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# Compressing Recurrent Neural Networks with Tensor Ring for Action Recognition

Recurrent Neural Networks (RNNs) and their variants, such as Long-Short Term Memory (LSTM) networks, and Gated Recurrent Unit (GRU) networks, have achieved promising performance in sequential data modeling. The hidden layers in RNNs can be regarded as the memory units, which are helpful in storing information in sequential contexts. However, when dealing with high dimensional input data, such as video and text, the input-to-hidden linear transformation in RNNs brings high memory usage and huge computational cost. This makes the training of RNNs unscalable and difficult. To address this challenge, we propose a novel compact LSTM model, named as TR-LSTM, by utilizing the low-rank tensor ring decomposition (TRD) to reformulate the input-to-hidden transformation. Compared with other tensor decomposition methods, TR-LSTM is more stable. In addition, TR-LSTM can complete an end-to-end training and also provide a fundamental building block for RNNs in handling large input data. Experiments on real-world action recognition datasets have demonstrated the promising performance of the proposed TR-LSTM compared with the tensor train LSTM and other state-of-the-art competitors.

## Authors

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## Introduction

Recurrent Neural Networks (RNNs) have achieved great success in analyzing sequential data in various applications, such as computer vision

[Byeon et al.2015, Liang et al.2016, Theis and Bethge2015]

, natural language processing, etc.. Thanks to the ability in capturing long-range dependencies from input sequences

[Rumelhart, Hinton, and Williams1988, Sutskever, Vinyals, and Le2014]. To address the gradient vanishing issue which often leads to the failure of long-term memory in vanilla RNNs, advanced variants such as Gate Recurrent Unit (GRU) and Long-Short Term Memory (LSTM) have been proposed and applied in many learning tasks [Byeon et al.2015, Liang et al.2016, Theis and Bethge2015].

Despite the success, LSTMs and GRUs suffer from the huge number of parameters, which makes the training process notoriously difficult and easily over-fitting. In particular, in the task of action recognition from videos, a video frame usually forms a high-dimensional input, which makes the size of the input-to-hidden matrix extremely large. For example, in the UCF11 [Liu, Luo, and Shah2009], a video action recognition dataset, an RGB video clip is a frame with a size of

, and the dimension of the input vector fed to the vanilla RNN can be over 57,000. Assume that the length of hidden layer vector is 256. Then, the input-to-hidden layer matrix has a load of parameters up to 14 millions. Although a pre-processing feature extraction step via deep convolutional neural networks can be utilized to obtain static feature maps as inputs to RNNs

[Donahue et al.2015], the over-parametric problem is still not fully solved.

A promising direction to reduce the parameter size is to explore the low-rank structures in the weight matrices. Inspired from the success of tensor decomposition methods in CNNs [Novikov et al.2015, Li et al.2017], various tensor decomposition methods have been explored in RNNs [Yang, Krompass, and Tresp2017, Ye et al.2018]. In particular, in [Yang, Krompass, and Tresp2017], the tensor train (TT) decomposition has been applied to RNNs in an end-to-end way to replace the input-to-hidden matrix, and achieved state-of-the-art performance in finding the low-rank structure in RNNs. However, the restricted setting on the ranks and the restrained order of core tensors makes TT-RNN models sensitive to parameter selection. In detail, the optimal setting of TT-ranks is that they are small in the border cores and large in middle cores, e.g., like an olive [Zhao et al.2016].

To address this issue, we propose to use the tensor ring decomposition (TRD) [Zhao et al.2016] to extract the low-rank structure of the input-to-hidden matrix in RNNs. Specifically, the input-to-hidden matrices are reshaped into a high-dimensional tensor and then factorized using TRD. Since this corresponding tensor ring layer automatically models the inter-parameter correlations, the number of parameters can be much smaller than the original size of the linear projection layer in standard RNNs. In this way, we present a new TR-RNN model with a similar representation power but with several orders of fewer parameters. In addition, since TRD can alleviate the strict constraints in tensor train decomposition via interconnecting the first and the last core tensors circularly [Zhao et al.2016], we expect TR-RNNs to have more expressive power. It is important to note that the tensor ring layer can be optimized in an end-to-end training, and can also be utilized as a building block into current LSTM variants. For illustration, we implement an LSTM with the tensor ring layer, named as TR-LSTM.222Note that the tensor ring layer can also be plugged in the vanilla RNN and GRU.

