

The objective of the topology estimator is to find such that a weighted sum of over several time instants is minimized. At time instant , the topology estimation problem can be formulated as:

 ^M=argminM∈MϵM, (16)

with

 ϵM=∑k′∈Kγk−k′∥ε[k′]∥, (17)

where is computed through (13) to (15). We refer to as the residual error associated with topology configuration .

Essentially, the topology estimator selects the topology under which the residual error is minimized, where the line parameters are estimated by the parameter estimator. The intuition here is that different topology configurations will impose different structural constraints on voltage sensitivity matrices, which consequently will impact the residual error. The true topology configuration is expected to result in the least residual error. The voltage sensitivity estimation algorithm is summarized in Algorithm 1.

### Iii-C Voltage Controller

The voltage controller aims to determine the optimal set-points for the DER active and reactive power injections while meeting all requirements discussed in Section II-B. Note that for a given set of power injections, the resulting voltage magnitude at each bus can be estimated using (1), where and are used instead of and . Define and , . Then, the voltage control problem can be formulated as the following convex program:

 minimizepg,qg c(pg,qg)

subject to

 v=^R(Cpg−pd)+^X(Cqg−qd)+v01N, (18a) p–g≤pg≤¯¯¯pg,q–g≤qg≤¯¯¯qg, (18b)

with

 c(qg)= (pg)⊤Wppg+(qg)⊤Wqqg +β1∥[v–−v]+∥2+β2∥[v−¯¯¯v]+∥2,

where , are non-negative diagonal matrices, returns a non-negative vector, and are non-negative weights. The first two terms of are the cost of active and power injections, while the last two terms penalize the violation of constraint C2.

Constraint (18a) is the LinDistFlow model, which is used to predict the voltage magnitudes for given power injections. Note that and are measured before solving the voltage control problem. Solving (18) gives the optimal set-points for the DER active and reactive power injections.

## Iv Voltage Sensitivity Identifiability

In this section, we first introduce the path matrix associated with a graph and then analyze the conditions under which the line parameters and correspondingly voltage sensitivities, can be estimated.

### Iv-a Path Matrix

Let denote the set of lines that form the path from bus —referred to as the root—to bus . Since the power distribution system is radial, then is unique (see Theorem 2.1.4 in [13]). Bus is a leaf if there is no line that starts from it. We say bus is closer to the root than bus if , where denotes the cardinality of a set. Let denote the path matrix of , with () if line is on and their directions agree (disagree), and all other entries equal to zero. We choose the sending end of line to be the bus that is closer to the root, then all entries in are in by definition since the direction of and any line on it will always agree. Under this setup, the relation between and is given by the following lemma.

###### Proof.

Consider the entry at the row and column in , which is .

1. Consider first the case where . If line is not connected to bus , then . If line starts from bus , then and . If line ends at bus , then and . Obviously, there is one line that ends at bus . Moreover, such a line is unique since otherwise there will be two paths from the root to bus . Therefore, , where line is the line that ends at .

2. Next consider the case where . Similar to the previous case, we only need to consider the lines that starts from or ends at bus .

1. If line ends at bus , then . If , then and . If , then . In the latter case, there must exist a unique line that starts from . Then . Therefore, .

2. If line starts from bus , then . If , then and . If , then . In the latter case, there must exist a unique line that ends at . Similar to the previous argument, .

In sum, equals to if and otherwise; therefore, . ∎

The path matrix will play an important role in the analysis of the identifiability of the voltage sensitivities, which is to be detailed in the next section.

### Iv-B Identifiability Analysis

Before presenting the main results for the identifiability of voltage sensitivities, we introduce the concept of downstream buses.

###### Definition 1.

If line , , i.e., line is on the path from the root to bus , then bus is a downstream bus of line . The set of downstream buses of line is denoted by .

As discussed in Section III-B, needs to have full rank, i.e., , in order to estimate according to (11). When does not have full rank, some of the line parameters cannot be estimated from the measurements. The main results for the voltage sensitivity identifiability is stated as follows:

###### Theorem 1.

The parameter of line , , is identifiable if and only if the following condition is satisfied for some :

 ∑i∈Nℓzℓpi[k′]+qi[k′]≠0. (19)
###### Proof.

Using the path matrix, can be written as . Note that is the column of and is a rank-one matrix. Let , then . Then,

 Ξℓ=2P⊤eℓe⊤ℓP=2πℓπ⊤ℓ, (20)

and can be written as

 Ψ[k] =2⎡⎢ ⎢ ⎢⎣π1π⊤1ρ1[k−m]⋯πLπ⊤LρL[k−m]⋮⋮⋮π1π⊤1ρ1[k]⋯πLπ⊤LρL[k]⎤⎥ ⎥ ⎥⎦. (21)

Let , where and are the sets of lines that meet and do not meet the conditions in (19), respectively. If line , then ,

 ∑i∈Nℓzℓpi[k′]+qi[k′]=0. (22)

Note that the entry in is if and only if bus is a downstream bus of line . Essentially, the non-zero entries in , which are ones, indicate the downstream buses of line . Therefore, it follows from (22) that, :

 π⊤ℓρℓ[k′]=0. (23)

Consequently, all entries in the column of are zero, and the value of does not affect the objective function in (10). Under such condition, cannot be identified. For all line , we can remove the column of , the entry of and , and obtain an estimation problem of reduced size.

Next we show that the line parameter can be identified as long as condition (19) is satisfied. Without loss of generality, we assume since otherwise we can remove the unidentifiable variables to obtain a reduced problem that satisfied this condition. Then, (23) is satisfied for all and for some . Assume , then there exist , which are not all zero, such that

 Ψ[k][a1,⋯,aL]⊤=0mL, (24)

where is an -dimensional all-zeros vector. Without loss of generality, assume are not zero, while are all zero, where . Then, it follows from (21) and (24) that

 L′∑l=1alπ⊤lρl[k′]πl=0L. (25)

Since are linear independent, then for . However, since for any there exists some such that , then for , which leads to a contradiction. Therefore, and the line parameters can be identified. ∎

###### Remark 1.

The voltage sensitivity matrices can be readily computed if all line parameters can be identified. If some line parameter cannot be identified, the resulting voltage sensitivity matrices may not be accurate. This, however, will not have any impact on estimating voltage magnitudes since in such cases the line parameter does not affect the voltage magnitudes anyway. Specifically, if follows from (1), (III-B1), and (7) that

 ~v[k′]=Rp[k′]+Xq[k′] =2L∑ℓ=1πℓπ⊤ℓρ[k′]xℓ, (26)

in which if cannot be identified. Therefore, for the purpose of achieving voltage control, the proposed voltage sensitivity estimation algorithm is effective.

If we think of as some “combined power” (in the sense that it is a combination of active and reactive power), then (19) essentially indicates that the sum of combined power injection at all downstream buses of line is nonzero, or equivalently, there exists some combined power flow on line . For any line whose receiving end is a leaf, its parameter can be identified as long as the combined power injection at the receiving end is nonzero. Condition (19) can be easily satisfied in actual power distribution systems.

## V Numerical Simulations

In this section, we validate the effectiveness of the proposed framework using a modified single-phase IEEE 123-bus distribution test feeder from [15], the topology of which is shown in Fig. 2. There are six switches in this feeder, four of which are normally closed and the other two open so as to maintain a radial structure of the system. Under Assumption A2, this feeder has nine possible topology configurations as listed in Table I, among which configuration is the nominal one.

The loads are simulated in the following way. First, historical hourly loads of a residential building in San Diego [16] are interpolated to increase the time granularity to second. A zero-mean Gaussian noise with a standard deviation of  p.u. is also added to the interpolated loads, which are then scaled to match the active and reactive power load levels in the feeder. Four DERs are added at buses 76, 97, 105, 112, respectively, with reactive power outputs within  kVAr. We set and , for . For simplicity, we assume the DERs do not output any active power, i.e., . The minimum and maximum voltage magnitudes are p.u. and p.u., respectively. In addition, . Unless otherwise specified, the discounter factor is set to , and the underlying topology configuration is configuration , i.e., the nominal one. While we assume the power distribution system is lossless for the analysis, in the simulation, we use a full nonlinear power flow model that is solved using Matpower [17].

### V-a Estimation Accuracy

Throughout this part, the DERs do not inject any reactive power into the power distribution system.

#### V-A1 Noise-free case

We first evaluate the accuracy of proposed estimation algorithm in the case where the measurements are noise-free. The algorithm is evaluated in Monte Carol runs under various loading conditions. In each simulation run, sets of measurements are used to compute the residual error. Residual errors are computed for each feasible topology configuration in , while the underlying true topology configuration is one of them. A box-plot of residual errors associated with each topology configuration when the underlying topology configuration is configuration , is shown in Fig. 3. Note that residual errors associated with topology configurations are at least one order of magnitude larger than those of the other configurations, and are hence not plotted. This is the case where the residual error differences between each topology configuration is the smallest. Yet, it is still obvious that the true topology configuration results in the minimum residual error, which is one order of magnitude smaller than those of other configurations.

The parameter estimation accuracy is evaluated using the mean absolute percentage error (MAPE) of the estimates. When set of measurements is utilized, a typical MAPE of is , and that of is , both of which are really small. We note that the loading conditions of the power distribution system does not affect the accuracy of the proposed algorithm. The -to- ratios of all lines are assumed to be known. Figure 4 shows that the MAPE is almost linear with respect to the -to- ratio errors. Therefore, relatively small error in the -to- ratios will not result in a significant increase in parameter estimator errors.

#### V-A2 Noisy case

To see the impacts of measurement noise, we add white Gaussian noise to measurements such that the signal-to-noise ratio (SNR) is  dB, which is adopted by authors in [18]. More measurements are required to obtain a good estimation accuracy in the presence of measurement noise. The algorithm is again evaluated in Monte Carol runs under the same setup as the noise-free case, except that sets of measurements—corresponding to measurements collected in minute—are used to compute the residual error. A box-plot of residual errors associated with each topology configuration is shown in Fig. 5. Note that residual errors associated with topology configurations are one order magnitude larger than those of the other configurations, and are hence not plotted. The true topology configuration, i.e., configuration , still results in the minimum residual error. We note that increasing more measurements generally lead to higher accuracy in identifying the topology configuration.