We have conducted empirical evaluations on two real-world action recognition datasets, i.e., UCF11 and HMDB51 [Kuehne et al.2011]. For a fair comparison with standard LSTM and TT-LSTM [Yang, Krompass, and Tresp2017] and BT-LSTM [Ye et al.2018], we conduct experiments in an end-to-end training. As shown in Figure 7, the proposed TR-LSTM has obtained an accuracy value of 0.869, which outperforms the results of standard LSTM (i.e., 0.681), TT-LSTM (i.e.,0.803), and BT-LSTM (i.e., 0.856). Meanwhile, the compression ratio over the standard LSTM is over 34,000, which is also much higher than the compression ratio given by TT-LSTM and BT-LSTM. Moreover, with the output by pre-trained CNN as the input to LSTMs, TR-LSTM has outperformed most previous competitors including LSTM, TT-LSTM, BT-LSTM, and others recently proposed action recognition methods. Since the TR layer can be used as a building block to other LSTM-based approaches, such as the two-stream LSTM [Gammulle et al.2017], we believe the proposed TR decomposition can be a promising approach for action recognition, by considering the tricks used in other state-of-the-art methods.

## Model

To handle the high dimensional input of RNNs, we introduce the tensor ring decomposition to represent the input-to-hidden layer in RNNs in a compact structure. In the following, we first present preliminaries and background of tensor decomposition, including graphical illustrations of the tensor train decomposition and the tensor ring decomposition, followed by our proposed LSTM model, namely, TR-LSTM.

### Preliminaries and Background

#### Notation

In this paper, a -order tensor, e.g., is denoted by a boldface Euler script letter. With all subscripts fixed, each element of a tensor is expressed as: . Given a subset of subscripts, we can get a sub-tensor. For example, given a subset , we can obtain a sub-tensor . Specifically, we denote a vector by a bold lowercase letter, e.g., , and matrices by bold uppercase letters, e.g., . We regard vectors and matrices as 1-order tensors and 2-order tensors, respectively. Figure 1 draws the tensor diagrams presenting the graphical notations and the essential operations.

#### Tensor Contraction

Tensor contraction can be performed between two tensors if some of their dimensions are matched. For example, given two 3-order tensors and , when , the contraction between these two tensors result in a tensor with the size of , where the matching dimension is reduced, as shown in Equation (1):

 (1)

#### Tensor Train Decomposition

Through the tensor train decomposition (TTD), a high-order tensor can be decomposed as the product of a sequence of low-order tensors. For example, a -order tensor can be decomposed as follows:

 Dl1,l2,…,ld TTD=G(1)l1G(2)l2…G(d)ld =∑r1,r2,…,rd−1G(1)r0,l1,r1…G(d)rd−1,ld,rd (2)

where each is called a core tensor. The tensor train rank, shorted as TT-rank, , for each , , corresponds to the complexity of tensor train decomposition. Generally, in tensor train decomposition, the constraint should be satisfied and other ranks are chosen manually. Figure 2 illustrates the form of tensor train decomposition.

#### Tensor Ring Decomposition

The main drawback in tensor train decomposition is the limit of the ranks’ setting, which hinders the representation ability and the flexibility of the TT-based models. At the same time, a strict order must be followed when multiplying TT cores, so that the alignment of the tensor dimensions is extremely important in obtaining the optimized TT cores, but it is still a challenging issue in finding the best alignment[Zhao et al.2016].

In the tensor ring decomposition(TRD), an important modification is interconnecting the first and the last core tensors circularly and constructing a ring-like structure to alleviate the aforementioned limitations of the tensor train. Formally, we set and , and conduct the decomposition as:

 Dl1,l2,…,ldTRD= ∑r0=rd,r2,…,rd−1 G(1)r0,l1,r1G(2)r1,l2,r2…G(d)rd−1,ld,rd (3)

For a -order tensor, by fixing the index where , the first order of the beginning core tensor and the last order of the ending core tensor , can be reduced to matrices. Thus, along each of the slices of , we can separate the tensor ring structure as a summation of of tensor trains. For example, by fixing , the product of has the form tensor train decomposition. Therefore, the tensor ring model is essentially the linear combination of different tensor train models. Figure 3 demonstrates the tensor ring structure, and the alternative interpretation as a summation of multiple tensor train structures.

### TR-RNN model

The core conception of our model is elaborated in this section. By transforming the input-to-hidden weight matrices in TR form, and applying them into RNN and its variants, we get our TR-RNN models.

#### Tensorizing x, y and W

Without loss of generality, we tensorize the input vector , output vector , and weight matrix into tensors , , and , shown in Equation (4):

 X∈RI1×I2×,…,×In,Y∈RO1×O2×,…,×Om W∈RI1×I2×,…,×In×O1×O2,…,×Om (4)

where

 n∏i=1Ii=I,   m∏j=1Oj=O

#### Decomposing W

For an -order input and -order output, we decompose the weight tensor into the form of TRD with core tensors multiplied one by one, each of which is corresponding to an input dimension or an output dimension, referring to Equation (5). Without loss of generality, the core tensors corresponding to the input dimensions and output dimensions are grouped respectively, as shown in Figure 4.

 TRD(W)i1,…,in,o1,…,om=∑r0,…,rn,rn+1,…,rn+m−1 G(1)r0,i1,r1…G(n)rn−1,in,rnG(n+1)rn,o1,rn+1…G(n+m)rn+m−1,om,r0 (5)

The tensor contraction from input to hidden layer in TR form is shown in Equation (6). We multiply the input tensor with input core tensors and output core tensors sequentially. The complexity analysis of forward and backward process is elaborated in the appendix.

 Yo1,o2,…,om=∑i1,…,inXi1,…,inTRD(W)i1,…,in,o1,…,om (6)

Compared with the redundant input-to-hidden weight matrix, the compression ratio in TR form is shown in Equation (7).

 CTRD=∏i=ni=1Ii∏j=mj=1Oj∑ni=1Ri−1IiRi+∑mj=1Rn+j−1OjRn+j (7)

#### Tensor Ring Layer (TRL)

After reshaping the input vector and the weight matrix into tensor, and decomposing weight tensor into TR representation, we can get the output tensor by manipulating and . The final output vector can be obtained by reshaping the output tensor into vector. Because the weight matrix is factorized with TRD, we denote the whole calculation from the input vector to output vector as tensor ring layer (TRL):

 y=TRL(W,x) (8)

which is illustrated in Figure 4.

#### Tr-Rnn

By replacing the multiplication between weight matrix and input vector with TRL in vanilla RNN. We get our TR-RNN model. The hidden state at time can be expressed as:

 (9)

where

denotes the sigmoid function and the hidden state is denoted by

. The input-to-hidden layer weight matrix is denoted by , and denotes the hidden-to-hidden layer matrix.

#### Tr-Lstm

By applying TRL to the standard LSTM, which is the state-of-the-art variant of RNN, we can get the TR-LSTM model as follows.

 gt=tanh(TRL(Wgx,xt)+Ughht−1+bg) ct=ft⊙ct−1+kt⊙gt ht=ot⊙tanh(ct), (10)

where , and denote the element-wise product, the sigmoid function and the hyperbolic function, respectively. The weight matrices (where can be or ) denote the mapping from the input to hidden matrix, for the input gate , the forget gate , the output gate , and the cell update vector , respectively. The weight matrice are defined similarly for the hidden state .