The number of measurements have a direct impacts on the estimation accuracy. As is shown in Fig. 6, the MAPE of drops quickly when increasing the number of measurements, approximately from with set of measurements to with sets of measurements. The MAPE of —which is what really matters—is relatively insensitive to the number of measurements, with an MAPE around . Indeed, this result illustrates the high effectiveness and efficiency of the proposed estimation algorithm.

Figure 7 shows the impacts of SNR on parameter estimation accuracy when sets of measurements are used. When the SNR is beyond  dB, the MAPE of the voltage sensitivity matrix is within , which is relatively small. In the rest of of the simulation, we assume a SNR of  dB for all measurements.

#### V-A3 Accuracy under topology reconfiguration

The proposed algorithm works well not only under a fixed topology configuration but also when topology reconfiguration occurs. To illustrate this, we simulate a case where the underlying topology configuration is changed from configuration to configuration at  s.  sets of measurements are used to compute the voltage sensitivities, i.e., . The discount factor is set to . The minimum residual error and the corresponding estimated topology configuration are shown in Figs. 8 and 9, respectively. A jump in the minimum residual error is observed when the topology is reconfigured. The new topology is successfully identified after  s. Correspondingly, the MAPE of is also reduced to less than after  s, as is shown in Fig. 10.

### V-B Voltage Control Performance

Next, we show the performance of the proposed voltage regulation framework in the same case as the one in the previous section with topology reconfiguration, where the underlying topology configuration is changed from configuration to configuration at  s. A mode-based voltage regulation scheme, which assumes the true voltage sensitivity matrices are known but is not aware of the topology reconfiguration, is used for the purpose of comparison. The voltage profiles with the model-based and the proposed voltage regulation schemes are presented in Figs. 11 and 12, respectively, and the DER reactive power injections are shown in Fig. 13. It is obvious that the proposed data-driven voltage regulation framework is very effective and efficient in restoring the voltage magnitudes to the desirable range. This illustrates the strong adaptivity of our voltage regulation framework to system condition changes such as topology reconfiguration.

## Vi Concluding Remarks

In this paper, we proposed a data-driven voltage regulation framework for DERs in a balanced radial power distribution system. This framework utilizes a linear model that approximates the nonlinear relation between the voltage magnitudes and power injections, and estimates its parameters—the so-called voltage sensitivities—indirectly by estimating the topology configuration and the corresponding line parameters. In particular, the proposed estimation algorithm for the voltage sensitivities requires much fewer data than existing ones by exploiting the structural characteristics of the power distribution system. Using the estimated voltage sensitivities, the optimal DER power injections can be readily determined by solving a convex optimization problem.

Theoretical analysis shows that the voltage sensitivities of interest are easily identifiable. The inherent data-driven nature of the framework makes it adaptive to changes in system conditions, such as topology reconfigurations. Numerical simulations illustrated that the voltage sensitivities can be estimated accurately using a few set of measurements even under topology reconfiguration; consequently, guaranteeing good voltage regulation performance.

## References

• [1] W. H. Kersting, Distribution system modeling and analysis.   CRC press, 2006.
• [2] B. A. Robbins, C. N. Hadjicostis, and A. D. Domínguez-García, “A two-stage distributed architecture for voltage control in power distribution systems,” IEEE Trans. Power Syst., vol. 28, no. 2, pp. 1470–1482, 2013.
• [3] B. A. Robbins and A. D. Domínguez-García, “Optimal reactive power dispatch for voltage regulation in unbalanced distribution systems,” IEEE Trans. Power Syst., vol. 31, no. 4, pp. 2903–2913, 2016.
• [4] K. Zhang, W. Shi, H. Zhu, and T. Basar, “Distributed equilibrium-learning for power network voltage control with a locally connected communication network,” in Proc. of Annual American Control Conference, June 2018, pp. 3092–3097.
• [5] M. E. Baran and F. F. Wu, “Network reconfiguration in distribution systems for loss reduction and load balancing,” IEEE Trans. Power Del., vol. 4, no. 2, pp. 1401–1407, 1989.
• [6] Y. C. Chen, A. D. Domínguez-García, and P. W. Sauer, “Measurement-based estimation of linear sensitivity distribution factors and applications,” IEEE Trans. Power Syst., vol. 29, no. 3, pp. 1372–1382, 2014.
• [7] J. Zhang, X. Zheng, Z. Wang, L. Guan, and C. Chung, “Power system sensitivity identification–inherent system properties and data quality,” IEEE Trans. Power Syst., vol. 32, no. 4, pp. 2756–2766, 2017.
• [8] H. Xu, A. D. Domínguez-García, and P. W. Sauer, “Adaptive coordination of distributed energy resources in lossy power distribution systems,” in Proc. of IEEE PES General Meeting, Aug. 2018, pp. 1–5.
• [9] C. Mugnier, K. Christakou, J. Jaton, M. De Vivo, M. Carpita, and M. Paolone, “Model-less/measurement-based computation of voltage sensitivities in unbalanced electrical distribution networks,” in Proc. of Power Systems Computation Conference.   IEEE, 2016, pp. 1–7.
• [10] J. Zhang, Z. Chen, C. He, Z. Jiang, and L. Guan, “Data-driven based optimization for power system var-voltage sequential control,” IEEE Trans. Ind. Informat., 2018.
• [11] J. Zhang, Z. Chen, and C. He, “Identification of var-voltage characteristics based on ambient signals,” IEEE Trans. Power Syst., vol. 33, no. 3, pp. 3202–3203, 2018.
• [12] H. Xu, A. D. Domínguez-García, and P. W. Sauer, “A data-driven voltage control framework for power distribution systems,” in Proc. of IEEE PES General Meeting, Aug. 2018, pp. 1–5.
• [13] D. B. West et al., Introduction to graph theory.   Prentice hall Upper Saddle River, 2001, vol. 2.
• [14] R. B. Bapat, Graphs and matrices.   Springer, 2010, vol. 27.
• [15] IEEE distribution test feeders. [Online]. Available: https://ewh.ieee.org/soc/pes/dsacom/testfeeders/
• [16] Commercial and residential hourly load profiles for all TMY3 locations in the United States. [Online]. Available: https://openei.org/doe-opendata/dataset
• [17] R. D. Zimmerman, C. E. Murillo-Sánchez, R. J. Thomas et al., “Matpower: Steady-state operations, planning, and analysis tools for power systems research and education,” IEEE Trans. Power Syst., vol. 26, no. 1, pp. 12–19, 2011.
• [18] L. Xie, Y. Chen, and P. R. Kumar, “Dimensionality reduction of synchrophasor data for early event detection: Linearized analysis,” IEEE Trans. Power Syst., vol. 29, no. 6, pp. 2784–2794, 2014.
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text:
# unibiv

unibiv has the purpose of detecting univariate and bivariate outliers

## Syntax

• fre=unibiv(Y)example
• fre=unibiv(Y,Name,Value)example

## Description

 fre =unibiv(Y) unibiv with all default options.

 fre =unibiv(Y, Name, Value) unibiv with optional arguments.

## Examples

expand all

### unibiv with all default options.

Run this code to see the output shown in the help file

    n=500;
p=5;
randn('state', 123456);
Y=randn(n,p);
[out]=unibiv(Y);

### unibiv with optional arguments.

Stack loss data.

    Y=load('stack_loss.txt');
% Show robust confidence ellipses
out=unibiv(Y,'plots',1,'textlab',1);

## Input Arguments

### Y — Input data. Matrix.

n x v data matrix; n observations and v variables. Rows of Y represent observations, and columns represent variables.

Missing values (NaN's) and infinite values (Inf's) are allowed, since observations (rows) with missing or infinite values will automatically be excluded from the computations.

Data Types: single|double

### Name-Value Pair Arguments

Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside single quotes (' '). You can specify several name and value pair arguments in any order as  Name1,Value1,...,NameN,ValueN.

Example:  'rf',0.99 , 'robscale',2 , 'plots',2 , 'textlab',0 , 'tag','new_tag' , 'madcoef',2

### rf —It specifies the confidence level of the robust bivariate ellipses.scalar.

0<rf<1.

The default value is 0.95 that is the outer contour in presence of normality for each ellipse should leave outside 5% of the values.

Example:  'rf',0.99

Data Types: double

### robscale —how to compute dispersion.scalar.

It specifies the statistical indexes to use to compute the dispersion of each variable and the correlation among each pair of variables.

robscale=1 (default): the program uses the median correlation and the MAD as estimate of the dispersion of each variable;

robscale=2: the correlation coefficient among ranks is used (Spearman's rho) and the MAD as estimate of the dispersion of each variable;

robscale=3: the correlation coefficient is based on Kendall's tau b and the MAD as estimate of the dispersion of each variable;

robscale=4: tetracoric correlation coefficient is used and the MAD as estimate of the dispersion of each variable;

otherwise the correlation and the dispersion of the variables are computed using the traditional (non robust) formulae around the univariate medians.

Example:  'robscale',2

Data Types: double

### plots —Plot on the screen.scalar.

It specifies whether it is necessary to produce a plot with univariate standardized boxplots on the main diagonal and bivariate confidence ellipses out of the main diagonal. If plots is equal to 1 a plot which contains univariate standardized boxplots on the main diagonal and bivariate confidence ellipses out of the main diagonal is produced on the screen. If plots is <> 1 no plot is produced. As default no plot is produced.

Example:  'plots',2

Data Types: double

### textlab —plot labels.scalar.

Scalar which controls the labels in the plots. If textlab=1 and plots=1 the labels associated to the units which are univariate outliers or which are outside the confidence levels of the contours are displayed on the screen.

Example:  'textlab',0

Data Types: double

### tag —plot tag.character.

It identifies the handle of the plot which is about to be created. The default is to use tag 'pl_unibiv'. Notice that if the program finds a plot which has a tag equal to the one specified by the user, then the output of the new plot overwrites the existing one in the same window else a new window is created.

Example:  'tag','new_tag'

Data Types: char

### madcoef —scaled MAD.scalar.

Coefficient which is used to scale MAD coefficient to have a robust estimate of dispersion. The default is 1.4815 so that 1.4815*MAD(N(0,1))=1.