Remark. As shown in Equation (6) and demonstrated in Figure 4, the multiplication between the input tensor data and the input core tensors (for ) will produce a hidden matrix in the size of . It is important to note that the size of the hidden matrix is much smaller than the original data size. In some sense, the “compressed” hidden matrix can be regarded as the information bottleneck [Tishby, Pereira, and Bialek2000, Shwartz-Ziv and Tishby2017], which seeks to achieve the balance between maximally compressing the input information and preserving the prediction information of the output. Thus the proposed TR-LSTM has high potentials to reduce the redundant information in the high-dimensional input while achieving good performance compared with the standard LSTM.

## Experiments

To evaluate the proposed TR-LSTM model, we first design a synthetic experiment to validate the advantage of tensor ring decomposition over the tensor train decomposition. Through two real-world action recognition datasets, i.e., UCF11(YouTube action dataset) [Liu, Luo, and Shah2009] and HMDB51 [Kuehne et al.2011], we evaluate our model from two settings: (1) end-to-end training, where video frames are directly fed into the TR-LSTM; and (2) pre-training to obtain features prior to LSTMs, where a pre-trained CNN was used to extract meaningful low-dimensional features and then forwarded these features to the TR-LSTM. For a fair comparison, we first compare our proposed method with the standard LSTM and previous low-rank decomposition methods, and then with the state-of-the-art action recognition methods.

### Synthetic Experiment

To verify the effectiveness of tensor decomposition methods in recover the original weights, we design a synthetic dataset. Given a low-rank weight matrix , which is illustrated in Figure 5(a)

. We first sample 3200 examples, and each dimension follows a normal distribution, i.e.,

where

is the identity matrix. We then calculate

according to for each where is a random Gaussian noise and

is the variance. Since the

is generated from , the recovered weight matrix should be similar to the ground truth. We use the root mean square error (RMSE) to measure the performance. Should be noted that since the purpose of this experiment is to provide a qualitative and intuitive comparison, we do not add any regularization to the models.

Based on the input data and responses, we estimate the weight matrix

by running the linear regression, tensor train decomposition, and tensor ring decomposition, respectively. For tensor train and tensor ring, we first reshape input data to a tensor of

, and reshape the weight matrix to a tensor of the same size. For illustration, Figure 5 shows one of the recovered (reshaped as a matrix) for the three models when the noise variance is set to 0.05 . Clearly, the proposed tensor ring model performs the best among the three models. As for the tensor train model, it is even worse than the linear regression model. We further illustrate the recovered error of with different levels of noises in Figure 6. It demonstrates that the weight recovered by the tensor ring model has the best tolerance with respect to various injected noises.

### Experiments on the UCF11 Dataset

The UCF11 dataset contains 1600 video clips of a resolution divided into 11 action categories (e.g., basketball shooting, biking/cycling, diving, etc.). Each category consist of 25 groups of video, within more than 4 clips in one group. It is a challenging dataset due to large variations in camera motion, object appearance and pose, object scale, cluttered background, and so on.

In this part, we conduct two experiments described as “End-to-End Training” and “Pre-train with CNN” on this dataset. In the “End-to-End Training”, we compare the proposed TR-LSTM model with other decomposition models (eg. TT-LSTM [Yang, Krompass, and Tresp2017] and BT-LSTM [Ye et al.2018]) to show the superior performance. And in another experiment, we apply decomposition on a more general model, achieving a better performance with less parameters.

##### End-to-End Training

Recent years, some tensor decomposition models are proposed to classify videos like TT-LSTM

[Yang, Krompass, and Tresp2017], BT-LSTM [Ye et al.2018] and others. For the reason that they use the end-to-end model for training, we set this experiment to compare with them. In this experiments, we scale down the original resolution to , and sample 6 frames from each video clip randomly as the input data. Since every frame is RGB, the input data vector at each step is , and there are 6 steps in every sample. We set the hidden layer as 256. So there should be a fully-connected layer of parameters to achieve the mapping for the standard LSTM.