Remark: if mad =median(y-median(y))=0 then the interquartile range is used. If also the interquartile range is 0 than the MD (mean absolute deviation) is used. In other words MD=mean(abs(y-mean(Y))

Example:  'madcoef',2

Data Types: double

## Output Arguments

### fre —Details about the univariate and bivariate outliers. n -by- 4 matrix

1st col = index of the units;

2nd col = number of times unit has been declared univariate outliers;

3rd col = number of times unit has been declared bivariate outlier;

4th col = pseudo MD as sum of bivariate MD.

## References

Riani, M., Zani S. (1997). An iterative method for the detection of multivariate outliers, Metron, vol. LV, pp. 101-117.

format: plain


text:
# Homomorphisms

## Definition and Examples

See this video and these notes.

Definition: A homomorphism $\phi:G\to H$ is a function that satisfies the property $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$ for all $g_1,g_2\in G$. An isomorphism is a homomorphism that is bijective.

A homomorphism is a map that gives a partial relation between the structure of $G$ and $H$.

Examples:

• Let $G=S_{n}$ and $H=\mathbb{Z}_{2}$. Define $\phi(\sigma) = \begin{cases} 0 & \hbox{if \sigma is an even permutation} \\ 1 & \hbox{if \sigma is an odd permutation} \end{cases}$ Then $\phi$ is a homomorphism.

• Let $G=\mathrm{GL}_{2}(\mathbb{R})$ and let $H=\mathbb{R}^{\times}$. Then $\phi(g)=\mathrm{det}(g)$ is a homomorphism.

• Let $H$ be any group and let $h\in H$ be an element. Define $\phi:\mathbb{Z}\to H$ by $\phi(n)=h^{n}$. Then $\phi$ is a homomorphism.

• Let $G=\mathbb{R}$ and $H=\mathbb{T}$, the group of complex numbers of norm $1$ with multiplication. Then the map $\phi(r)=\mathrm{cis}(r)=\cos(r)+i\sin(r)$ is a homomorphism.

## Key Properties of Homomorphisms

See this video and these notes.

Proposition: (Proposition 11.4 of the text) Let $\phi: G_1\to G_2$ be a homomorphism. Then:

• If $e_1$ is the identity of $G_1$, $\phi(e_1)$ is the identity of $G_2$.
• If $g_1$ is an element of $G_1$, then $\phi(g_1^{-1})$ is the inverse $\phi(g_1)^{-1}$ of $\phi(g_1)$.
• If $H_1$ is a subgroup of $G_1$, then the image $\phi(H_1)$ is a subgroup of $G_2$.
• If $H_2$ is a subgroup of $G_2$, then the preimage $\phi^{-1}(H_2)$ is a subgroup of $G_1$.
• If $H_2$ is a normal subgroup of $G_2$, then the preimage $\phi^{-1}(H_2)$ is a normal subgroup of $G_1$.

## Kernels

See this video and these notes

Definition: The kernel of a homomorphism $\phi:G\to H$ is the preimage of the identity of $H$: $\mathrm{Ker}(\phi)=\phi^{-1}(\{e_H\}).$

Proposition: (Theorem 11.5 of the text) The kernel of a homomorphism $\phi:G\to H$ is a normal subgroup of $G$.

• The kernel of the determinant is $\mathrm{SL}_{2}(\mathbb{R})$.
• The kernel of the map $\phi: \mathbb{Z}\to G$ given by $\phi(n)=g^{n}$ is either ${0}$, if $g$ has infinite order, or the subgroup $k\mathbb{Z}$ where $k$ is the order of $g$ in $G$.
• The kernel of the map $\mathrm{cis}:\mathbb{R}\to\mathbb{T}$ are the integer multiples of $2\pi$ in $\mathbb{R}$.

## Isomorphism Theorems

See this video and these notes.

Let $G$ be a subgroup and $H$ be a normal subgroup of $G$. Then the map $\phi: G\to G/H$ defined by $\phi(g)=gH$ is a homomorphism called the natural homomorphism or the canonical homomorphism.

Theorem: (First isomorphism theorem) Suppose $\psi:G\to H$ is a group homomorphism and $K$ is the kernel of $\psi$. Let $\phi:G\to G/H$ be the canonical homomorphism. Then there exists a unique isomorphism $\eta:G/K\to \psi(G)$ such that $\psi=\eta\phi$.

In words: every homomorphism $\psi$ from $G$ to $H$ is a composition of two steps:

• first, you take the natural map from $G$ to the quotient $G/K$ where $K$ is the kernel of $\psi$;
• then you have an isomorphism from $G/K$ to a subgroup of $H$.

Or, put another way, the image of a homorphism of a group $G$ is isomorphic to a quotient of $G$.

Theorem: (Second isomorphism theorem) Let $H$ be a subgroup of $G$, not necessarily normal in $G$, and $N$ a normal subgroup of $G$. Then $HN$ is a subgroup of $G$, $H\cap N$ is a normal subgroup of $H$, and there is an isomorphism $\phi$ $\phi: H/(H\cap N) \to HN/N.$
Theorem: (Correspondence Theorem) Let $N$ be a normal subgroup of $G$. Then $H\mapsto H/N$ is a bijection between the set of subgroups of $G$ that contain $N$ and the set of subgroups of $G/N$. In this correspondence, normal subgroups correspond to normal subgroups.
Theorem: (Third isomorphism theorem) Let $G$ be a group and $N$ and $H$ be normal subgroups of $G$ with $N\subset H$. Then $G/H\stackrel{\sim}{=}\frac{G/N}{H/N}.$


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# Force and Simple Dynamics: Pushy Astronaut

1. Jan 28, 2009

### mrnastytime

1. The problem statement, all variables and given/known data
An astronaut of mass 65 kg pushes off a satellite of mass 420 kg, exerting a force of 120 N for the 0.94 s they are in contact.

2. Relevant equations
What is the speed of the astronaut after they have separated?
Ans: 1.84 m/s
What is the speed of the satellite after they have separated?
Ans: .268 m/s
How far apart are they after 1 minute?
d=??? m

3. The attempt at a solution
This is my attempt. I used the position as a function of time formula:
$$x = x_0 + v_0 t + (1/2) a t^2$$
For the astronaut: x= 0+0+.5(1.84 m/s^2)(60 s)^2= 3312 m
For the satelite: x= 0+0+.5(.268 m/s^2)(60)^2= 513 m
In order to get the separation, i subtracted the two and got 2799 m. But this is incorrect according to my online homework program. did i make a mistake or use the wrong formula?

2. Jan 28, 2009

### rl.bhat

What is the speed of the astronaut after they have separated?
Ans: 1.84 m/s
What is the speed of the satellite after they have separated?
Ans: .268 m/s

How did you get these two values?

3. Jan 29, 2009

### mrnastytime

accelration of satelite=F/m=120N/420kg=0.285m/s^2
accelration of astronaut=F/m=120N/65kg=1.84 m/s^2

To find velocity, i used the velocity as a function of time equation:
V=V0+at

astronaut: V=0+1.84 m/s^2(0.94s)=1.74 m/s
satelite: V=0+0.285 m/s^2(0.94s)=0.268 m/s


text:
Home » » Integrate $$4x^{3} + \frac 1 x$$ with respect to $$x$$

# Integrate $$4x^{3} + \frac 1 x$$ with respect to $$x$$

### Question

Integrate $$4x^{3} + \frac 1 x$$ with respect to $$x$$

### Options

A) $$\ln x$$ + $$x^{4}$$ + K

B) $$x^{-1}4$$ + $$x^{4}$$ + K

C) $$12x^{2}$$ - $$x^{-2}$$ + K

D) $$\frac 1 5x^{5}$$ + $$x^{-2}$$ + K

### Explanation:

$$f(x) = 4x^3 + \cfrac{1}{x}$$
$$\int \left( 4x^3 + \cfrac{1}{x} \right)dx = \cfrac{4x^{3+1}}{3+1} + \ln x + c$$
$$=\cfrac{4x^4}{4}+\ln x + c$$
$$=x^4+\ln x + C$$

## Dicussion (1)

• $$f(x) = 4x^3 + \cfrac{1}{x}$$
$$\int \left( 4x^3 + \cfrac{1}{x} \right)dx = \cfrac{4x^{3+1}}{3+1} + \ln x + c$$
$$=\cfrac{4x^4}{4}+\ln x + c$$
$$=x^4+\ln x + C$$


text:
## Bananas1234 one year ago Aaron is solving the equation x2 – x = –2 with the quadratic formula. What values should he use for a, b, and c? a = 1, b = –1, c = –2 a = –1, b = 1, c = –2 a = 1, b = –1, c = 2 a = –1, b = 1, c = 2

1. Nnesha

first of all set that equation equal to 0

2. Bananas1234

how do i do that?

3. Nnesha

transfer -2 to the left side how would you cancel from the right side ?

4. Bananas1234

5. Nnesha

6. Bananas1234

ok

7. Nnesha

what do you get ?

8. Bananas1234

2x - x + 2 + 0

9. Bananas1234

= 0 i ment

10. Nnesha

yes right now quadratic e question is $\huge\rm \color{reD}{A}x^2+\color{Red}{B}x+\color{reD}{C}=0$

11. Bananas1234

ok

12. Nnesha

so a= ? b=? c=?

13. Bananas1234

a = 1, b = –1, c = 2

14. Nnesha

yes right

15. Nnesha

btw $$\color{blue}{\text{Originally Posted by}}$$ @Bananas1234 2x - x + 2 + 0 $$\color{blue}{\text{End of Quote}}$$ this is not right it suppose to be x^2 -x +2

16. Bananas1234

ok, thank you

17. Nnesha

yo

18. Nnesha

yw*

Find more explanations on OpenStudy


text:
# 2.1 Properties of matter

 Page 1 / 1

## The properties of matter

Let us now look at what we have learned about chemical bonds, intermolecular forces and the kinetic theory of matter, and see whetherthis can help us to understand some of the macroscopic properties of materials.