We compare our model with BT-LSTM and TT-LSTM, while using a standard LSTM as a baseline. The hyper-parameters in BT-LSTM and TT-LSTM are set as announced in their papers. Figure 7(c) shows all decomposition methods converging faster than the LSTM. The accuracy of BT-LSTM is 0.856 which is much higher than TT-LSTM with 0.803 while the LSTM only gain an accuracy of 0.681. In our TR-LSTM, the shape of input tensor is , the output tensor’s shape is set as and all the TR-ranks are set as 5 except . Results are compared in Table 1. With 1725 parameters in our model, about half of TT-LSTM and BT-LSTM with parameters 3360 and 3387 respectively. We gain the top accuracy 0.869, showing the outstanding performance of our model in this experiment.

##### Pre-train with CNN

Recently, some methods based on RNNs achieved higher accuracy by using the extracted feature as input vectors in computer vision [Donahue et al.2015]. Compared with using frames as input data, extracted features are more compact. But there is still some room for improving the ability of the models. The over-parametric problem is just partial solved. To get better performance, we use extracted features via the CNN model Inception-V3 as input data to LSTM.

We set the size of the hidden layer as , which is consistent with the size of the output via Inception-V3. After using the extracted feature as the inputs of LSTM, the accuracy of the vanilla LSTM attains 0.923. At the same time, the accuracy of our TR-LSTM model whose ranks are set as achieves 93.8. By replacing the standard LSTM with our model, a compression ratio of 25 can be obtained. We compare some state-of-the-art methods in Table 2 on UCF11. The Two Stream LSTM[Gammulle et al.2017] with highest accuracy has more than 141M parameters. The TR-LSTM can be used to replace the vanilla LSTMs in the Two Stream LSTM model to reduce the parameters.

### Experiments on the HMDB51 Dataset

The HMDB51 dataset is a large collection of realistic videos from various sources, such as movies and web videos. The dataset is composed of 6766 video clips from 51 action categories.

In this experiment, we still use extracted features as the input vector via Inception-V3 and reshape it into . We sample 12 frames from each video clip randomly and be processed through the CNN as the input data. The shape of hidden layer tensor is set as . The ranks of our TR-LSTM are . Some of the state-of-the-art models like I3D [Carreira and Zisserman2017] are presented in Table 3. The I3D model with highest accuracy based on 3D ConvNets, which is not RNN-based method, has 25M parameters, while the TR-LSTM model only has 0.7M parameters. The TR-LSTM gains a higher accuracy of 63.8% than the standard LSTM with a compressing ratio of 25.

## Related Work

In the past decades, a number of variants of recurrent neural networks (RNNs) were proposed to capture sequential information more accurately [Liu et al.2018b]. However, when dealing with the large input data, in the field of computer vision, the input-to-hidden weight matrix owns loads of parameters. The limitation of computation resources and the severe over-fitting problem are emerging. Some methods used CNNs as feature extractors to pre-processing the input data into a more compact way [Donahue et al.2015]. These methods have improved the classification accuracy, but the over-parametric problem is just still partially solved. In this work, we focus on designing low-rank structure to replace the redundant input-to-hidden weight matrix in RNNs, while compressing the whole model and maintaining the model performance.

The most straight-forward way to apply low-rank constraint is implementing matrix decomposition on weight matrices. Singular Value Decomposition (SVD) has been applied in convolutional neural networks to reduce parameters

[Denton et al.2014] but incurred a loss in model performance. Besides, the compression ratio is limited because of the rank in matrix decomposition still relatively large.

Compared with matrix decomposition, tensor decomposition [Xu, Yan, and Qi2015, Zhe et al.2016, Hao et al.2018, He et al.2018, Liu et al.2018a] conducts data decomposition in a higher dimension, capturing higher-order correlations while maintaining several orders of fewer parameters [Kim et al.2015, Novikov et al.2015]. Among these methods, [Lebedev et al.2014] utilized CP decomposition to speed up convolution computation, which has the similar design philosophy with the widely-used depth-wise separable convolutions [Howard et al.2017]. However, the instability issue [de Silva and Lim2008] hinders the low-rank CP decomposition from solving many important computer vision tasks. [Kim et al.2015] used Tucker decomposition to decompose both the convolution layer and the fully connected layer, reducing run-time and energy significantly in mobile applications with minor accuracy drop. Block-Term tensor decomposition combines the CP and Tucker by summing up multiple Tucker models to overcome their drawbacks and has obtained a better performance in RNNs [Ye et al.2018]. However, the computation of the core tensor in the Tucker model is highly inefficient due to the complex tensor flatten and permutation operations. Recent years, the tensor train decomposition also used to substitute the redundant fully connected layer in both CNNs and RNNs [Novikov et al.2015, Yang, Krompass, and Tresp2017], preserving the performance while reducing the number of parameters significantly up to 40 times.

But tensor train decomposition has some limitations: 1) certain constraints for TT-rank, i.e., the ranks of the first and last factors are restricted to be 1, limiting its representation power and flexibility. 2) A strict order must be followed when multiplying TT cores, so that the alignment of the tensor dimensions is extremely important in obtaining the optimized TT cores, but it is still a challenging issue in finding the best alignment. In this paper, we use the Tensor Ring(TR) decomposition [Zhao et al.2016] to overcome the drawbacks in TTD, while achieving more computation efficiency than BT decomposition.

## Conclusion

In this paper, we applied TRD to plain RNNs to replace the over-parametric input-to-hidden weight matrix when dealing with high-dimensional input data. The low-rank structure of TRD can capture the correlation between feature dimensions with fewer orders of magnitude parameters. Our TR-LSTM model achieved best compression ratio with the highest classification accuracy on UCF11 dataset among other end-to-end training RNNs based on low-rank methods. At the same time, when processing the extracted feature through InceptionV3 as the input vector, our TR-LSTM model can still compress the LSTM while improving the accuracy. We believe that our models provide fundamental modules for RNNs, and can be widely used to handle large input data. In future work, since our models are easy to be extended, we want to apply our models to more advanced RNN structures [Gammulle et al.2017] to get better performance.

## Acknowledgments

We thank the anonymous reviewers for valuable comments to improve the quality of our paper. This work was partially supported by National Natural Science Foundation of China (Nos.61572111 and 61876034), and a Fundamental Research Fund for the Central Universities of China (No.ZYGX2016Z003).

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## Appendix

### Complexity Analysis

#### Complexity in Forward Process

Since the weight tensor has been decomposed into the form of TRD with core tensors, the order of multiplication among input tensor and core tensors in Equation (6) determines the computational cost. In our implementation, we multiply the input tensor with input core tensors and output core tensors sequentially. We can rewrite the Equation (6) as:

 Y= X×12G(1)×1,n+12,1G(2)…×1,n−N+32,1G(N)… ×1,32,1G(n)×21G(n+1)…×M+11G(n+M)… ×m1G(n+m−1)×1,m+13,1G(n+m) (11)

In Equation (11), the symbols and denote the Tensor Contraction Operation as described in [Cichocki et al.2016], which is a fundamental and important operation in tensor networks. Tensor contraction can be considered as a higher-dimensional analogue of matrix multiplication, inner product, and outer product. Note the -th component can be formulated as and the -th component can be formulated as .

Our model is trained via Back Propagation Through Time. In Equation (11), according to the left-to-right multiplication order, our forward computational complexity reaches while the forward space complexity is , where all the ranks in our model are set to .