1. Melting point
Melting point
The temperature at which a solid changes its phase or state to become a liquid . The process is called melting and the reverse process (change in phase from liquidto solid) is called freezing .
In order for a solid to melt, the energy of the particles must increase enough to overcome the bonds that are holding the particlestogether. It makes sense then that a solid which is held together by strong bonds will have a higher melting point than one where the bonds are weak, because more energy (heat) is needed to breakthe bonds. In the examples we have looked at metals, ionic solids and some atomic lattices (e.g. diamond) have high melting points, whereas the meltingpoints for molecular solids and other atomic lattices (e.g. graphite) are much lower. Generally, the intermolecular forces between molecular solids are weaker than those between ionic and metallic solids.
2. Boiling point
Boiling point
The temperature at which a liquid changes its phase to become a gas . The process is called evaporation and the reverse process is called condensation
When the temperature of a liquid increases, the average kinetic energy of the particles also increases and they are able to overcomethe bonding forces that are holding them in the liquid. When boiling point is reached, evaporation takes place and some particles in the liquid become a gas. In other words, the energy of theparticles is too great for them to be held in a liquid anymore. The stronger the bonds within a liquid, the higher the boiling point needs to be in order tobreak these bonds. Metallic and ionic compounds have high boiling points while the boiling point for molecular liquids is lower.The data in [link] below may help you to understand some of the concepts we have explained. Not all of the substances in the table aresolids at room temperature, so for now, let's just focus on the boiling points for each of these substances. What do you notice?
 Substance Melting point ( ${}^{°}\mathrm{C}$ ) Boiling point ( ${}^{°}\mathrm{C}$ ) Ethanol ( ${\mathrm{C}}_{2}{\mathrm{H}}_{6}\mathrm{O}$ ) - 114,3 78,4 Water 0 100 Mercury -38,83 356,73 Sodium chloride 801 1465
You will have seen that substances such as ethanol, with relatively weak intermolecular forces, have the lowest boiling point, while substances withstronger intermolecular forces such as sodium chloride and mercury, must be heated much more if the particles are to have enough energy to overcome theforces that are holding them together in the liquid. See the section below for a further exercise on boiling point.
3. Density and viscosity
Density and viscosity is not in CAPS - Included for Completeness
Density
Density is a measure of the mass of a substance per unit volume.
The density of a solid is generally higher than that of a liquid because the particles are held much more closely together and therefore thereare more particles packed together in a particular volume. In other words, there is a greater mass of the substance in a particular volume. In general, densityincreases as the strength of the intermolecular forces increases.
Viscosity
Viscosity is a measure of how resistant a liquid is to flowing (in other words, how easy it is to pour the liquid from one container toanother).
Viscosity is also sometimes described as the 'thickness' of a fluid. Think for example of syrup and how slowly it pours from one container into another. Now compare this to how easy it is to pour water. The viscosity ofsyrup is greater than the viscosity of water. Once again, the stronger the intermolecular forces in the liquid, the greater its viscosity.

It should be clear now that we can explain a lot of the macroscopic properties of matter (i.e. the characteristics we can see or observe) by understanding their microscopic structure and the way in which the atoms and molecules that make up matter are held together.

## Exercise: forces and boiling point

The table below gives the molecular formula and the boiling point for a number of organic compounds called alkanes (more on these compounds in grade 12). Refer to the table and then answer the questions that follow.

 Organic compound Molecular formula Boiling point ( ${}^{°}\mathrm{C}$ ) Methane ${\mathrm{CH}}_{4}$ -161.6 Ethane ${\mathrm{C}}_{2}{\mathrm{H}}_{6}$ - 88.6 Propane ${\mathrm{C}}_{3}{\mathrm{H}}_{8}$ -45 Butane ${\mathrm{C}}_{4}{\mathrm{H}}_{10}$ -0.5 Pentane ${\mathrm{C}}_{5}{\mathrm{H}}_{12}$ 36.1 Hexane ${\mathrm{C}}_{6}{\mathrm{H}}_{14}$ 69 Heptane ${\mathrm{C}}_{7}{\mathrm{H}}_{16}$ 98.42 Octane ${\mathrm{C}}_{8}{\mathrm{H}}_{18}$ 125.52
Data from: http://www.wikipedia.com
1. Draw a graph to show the relationship between the number of carbon atoms in eachalkane and its boiling point. (Number of carbon atoms will go on the x-axis and boiling point on the y-axis).
2. Describe what you see.
3. Suggest a reason for what you have observed.
4. Why was it enough for us to use 'number of carbon atoms' as a measure of the molecular weight of the molecules?

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
The fundamental frequency of a sonometer wire streached by a load of relative density 's'are n¹ and n² when the load is in air and completly immersed in water respectively then the lation n²/na is
Properties of longitudinal waves


text:
## Complex Analysis – Prove the integral of \$ f (z) = frac {1-e ^ {2iz}} {4z ^ 2} \$ tends to \$ 0 \$ on the curve \$ gamma = Re ^ {it} \$ as \$ R rightarrow { infty} \$

I am asked to prove that the integral of $$f (z) = frac {1-e ^ {2iz}} {4z ^ 2}$$ on the curve $$gamma = Re ^ {it}$$ $$0 leq t leq pi$$, when $$R rightarrow { infty}$$, go to zero. To do this I've tried $$int_ gamma f (z) dz = int_0 ^ { pi} f ( gamma) gamma dt = int_0 ^ { pi} frac {1-e ^ {2iRe ^ {it }}} {4 (Re ^ {it}) ^ 2} Rie ^ {it} dt = int_0 ^ { pi} frac {1-e ^ {2iRe ^ {it}} {4R ^ 2e ^ { it}} idt$$
And now I need to see that $$lim_ {R to infty} int_0 ^ { pi} frac {1-e ^ {2iRe ^ {}}} {4} ^ {it}} idt = 0$$
However, I do not know how to solve this integral, so I can not continue.

## calculation – Proof of the Dirac delta screening property with Integral Volume

The delta function of Dirac has the sieving property which says,

$$int _ {a} ^ {b} f (x) delta (x-x) dx = begin {case} f (x) & a

I suspect by analogy that it also has a similar property for volume integrals,

$$int _ {V} f ( mathbf {r}) delta ( mathbf {r} – mathbf {r} & # 39;) d ^ {3} mathbf {r} & # 39; = begin {case} f ( mathbf {r}) & mathbf {r } in V \ 0 & otherwise end {cases}.$$

Can any one confirm this property – how could I prove it?

## abstract algebra – Is a R module free from torsion and finite on a free base free on R if R is an integral domain?

Let $$R$$ to be a Dedekind domain, which is not a prime ideal domain, for example
the ring of integers in a number field such that $$R = Bbb Z[sqrt{-6}]$$.
then $$R$$ will have a non-main ideal, $$I$$ say. Then, as $$R$$-module, $$I$$ is finely generated,
without torsion, but not main. In the example above, we could take $$I = 2R + sqrt {-6} R$$.

## integration – The curve integral of the second type (can I change the order from \$ dz \$ to \$ dx \$ to write?)

Given an oriented curve $$Sigma$$ which is smooth, suppose that the function $$f (x, y, z)$$ is continuous on $$Sigma$$can the curve integral of the second type be
$$iint _ { Sigma} f (x, y, z) dzdx$$
can be written as
$$iint _ { Sigma} f (x, y, z) dxdz.$$
I mean I can change the order of $$dz$$ and $$dx$$ in writing?

Thank you.

## Integral sum of n radicals

For positive integers n, S (n) denotes the minimum value of the sum (sum) of k = 1 to n ((2k-1) ^ 2 + (a (k)) ^ 2)) ^ (1/2 where a (1), a (2), …. a (n) are positive real numbers whose sum is 17. if there exists a unique positive integer n for which S (n) is also an integer, then (n / 2) is

## Disappearance of the Integral Sound Function and Its Derivative

assume $$f in C ^ 1 (R ^ +)$$, $$int_0 ^ infty | f | ^ a < infty$$ and $$| f & # 39; < M$$ for some positive $$M$$ and $$a$$. Is it true that $$f (x) to0$$ as $$x to infty$$?

I've tried to use the integration by parts $$int_0 ^ x f (t) = xf (x) – int_0 ^ x tf (t) , dt$$
but in vain.

## Why is the \$ x ^ 2 \$ integral of \$ 0 to \$ 5 not equal to the \$ sqrt x \$ integral of \$ 0 to \$ 25?

My calculator says the first is 41.67 and the last is 83.34. Why is that? Should not they be equal?

## calculation and analysis – Integral limit exchange problems

I have an integrand that is a function of $$x in (t1, t2), y in (0, t1)$$

``````P6[x_, y_, t1_, t2_] = (x * y * baby cot[(t1 - t2)/2]* Crib[(x - y)/2]^ 2 * Csc[(x - y)/2]^ 2) / 2
``````

Now I integrate with x first, then with y and the plot,

``````I6x[x_, y_, t1_, t2_] = Integrate[P6[x, y, t1, t2], X]I6yx[x_, y_, t1_, t2_] = Integrate[I6x[x, y, t1, t2], y]Manipulate[
Plot[{Re[I6yx[x, y, [Pi]/ 3, 2 [Pi]/ 3]],
I am[I6yx[Xy[I6yx[Xy[I6yx[xy[I6yx[xy[Pi]/ 3, 2 [Pi]/ 3]]}, {X, [Pi]/ 3, 2 [Pi]/ 3},
PlotRange -> {{-2, 2}, {-50, 50}}], {y, 0, [Pi]/ 3}]
``````

Now, I exchange the integration command,

``````I6y[x_, y_, t1_, t2_] = Integrate[P6[x, y, t1, t2], y]I6xy[x_, y_, t1_, t2_] = Integrate[I6x[x, y, t1, t2], X]Manipulate[
Ground[{Re[I6xy[Xy[{Re[I6xy[Xy[{Ré[I6xy[xy[{Re[I6xy[xy[Pi]/ 3, 2 [Pi]/ 3]],
I am[I6xy[Xy[I6xy[Xy[I6xy[xy[I6xy[xy[Pi]/ 3, 2 [Pi]/ 3]]}, {X, [Pi]/ 3, 2 [Pi]/ 3},
PlotRange -> {{-2, 2}, {-50, 50}}], {y, 0, [Pi]/ 3}]
``````

Do I get different behavior from plots? Why is it? More importantly, I incorporate real functions on real values, why do I have an imaginary value?

## Quadratic equation with integral coefficient

Let $$a, b, c$$ to be natural numbers, such as the roots of the equation $$ax ^ 2 + bx + c = 0$$ are distinct and both are in the meantime

1. (0.1)
2. (1,2)
3. (2,3)

Find the minimum possible value of $$a, b, c.$$

For my part, I resolved for part 1, that is to say for separate roots between (0,1). But for the next two games, things get a bit too complicated.

Although there may be a similarity for a given part (0,1) in the stack exchange, there is no generalized method for solving other intervals of time. this type.

So please, help, I am new to stack the exchange.

## Fourier series of exp (x) and its integral

I have to find the Fourier coefficients of $$f (x) = text {exp} (x)$$ for $$-1 and evaluate the value of this series to $$x = 2$$.

I've calculated the coefficients to
$$a_n = frac {(e ^ 2-1)} {e pi ^ 2 n ^ 2 + e} (-1) ^ n$$ for the cosine-terms and
$$b_n = frac {(1-e ^ 2) pi n} {e pi ^ 2 n ^ 2 + e} (- 1) ^ n$$
for the sinus-terms.

My first question would be, if these are correct?

Assuming they are correct, I get the next set

$$f (x) = frac {a_0} {2} + sum_ {n geq1} frac {(- 1) ^ n} {e pi ^ 2 n ^ 2 + e} left ((e ^ 2-1) cos ( pi n x) + (1-e ^ 2) pi n sin ( pi n x) right)$$

Now, how can I calculate its value for $$x = 2$$? In this case, the series becomes a little easier, but I do not know how to calculate its value.

And finally, when I integrate the term run by term, I do not really recover the old series anymore.

I am very grateful for any kind of help!

Thank you!


text:
# Simple trig functions

1. Jan 6, 2010

### tomwilliam

1. The problem statement, all variables and given/known data
(In radians I assume) Using the exact values of sin, cos, tan of 1/3 pi and 1/6 pi, and the symmetry of the graphs of sin, cos and tan, find the exact values of sin(-1/6pi), cos (5/3pi) and tan(4/3pi).

2. Relevant equations
cos(x)=sin(x+pi/2)
sin(x) = sin(x+2pi)
sin^2(x)+cos^2(x)=1

3. The attempt at a solution

I have a little difficulty understanding what is required here. I have a list of trig functions but these don't seem to be here. I imagine the answer is some square root with a coefficient, but I'd like to understand it better. Can someone point me in the right direction without just giving me the answer? Thanks

2. Jan 6, 2010

### rock.freak667

Try the numbers as combination of π, for example to find tan(5π/4)=tan(π+π/4). Like that.

3. Jan 6, 2010

4. Jan 6, 2010

### vela

Staff Emeritus
The idea is to look at the graphs of functions to figure out properties that would be useful in solving the problems. For example, if you graph $$\sin x$$, you'll see that it's an odd function; in other words, it has the property $$\sin(x) = -\sin(-x)$$. If $$x=-\pi/6$$, you can say $$\sin(-\pi/6) = -\sin[-(-\pi/6)] = -\sin(\pi/6)$$.

5. Jan 6, 2010

### tomwilliam

Thanks guys. I guess I'm being slow, but I've had a few tries now and can't equate, e.g. sin(-pi/6) with either sin, cos, or tan of 1/3 or 1/6 pi.
For example:

sin(-pi/6) = -sin(pi/6)
fine.
cos(pi/6 + pi/2) = -sin(pi/6)
so
cos(4/6 pi) = sin(-pi/6)

but what I really want to equate the LHS to is
-cos(1/3 pi).
I know I'm being really slow, but I can't seem to get there.

6. Jan 6, 2010

### tomwilliam

The graph does help a lot.
I can see that sin(-1/6 pi) is equal to cos exactly half pi earlier in the graph. So is there an identity that states sin(x - pi/2)=cos(x)?
If so, I've cracked it.

7. Jan 6, 2010

### l'Hôpital

Sorta but not quite.

Notice if you put x = 0, you'd get
sin (-pi/2) = cos 0 = 1 which is wrong. The right identity.

It'd be
$$\sin (\frac{\pi}{2} - x ) = \cos x$$

8. Jan 6, 2010

### rock.freak667

should be sin(x-π/2)= -cos(x)

9. Jan 6, 2010

### vela

Staff Emeritus
Actually, I think Tom discovered the identity

$$sin(x+\frac{\pi}{2}) = cos x$$

which just says that the graph of cosine is the sine shifted by $$\pi/2$$. It's easy to screw up the sign of the phase shift because, to me, it seems like it should be the opposite sign intuitively. I always have to double-check to make sure I get it right.

10. Jan 6, 2010

### vela

Staff Emeritus
You probably can't because that's actually kind of hard to do. :)

The problem wants you to figure out what $$\cos(5\pi/3)$$ equals in terms of $$\cos(\pi/6)$$ or $$\cos(\pi/3)$$. Same with $$\tan(4\pi/3)$$ and $$\tan(\pi/6)$$ or $$\tan(\pi/3)$$.

11. Jan 6, 2010

### tomwilliam

Thanks all.
Well I think I'm getting there.
I think for the first one I can just say

sin(pi/6)=0.5
sin (-x) = -sin(x), therefore
sin(-pi/6)=-sin(pi/6)=-0.5

Simple.

For the second one, I'm guessing I have to use:
cos(pi/3)=0.5
and somehow show that it is equal to:
cos(5pi/3).

I can see the graph very clearly, and understand why this is the case. But I'm not sure how I can show it. I know that as the angle goes from 0 to pi, the cosine of the angle goes from 1 to -1, and then as the angle goes from pi to 2pi, the cosine goes back round from -1 to 1. So of course, the cosine at 1/3 of the way round to pi will be the same as it is at 2pi - 1/3pi. But how to show that mathematically?

Thanks again, and sorry for being slow-witted.

12. Jan 6, 2010

### rock.freak667

5π/3 = 2π- π/3

Use this.

13. Jan 6, 2010

### vela

Staff Emeritus
So essentially what you've figured out is that $$\cos x = \cos(2\pi - x)$$. This another one of the many trig identities. You can just say

$$\cos 5\pi/3 = \cos(2\pi-\pi/3) = \cos \pi/3 = ...$$


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## Saturday, August 22, 2020

### I need a half-tablespoon. So I have absolutely no choice but to accept complex numbers.

You never know what math problems will show up in applications. I once had a passing thought to include the following problem on a calculus test, but I sure as heck am glad I didn't (I have included variations which seemed just slightly different from this, but solving for a different variable). Anyway so here's the story. I use measuring spoons (and cups) when I cook (I may already cause offense by doing so, lol). Notably, my measuring spoons are hemispherical. I needed a half tablespoon of something. A bit of soy sauce, being the Real Asian™ that I am. I'm missing a few teaspoon measures, due to a mishap in the garbage disposer (oops). So not having my usual measures for 1½ teaspoons (or 3/2 teaspoons, since mathematicians like improper fractions), involving very easy math, like multiplication by 3, I decided, well, why don't I just fill the hemisphere to the exact height needed to get a volume of half a hemisphere? Easy peasy. Shouldn't be much harder than multiplying by 3, right? Well... turns out to involve the number 3 in something way, way more complicated.

Now the calculus problem that I have definitely put on exams is: if you fill a hemisphere (of radius $R$) to exactly half the height ($\frac{1}{2} R$), what's the volume of such a thing? Now of course, because the curve of the hemisphere changes with the height, filling to the halfway height point is going to give you less than half the volume. Realizing the volume as a stack of disks, the answer is

$\int_{R/2}^{R} \pi(R^2 - z^2) \; dz = \frac{2\pi}{3} R^3 - \int_0^{R/2} \pi(R^2 - z^2) \; dz = \frac{2\pi}{3} R^3 - \pi R^2 z + \frac{\pi}{3} z^3 = \frac{2\pi}{3}R^3 - \frac{\pi}{2} R^3 +\frac{\pi}{3\cdot 8} R^3 = \frac{5\pi}{24} R^3.$

Oh. I thought it involved $\sqrt{2} - 1$ or something. Oh well. My memory is not what it used to be, I guess. I think that may be the answer to my volume-of-a-Sno-Cone problem, lol, which is the volume of the segment of spherical cap cut off by a cone. I don't feel like busting out the spherical coordinates today, anyway, to do that problem, lol. Some other time!

Anyway, this is of course, not half a hemisphere's worth, which is $\frac{1}{3} \pi R^3$. To get that, we have to solve for the height. It is easier to measure from the top of the spoon ($xy$-plane) than from the bottom, so we solve

$\frac13 R^3 = \int_0^h \pi(R^2 - z^2) \; dz = hR^3-\frac{1}{3} \pi h^3.$

for $h$. Simple, right? Clearing denominators and moving stuff to the other side, and canceling off the pesky $\pi$, we end up with

$h^3 - 3R^3 h + R^3 = 0$.

A nontrivial cubic polynomial. As it happens, this is the case of the depressed cubic (insert many jokes/puns here and catch flack from the sensitivity police for making light of serious, real mental illnesses). Of course, using Mathematician's Privilege (dimensional consistency is such a ... physicist ... thing), we can just set our units to $1$ and be done with it; this gives us the polynomial

$h^3 -3h + 1 = 0$.

Now I could have also tried to have my college algebra students solve this one. Rational roots theorem and all that. Nope, not gonna work, throwing $1$ and $-1$ at it. It turns out that this one is a true casus irreducibilis. It being a depressed cubic, I don't have to deal with the Germans, and can skip directly to Italians with fancy cars (perhaps I'll "retcon" my story and say I was cooking a nice risotto and needed to use half a hemispherical teacup), the solution to this is

$\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} - 1}}$,

$\zeta\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \bar\zeta\sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} - 1}}$,

$\bar \zeta\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \zeta \sqrt[3]{-\frac{1}{2} -\sqrt{\frac{1}{4} - 1}}$, where $\zeta$ is a cube root of unity (please print these formulas out onto posters and hold them up to enforce social distancing). It happens that $\zeta = -\frac{1}{2} + \frac{\sqrt{3}}{2}\mathsf i$, which is exactly what is under the cube root, so we're dealing with ninth roots of unity here! (who would have thought??). The way to see that these formulas work, generally from the depressed cubic, is basically to substitute $x = w - \frac{p}{3w}$, which results in a quadratic for $w^3$. That it's quadratic explains the nearly identical features of the stuff under the cube roots, but for a sign, and that it's $w^3$ that gets solved for is responsible for the cube roots outside, as well as mutiplying by roots of unity.

 The cubic $y = x^3 - 3x + 1$. It has three real roots, but they only can be expressed algebraically by the sum of complex numbers!

So the three solutions are $\sqrt[3]{\zeta} + \sqrt[3]{\bar{\zeta}} = \zeta^{1/3} + \zeta^{-1/3}$, $\zeta^{4/3} + \zeta^{-4/3}$, and $\zeta^{-2/3} + \zeta^{2/3}$. Taking the cube root of a complex number is not a trivial matter. It happens that the imaginary parts all cancel out and you get three real solutions, but this cannot be done without admitting complex numbers in some form (after now consulting with the French). Well, almost, anyway. Let's call in trigonometry to save the day (but that's cheating, because trigonometry really is $e^{\mathsf i x}$). We'll take $\zeta = e^{2\pi\mathsf i/3}$ and the "principal" complex cube root to take minimum argument. Actually this is irrelevant, because the extra multiplications of the $\zeta$s give you the other cube roots anyway, but for definiteness... $\zeta^{1/3}$ then is $e^{2\pi\mathsf i / 9}$ and the possible solutions are

$e^{2\pi\mathsf i / 9} + e^{-2\pi\mathsf i / 9} = 2\cos(2\pi/9)$
$e^{8\pi\mathsf i / 9} + e^{-8\pi\mathsf i / 9} = 2\cos(8\pi/9)$, and
$e^{-4\pi\mathsf i / 9} + e^{4\pi\mathsf i / 9} = 2\cos(4\pi/9)$.

Ok, so which one of these solutions is the correct one? The solution has to be between $-1$ and $1$ (in our units; the original units would be between $-R$ and $R$), and as such, only $2 \cos(4\pi/9) \approx 0.3472$ (the middle one in the graph of the full cubic above) works. So to answer the question, you need to fill up the tablespoon measure so that it is 34.72% of the way down from the top. Here's another picture and zoomed in:

 A cross section of the hemisphere in question, together with a zoomed in view of the cubic curve (green). The vertical line (cyan) drawn through the cubic at its root will intersect the circle (red) where it needs to be filled (here you'd fill everything to the right of this line)

Again the fact that solving a cubic has to go through the complex numbers is why complex numbers were finally accepted. It is not, as commonly believed, due to trying to solve for the quadratic, since in that case, complex numbers only appear when the parabola fails to intersect the $x$-axis, meaning the complex solutions are of the impossible, nonexistent intersection points, so, the various Renaissance mathematicians thought, obviously should not be taken seriously. But the cubic forces it to happen no matter what, because a cubic curve always crosses the $x$-axis. So there it is: half a tablespoon? You'll need complex numbers for that. Some more views (the title image sort of is looking from above, so might obscure the fact you must fill it more than halfway from the bottom):

 View from below the tablespoon. The little extra disk sticking out is the fill line.

Finally, I'll end with more social distancing formulas: the full cubic formula $ax^3 + bx^2 + cx + d = 0$:

For those who are interested, here's a nice video explaining this formula, and why we don't teach it in high school...

It's not taught because the solutions you get may not be easily reducible to something you can recognize, even in simple cases. Conceptually, however, it is a real adventure, and solving it by breaking it into a 2-step process makes it tractable and interesting. Our tablespoon example already skips step 1 ... all that about Germans and Italians.

Enjoy and stay safe!


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# Tag Info

15

Here are the fish you wanted, that I caught with texdoc chemfig. \documentclass{article} \usepackage{chemfig} \begin{document} \setatomsep{2em} \setdoublesep{.3em} \renewcommand*\printatom[1]{\ensuremath{\mathsf{#1}}} \chemfig[line width=1pt] { HO-*6(-=-(-(-[::90]CH_3)(-[::-90]CH_3)-*6(-=-(-OH)=-=))=-=) } \end{document} Now your turn to fetch the ...

10

I introduce \lewis with 7 arguments, and I apologize that I don't know the precise naming conventions for chemistry. (EDITED to specify valence in only one place). Arguments: #1 Core atom #2 Top electrons #3 Right electrons #4 bottom electrons #5 left electrons #6 valence #7 inner electron shells \documentclass{article} ...

10

An example for this is given in the documentation of chemfig (part III section 12.5 Draw polymer element). There a macro \makebraces(<dim up>,<dim down>){<subscript>}{<opening node name>}{<closing node name>} is defined that exploits chemfig's @{<node name>,<pos>} syntax inside formulas to position delimiters on a ...

9

Here are a different versions using different chemistry packages. Which one you want to use is up to you... chemfig, provides \startscheme, \stopscheme, \chemfig, \lewis, \Lewis, \chemname ... ; mhchem, provides \ce{}; chemformula (part of the chemmacros bundle), provides \ch{} with the !(<below>)(<formula>) syntax and \chlewis; bohr, provides ...

8

chemfig allows to add explizit node names to either bonds or atoms in its formulae by using the @{<name>} syntax. These names can be used in a tikzpicture with the options remember picture, overlay to draw the curved arrows. chemfig provides the wrapper \chemmove for this. So a combination of chemfig and TikZ can be used to draw the schemes. (BTW: the ...

8

This is not fully automatic (you need to specify the rotation for the molecule and the corresponding \chemmove command) and the TikZ code can most likely be improved but it may be a start. It places two invisible bonds where I've marked the respective ends with chemfigs @{<node name>} syntax. They are used to draw the rectangle later. ...

7

\chemabove and \chembelow may help : \documentclass{article} \usepackage{chemfig} \newcommand*\circleatom[1]{\tikz\node[circle,draw,fill=green!30]{\printatom{#1}};} \begin{document} \setlewisdist{4pt} \schemestart \chemname{% \chemfig{% H_{2} -[:-90]\lewis{0:,N} -[:225]H_{2}C -[:-45]C(=[:-90]O) ...

7

\centerline{ isn't really a latex command (it is in the latex format but just escaped from plain TeX). It makes a box \hsize wide but does not know about latex list structures and the indentation they introduce. So you line is too wide by 25pt which will be the left margin of the list item. Just remove \centerline and replace it with a ...

6

I'll answer under the assumption that you're the one who emailed me the other day about Missing bond type in chemical rendering packages? and to whom I suggested the following definition of the submol {Rsubst}: \tikzset{ subst/.style={shorten <= 10pt,preaction={draw=white,line width=6pt}} } \definesubmol{subst}{-[,-1,,,draw=none]-[::-30,1.5,,,subst]} ...

6

My suggestions for the three points: just use math mode instead of a \chemfig command. write the N before you start the ring: N*5(-----) I would pass a suitable TikZ style to the node containing the text. This can be achieved with the \arrow command: \arrow(<start node>[<options>]--<end node>[<options>]). ...

5

Here is a pure chemfig solution : \documentclass{article} \usepackage{chemfig} \begin{document} \definesubmol{eau}{O(-[::-45]H)-[::45]H} \def\dist{.75}% set the distance between C+ and water \chemfig{C^{+}(-[:180,\dist,,,draw=none]!{eau})(-[:60,\dist,,,draw=none]!{eau})-[:-60,\dist,,,draw=none]!{eau}} \end{document}

5

All of chemfig's bonds have optional arguments which accept the following input: <bond symb>[<angle spec>,<length factor>,<dpt atom>,<arr atom>,<tikz spec>] The third, <dpt atom>, and fourth ,<arr atom>, are the important ones here: they specify from which atom of a group of atoms the bond departs and at ...

5

There seems to be some problem with chemfig and easytable; loading chemfig after easytable solves the problem (I commented out the line loading the morehelp package, since my LaTeX system doesn't have it): \documentclass[12pt,doublespacing,letterpaper]{report} \usepackage{etex} \usepackage{cite} %\usepackage{morehelp} \usepackage{setspace} ...

5

The use of \rlap and \llap seems much easier in this case since there is no need to copy and adapt a complex macro: \def\leftparen{\llap{$\left(\strut\right.$\kern5pt }} \def\rightparen{\rlap{$\kern5pt \left.\strut\right)_n$}} \chemfig{ CH_2=[:-30]CH-[:30]C(=[:90]O)-[:-30]O -[:30]\leftparen CH_2CH_2|\rightparen ...

5

We can set the opacity of the bonds we do want to hide to zero and make them visible again with \pgfkeysalso. Sadly, I have not found a simple way to pass TikZ options to the nodes (the atoms like C and H_3) otherwise it would work like the same with visible on=<slides>. I propose two solutions: Redefinition of \printatom with checks a set of ...

5

You can use \noindent to remove the paragraph indentation which saves half the over-hang, and for the rest you can shrink the arrows a bit (I hope that doesn't change the meaning, my chemistry is a bit rusty) \documentclass[12pt,twocolumn]{article} \usepackage{chemfig} \usetikzlibrary{decorations.pathmorphing} \usepackage{lipsum} \begin{document} ...

4

You could place the arrow's label's contents in a \parbox: \documentclass{article} \usepackage{chemfig} \begin{document} \schemestart A \arrow{->[\parbox{5cm}{\centering a\\b}][c]} B \schemestop \end{document} or for convenience define a command: \documentclass{article} \usepackage{chemfig} \makeatletter ...

4

You can load the TikZ-library arrows and use the arrow tips left to and right to. Here's a quick example: \documentclass{article} \usepackage{chemfig} \setatomsep{1.8em} \usetikzlibrary{arrows} \begin{document} \schemestart \chemfig{Cl|@{Cl}\Lewis{0.,\strut}} % a long invisible bond should suffice but this demonstrates % that also scheme nodes can ...

4

The latest example in section "12 The + sign" of the documentation of chemfig shows \arrow(--[yshift=-10pt]){...}. This also works for xshift: \documentclass[12pt,doublespacing,letterpaper]{report} \usepackage{chemfig} \begin{document} \setatomsep{2em} \setbondoffset{1pt} \setdoublesep{3pt} \setbondstyle{line width=1pt} ...

4

Chemfig's bonds have an optional argument that takes several parameters one of them being a factor to scale the bond length: <bond>[<angle spec>,<length factor>,<other parameters>] So you can just add that option to the bonds you want shorter: \documentclass[a4paper,11pt]{scrartcl} \usepackage[utf8]{inputenc} ...

4

You could mark two nodes with the @{<name>} syntax and then draw a rectangle with fill opacity=.2 (or whatever suits you best): \documentclass{article} \usepackage{chemfig,tikz} \begin{document} \chemfig{R_1-@{C}C(=[:90]@{O}O)-R_2} \chemmove{ \draw[ fill=purple, draw=purple, fill opacity=.2, rounded corners=2pt ] ...

4

I don't normally advertize the tabu package (see this post for reasons why) but in this case it is very handy. It scales the table horizontally to textwidth per default if you use its X type columns. Those columns have an optional argument for specifying the ratio of one column to another: X[1]X[2] would mean that the second column has a width twice as much ...

4

Perhaps something like this is what you want: \documentclass[12pt]{report} \usepackage{chemfig} %these for bonds in chemfig \setdoublesep{0.35700 em} % 'Bond Spacing' \setatomsep{1.78500 em} % 'Fixed Length' \setbondoffset{0.18265 em} % 'Margin Width' \newcommand{\bondwidth}{0.06642 em} % 'Line Width' \setbondstyle{line width = \bondwidth} %Drawing ...

4

You typed \chemfig{...Cl[@{op,.25}]...} which implies an optional argument to the »Cl« atom. There are no optional arguments to atoms, though! This means chemfig interprets the quare brackets as atoms, i.e., it typesets them. This leads to the “little box” you see. There are two legal places for the @{...} syntax: a zero size node on a bond using the ...

3

Since you can use \chemfig inside a tikzpicture environment, you can place the three compounds (forgive me if that's not the right name) inside \nodes and then use \draw to draw the arrows: \documentclass{article} \usepackage{chemfig} \usetikzlibrary{positioning,calc} \begin{document} \definesubmol\Me[H_3C]{CH_3} \begin{tikzpicture}[node distance=0cm and ...

3

- is interpreted as a single bond. It seems as if it vanishes because the width of the “atom” BF{_3^-}\emph{n} is so big. You actually don't want the hyphen to be parsed as a bond so you need to hide it. In the comments it was proposed to use \emph{n-} which works as it hides the hyphen from chemfig's parser. I have another suggestion: since the atoms in ...

3

I must confess the question is a bit unclear to me but it seems like you simply want a {figure} environment. If by »refer to the [...] molecule« you mean the practice of refering to molecules with numbers in the text you may want to have a look at the chemnum package. \documentclass{article} \usepackage{chemfig} \usepackage{chemnum} ...

3

\smash or (maybe even better) amsmath's \smash[b] can help here \chemfig{A-X-\smash[b]{\printatom{Mg}}}. For the sake of not having to type \smash and \printatom I define a \smashatom below: \documentclass{article} \usepackage{chemfig} \usepackage{amsmath} \newcommand*\smashatom[2][]{% \if\relax\detokenize{#1}\relax \smash{\printatom{#2}}% \else ...

3

Your code is missing an opening brace for the \parbox and a \fi in the arrow definition. After adding those I changed the definition of the arrow as follows: \makeatletter \definearrow4{s>}{% \ifx\@empty#3\@empty \expandafter\draw\expandafter[\CF@arrow@current@style,-CF@full] \CF@arrow@start@node)--(\CF@arrow@end@node);% \else ...

3

Here is an idea for a new arrow type ->> that allows to input both arrows as one. The arrow type will get 6 arguments: \arrow{->>[above first] [below first] [above second] [below second] [ratio of arrow lengths (first:both)] [label offset]} The input \schemestart ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# How many right angles does a rhombus have?

##### 1 Answer
Dec 25, 2015

If a rhombus is a square, all four of its angles are right.
Otherwise, all angles are either acute or obtuse, but not right.

#### Explanation:

As a parallelogram, rhombus has a sum of two interior angles that share a side equal to ${180}^{\circ}$.
Therefore, only if all angles are equal, they all are equal to ${90}^{\circ}$.
That equality of all angles makes a square out of rhombus.


text:
The sharp upper bound of the lifespan of solutions to critical semilinear wave equations in high dimensions

Preprint Series # 969
Takamura, Hiroyuki and Wakasa, Kyouhei The sharp upper bound of the lifespan of solutions to critical semilinear wave equations in high dimensions. (24 September 2010); (Submitted)

 PDF - Requires a PDF viewer such as GSview, Xpdf or Adobe Acrobat Reader127Kb

Abstract

The final open part of Strauss' conjecture on semilinear wave eqautions was the blow-up theorem for the critical case in high dimensions. This problem was solved by Yordanov and Zhang in 2006, or Zhou in 2007 independently. But the estimate for the lifespan, the maximal existence time, of solutions was not clarified in both papers. \par In this paper, we refine their theorems and introduce a new iteration argument to get the sharp upper bound of the lifespan. As a result, with the sharp lower bound by Li and Zhou in 1995, the lifespan $T(\e)$ of solutions of $u_{tt}-\Delta u=u^2$ in $\R^4\times[0,\infty)$ with the initial data $u(x,0)=\e f(x),u_t(x,0)=\e g(x)$ of a small parameter $\e>0$, compactly supported smooth functions $f$ and $g$, has an estimate $\exp\left(c\e^{-2}\right)\le T(\e)\le\exp\left(C\e^{-2}\right),$ where $c$ and $C$ are positive constants depending only on $f$ and $g$. This upper bound has been known to be the last open optimality of the general theory for fully nonlinear wave equations.

Item Type: Preprint lifespan, semilinear wave equation, critical exponent, high dimensions 35-xx PARTIAL DIFFERENTIAL EQUATIONS 2118


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# Wydawnictwa / Czasopisma IMPAN / Fundamenta Mathematicae / Wszystkie zeszyty

## Continuation of the connection matrix for singularly perturbed hyperbolic equations

### Tom 196 / 2007

Fundamenta Mathematicae 196 (2007), 253-273 MSC: Primary 37B30, 35B25; Secondary 37L05, 37L65. DOI: 10.4064/fm196-3-3

#### Streszczenie

Let ${\mit\Omega}\subset \mathbb R^N$, $N\le 3$, be a bounded domain with smooth boundary, $\gamma\in L^2({\mit\Omega})$ be arbitrary and $\phi\colon \mathbb R\to \mathbb R$ be a $C^1$-function satisfying a subcritical growth condition. For every $\varepsilon\in]0,\infty[$ consider the semiflow $\pi_\varepsilon$ on $H^1_0({\mit\Omega})\times L^2({\mit\Omega})$ generated by the damped wave equation \eqalign{ \varepsilon \partial_{tt}u+\partial_t u&={\mit\Delta} u+\phi(u)+\gamma(x), \quad\ x\in{\mit\Omega},\,t>0,\cr u(x,t)&=0,\quad\ x\in \partial {\mit\Omega},\,t>0. } Moreover, let $\pi'$ be the semiflow on $H^1_0({\mit\Omega})$ generated by the parabolic equation \eqalign{ \partial_t u&={\mit\Delta} u+\phi(u)+\gamma(x), \quad\ x\in{\mit\Omega}, \,t>0,\cr u(x,t)&=0,\quad\ x\in \partial {\mit\Omega},\,t>0. } Let ${\mit\Gamma}\colon H^2({\mit\Omega})\to H^1_0({\mit\Omega})\times L^2({\mit\Omega})$ be the imbedding $u\mapsto (u,{\mit\Delta} u+\phi(u)+\gamma)$. We prove that whenever $K'$ is a compact isolated $\pi'$-invariant set and $(M_p')_{p\in P}$ is a partially ordered Morse decomposition of $K'$ then the imbedded sets $K={\mit\Gamma}(K')$ and $M_{p,{0}}={\mit\Gamma}(M_p')$, $p\in P$, continue, for $\varepsilon>0$ small, to an isolated $\pi_\varepsilon$-invariant set $K_\varepsilon$ a Morse decomposition $(M_{p,\varepsilon})_{p\in P}$ of $K_\varepsilon$, relative to $\pi_\varepsilon$, such that the homology index braid of $(\pi_\varepsilon,K_\varepsilon, (M_{p,\varepsilon})_{p\in P})$ is isomorphic to the homology index braid of $(\pi',K',(M_p')_{p\in P})$. This, in particular, implies a connection matrix continuation principle.

#### Autorzy

• Maria C. CarbinattoDepartamento de Matemática, ICMC-USP
Caixa Postal 668
13.560-970 São Carlos SP, Brazil
e-mail
• Krzysztof P. RybakowskiInstitut für Mathematik
Universität Rostock
Universitätsplatz 1
18055 Rostock, Germany
e-mail

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1.163 Acceleration of rocket

Problem 1.163:A small test rocket launched from rest travels $14.4 \:m$ upward in $2.0 \:s$. Determine acceleration of the rocket.

Solution:

The rocket was initially at rest. Then it speeds upward and covers a distance of $14.4 \:m$ in $2.0 \:s$.

We will assume that it was uniformly accelerated.

We can use kinematic equation $S\:=\:ut\:+\:\frac{1}{2}at^2$ to solve the problem.

For the rocket, initial velocity, $u\:=\:0$.

Time, $t\:=\:2.0\:s$.

Displacement, $S\:=\:14.4\:m$.

Then, $14.4\:=\:0\:+\:\frac{1}{2}\times a \times (2)^2$.

Or, $14.4\:=\:\frac{1}{2}\times a \times 2 \times 2$

Or, $14.4\:=\: a \times 2$.

From this acceleration, $a\:=\:\frac{14.4}{2}\:=\:7.2\:ms^{-2}$.

So the rocket has upward acceleration of $7.2\:ms^{-2}$.

1.162 Acceleration and velocity vector in projectile motion

Problem 1.162: A projectile is fired into the air at an angle with the ground. It follows the parabolic path as shown in the diagram, landing on the right. Neglect air resistance. At any instant, the projectile has a velocity $\vec{v}$ and an acceleration $\vec{a}$  . Which one or more of the drawings could not represent the directions for $\vec{v}$and $\vec{a}$  at any point on the trajectory?

Solution:

This is projectile motion. When air resistance is neglected, only one force acts on the body. It is the gravitational pull of earth which is always towards the ground in a vertical direction. So the acceleration of the projectile must be always in the downward vertical direction.

Velocity at any instant will be along the direction of motion of the projectile. Therefore it must be tangential to the path.

Keeping these facts in mind, let us examine each of the four cases given to us.

In diagram $1$.

Velocity is inclined and upward. This must represent the upward rise of the projectile just as it starts its motion. See the acceleration vector is vertically downward. So this is a possible situation.

In diagram $2$.

Velocity is perfectly horizontal. This is is possible at the highest point on the flight path. Acceleration vector is vertically downward. So this is a also possible .

In diagram $3$.

Velocity is inclined and downward. This  represents  the downward motion  of the projectile just as it starts falling. Ahe acceleration vector is vertically downward.. This is shown in the following diagram.

In diagram $4$.

In this the velocity vector is horizontal which is possible at the highest point on the path. But acceleration is shown pointed upward. This is not a possible cases.

Out of the four situations, option $4$ is not practically correct.

4.101 Speed of car – Doppler effect

Problem 4.101: A policeman on duty detects a drop of $30\: \%$ in the pitch of the horn of a motor-car as it crosses him. Calculate the speed of car, if the velocity of sound is $330m/s$.

Solution:

How to solve this problem? First try to visualize….There is a car and a policeman. The car is honking its horn. The policeman naturally hears the sound. But the car is moving towards him. He will hear an increase in pitch or frequency. This is due to Doppler effect. Now the car will go past him and will move away from him. He hears the sound with a lower frequency again due to Doppler effect.

The problem says  there is a drop of $30\%$ in the pitch. This is the percentage difference between the high frequency heard while the car approached him and the lower frequency heard while the car move past him.

Let $f$ be the actual frequency of the sound. Policeman hear this as frequency $f_1$ as th car moves towards him and as $f_2$ as the car moves away from him. Then second frequency is $30\:\%$ less than the first frequency. It is only $70\:\%$ of $f_1$.

Mathematically, $f_2$ =  $70\:\%$ of $f_1$.

Or, $f_2$ = $\frac{70}{100}\:f_1$.

Then, $\frac{f_2}{f_1}\:=\:\frac{70}{100}\:=\:\frac{7}{10}$------$(1)$.

Now we need to calculate $f_1$ and $f_2$ separately.

Calculation of $f_1$.

In Doppler effect the apparent frequency $f_1$ of a source moving towards a stationary observer is given by , $f_1\:=\:f\frac{V}{V\:-\:V_s}$------$(2)$.

Calculation of $f_2$.

The apparent frequency $f_2$ of a source moving away a stationary observer is given by , $f_2\:=\:f\frac{V}{V\:+\:V_s}$------$(3)$.

From $(1)$, $(2)$ and $(3)$,

$\frac{f_2}{f_1}\:=\:\frac{f\frac{V}{V\:+\:V_s}}{f\frac{V}{V\:-\:V_s}}\:=\:\frac{7}{10}$

Or, $\frac{V\:-\:V_s}{V\:+\:V_s}\:=\:\frac{7}{10}$.

Then, $10V\:-\:10V_s\:=\:7V\:+\:7V_s$.

Or, $7V_s\:+\:10V_s\:=\:10V\:-\:7V$.

That is, $17V_s\:=\:3V$

From this, $V_s\:=\:\frac{3V}{17}$.

Given that speed of sound in air is $330\:ms^{-1}$.

Therefore, speed of car,

$V_s\:=\:\frac{3\times 330}{17}\:=\:\frac{990}{17}\:=\:58.24\:ms^{-1}$.

So the car was traveling with a speed of $58.24\:ms^{-1}$.

Problem 1.161:A system consists of three blocks of masses $m_1$, $m_2$ and $m_3$ connected by a string passing over a pulley $P$. The mass $m_1$ hangs freely and $m_2$ and $m_3$ are on a rough horizontal table $($the coefficient of friction = $\mu$$). The pulley is frictionless and of negligible mass. Find the downward acceleration of mass m_1. Assume m_1\:=\:m_2\:=\:m_3\:=m. Solution: When released from rest, the hanging block will speed up downward, pulling the other to with it. This is possible due to the force exerted by the string which is called tension. There will be frictional force from the rough surface, opposing the sliding motion of the other two blocks.Different forces acting on the system are as shown in the diagram below. Since the blocks are connected by strings, they will have acceleration of same magnitude. Unbalanced force in any block will accelerate it. For first block Normal force from the floor, N\:=\:mg. Force of friction, f\:=\:\mu\:mg. The block is moving to the right side. Therefore force T_2 must be greater than force of friction. So T_2\:-\:f\:=\:ma. Or, T_2\:-\:\mu mg\:=\:ma.....(1) For second block Normal force from the floor, N\:=\:mg. Force of friction, f\:=\:\mu\:mg. The block is moving to the right side. Therefore force T_1 must be greater than the sum of force of friction and tension T_2. So T_1\:-\:(T_2\:+\:f)\:=\:ma. Or, T_1\:-\:T_2\:-\:\mu mg\:=\:ma......(2) For third block The block is moving downward. Therefore weight mg must be greater than tension T_1. So mg\:-\:T_1\:=\:ma.......(3) Adding equations (1), (2) and (3) we get, T_2\:-\:\mu mg\:+\:T_1\:-\:T_2\:-\:\mu mg\:+\:\:mg-\:T_1\:=\:ma\:+\:ma\:+\:ma. Or, mg\:-\:2\mu mg =\:3ma From this acceleration, a\:=\:\frac{mg\:-\:2\mu mg}{3m}. Or, a\:=\:\frac{(1\:-\:2\mu)mg}{3m}. Then, a\:=\:\frac{(1\:-\:2\mu)g}{3}. Tension T_2. We have, from equation (1) T_2\:-\:\mu mg\:=\:ma Or, T_2\:=\:ma\:+\:\mu mg Using this value of acceleration, a\:=\:\frac{(1\:-\:2\mu)g}{3}, We get, T_2\:=\:m(\frac{(1\:-\:2\mu)g}{3})\:+\:\mu mg Or, T_2\:=\:mg(\frac{(1\:-\:2\mu)}{3}\:+\:\mu) Or, T_2\:=\:mg\frac{(1\:-\:2\mu+\:3\mu)}{3} Then Tension, T_2\:=\:mg\frac{(1\:+\mu)}{3}\:. Tension T_1 We have, from equation (3) mg\:-\:T_1\:=\:ma From this, mg\:-\:ma\:=\:T_1 Or T_1\:=\:mg\:-\:ma. Using this value of acceleration, a\:=\:\frac{(1\:-\:2\mu)g}{3}, We get, T_1\:=\:mg\:-\:m(\frac{(1\:-\:2\mu)g}{3}). Or, T_1\:=\:mg(1\:-\:\frac{(1\:-\:2\mu)}{3}) Then, T_1\:=\:mg\frac{(3\:-\:1\:+\:2\mu)}{3} or, T_1\:=\:mg\frac{(2\:+\:2\mu)}{3} Then Tension T_1\:=\:2mg\frac{(1\:+\:\mu)}{3} Tension in the vertical string is twice that in the horizontal string. 1.160: Acceleration of connected blocks Problem 1.160:A system consists of three blocks of masses m_1, m_2 and m_3 connected by a string passing over a pulley P. The mass m_1 hangs freely and m_2 and m_3 are on a rough horizontal table (the coefficient of friction = \mu$$)$. The pulley is frictionless and of negligible mass. Find the downward acceleration of mass $m_1$. Assume $m_1\:=\:m_2\:=\:m_3\:=m$. Take acceleration due to gravity as $10\:ms^{-2}$.

Solution:

When released from rest, the hanging block will speed up downward, pulling the other to with it. This is possible due to the force exerted by the string which is called tension.

There will be frictional force from the rough surface, opposing the sliding motion of the other two blocks.

Different forces acting on the system are as shown in the diagram below.

Since the blocks are connected by strings, they will have acceleration of  same magnitude.

Unbalanced force in any block will accelerate it.

For first block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_2$ must be greater than force of friction.

So $T_2\:-\:f\:=\:ma$.

Or, $T_2\:-\:\mu mg\:=\:ma$.....$(1)$

For second  block

Normal force from the floor, $N\:=\:mg$.

Force of friction, $f\:=\:\mu\:mg$.

The block is moving to the right side. Therefore force $T_1$ must be greater than the sum of  force of friction and tension $T_2$.

So $T_1\:-\:(T_2\:+\:f)\:=\:ma$.

Or, $T_1\:-\:T_2\:-\:\mu mg\:=\:ma$......$(2)$

For third block

The block is moving downward. Therefore weight $mg$ must be greater than tension $T_1$.

So $mg\:-\:T_1\:=\:ma$.......$(3)$

Adding equations $(1)$, $(2)$ and $(3)$ we get,

$T_2\:-\:\mu mg\:+\:T_1\:-\:T_2\:-\:\mu mg\:+\:\:mg-\:T_1\:=\:ma\:+\:ma\:+\:ma$.

Or, $mg\:-\:2\mu mg =\:3ma$

From this acceleration, $a\:=\:\frac{mg\:-\:2\mu mg}{3m}$.

Or, $a\:=\:\frac{(1\:-\:2\mu)mg}{3m}$.

Then, $a\:=\:\frac{(1\:-\:2\mu)g}{3}$.

All blocks will accelerate at the same rate.

Problem 3.102 Heating a gas at constant volume

Problem 3.102:A cylinder of fixed capacity $44.8$ liters contains helium gas at standard temperature and pressure. What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15\:^oC$. Take $R$ as $8.314\:Jmol^{-1}$

Solution:

The gas is in a cylinder of fixed capacity. So the process is isochoric in nature.

The cylinder has a capacity of $44.8$ liters.

The gas is at standard temperature and pressure.

From Avogadro's law, $22.4$ liters of gas at standard temperature and pressure will be one mole.

So the given gas has $2$ mole.

Helium is monoatomic with degree of freedom $f\:=\:3$.

Specific heat capacity at constant volume, $C_v\:=\:\frac{f}{2}R$.

With $f\:=\:3$, $C_v\:=\:\frac{3}{2}R$.

The required change in temperature of the gas in the cylinder is $15\:^oC$

This is equal to change of $15\:K$.

Heat required for this constant volume process can be estimated using the relation $Q\:=\:nC_v\delta T$.

Then, $Q\:=\:2\times\frac{3}{2}R\times 15$.

Or, $Q\:=\:3\times R\times 15\:=\:3\times 8.314\times 15\:=\:374.13\:J$.

Therefore, the amount of heat needed to raise the temperature of  $44.8$ liters helium gas without changing its volume is $374.13\:J$.