#### Complexity in Backward Process

We implement our back propagation with considering the backward complexity analyze of TT [Novikov et al.2015]. The backward computational complexity of different cores is different due to the multiplication order, so we choose , to represent the backward complexity of each core, which has the highest computational and space complexity in all the cores:

 ∂L∂G(k)= X×12G(1)×1,n+12,1G(2)…×1,n−N1+32,1G(N1)… ×1,n−k+42,1G(k−1)×22G(k+1)×2,n−k+42,1G(k+2)… ×2,n−k−N2+62,1G(k+N2)…×2,62,1G(n) ×51G(n+1)…×4+M1G(n+M)…×m+31G(n+m−1) ×2,m+43,1G(n+m)×4,5…,m+31,2…,m∂L∂Y (12)

In Equation (12), the notations , and are the same as and in Equation (11) and here , , . The backward computational complexity is . Note that the number of cores is in our model, the total backward computational complexity reaches to while the backward space complexity is . The statistics of comparison with some other compressing methods are shown in Table 4.


text:
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Dynamical behaviors of fuzzy reaction-diffusion periodic cellular neural networks with variable coefficients and delays. (English) Zbl 1185.35129
Summary: When modeling neural networks in a real world, not only diffusion effect and fuzziness cannot be avoided, but also self-inhibitions, interconnection weights, and inputs should vary as time varies. In this paper, we discuss the dynamical behaviors of delayed reaction-diffusion fuzzy cellular neural networks with varying periodic self-inhibitions, interconnection weights as well as inputs. By using Halanay’s delay differential inequality, $M$-matrix theory and analytic methods, some new sufficient conditions are obtained to ensure the existence, uniqueness, and global exponential stability of the periodic solution, and the exponentially convergent rate index is also estimated. In particular, the traditional assumption on the differentiability of the time-varying delays is no longer needed. The methodology developed in this paper is shown to be simple and effective for the exponential periodicity and stability analysis of neural networks with time-varying delays. Two examples are given to show the usefulness of the obtained results that are less restrictive than recently known criteria.
##### MSC:
 35K57 Reaction-diffusion equations 37N25 Dynamical systems in biology 92B20 General theory of neural networks (mathematical biology)
##### References:
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Neural networks 16, No. 6, 1340-1351 (2005) [6] Senan, S.; Arik, S.: New results for exponential stability of delayed cellular neural networks, IEEE trans. Circ. syst. II 52, No. 3, 154-158 (2005) [7] Liao, X. X.; Wang, J.; Zeng, Z. G.: Global asymptotic stability and global exponential stability of delayed cellular neural networks, IEEE trans. Circ. syst. II 52, No. 7, 403-409 (2005) [8] Liang, J.; Wang, Z.; Liu, Y.; Liu, X.: Global synchronization control of general delayed discrete-time networks with stochastic coupling and disturbances, IEEE trans. Syst., man, and cyber. – part B 38, No. 4, 1073-1083 (2008) [9] Huang, L. H.; Huang, C. X.; Liu, B. W.: Dynamics of a class of cellular neural networks with time-varying delays, Phys. lett. A 345, No. 4 – 6, 330-344 (2005) [10] Zhao, H.: Global exponential stability and periodicity of cellular neural networks with variable delays, Phys. lett. 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Q.; Tang, W. S.: Exponential stability of fuzzy cellular neural networks with constant and time-varying delays, Phys. lett. A 323, No. 3 – 4, 224-233 (2004) · Zbl 1118.81400 · doi:10.1016/j.physleta.2004.01.064 [17] Yuan, K.; Cao, J. D.; Deng, J. M.: Exponential stability and periodic solutions of fuzzy cellular neural networks with time-varying delays, Neurocomputing 69, No. 13 – 15, 1619-1627 (2006) [18] Huang, T. W.: Exponential stability of fuzzy cellular neural networks with distributed delay, Phys. lett. A 351, No. 1 – 2, 48-52 (2006) [19] Huang, T. W.: Exponential stability of delayed fuzzy cellular neural networks with diffusion, Chaos, solitons fract. 31, No. 3, 658-664 (2007) · Zbl 1138.35414 · doi:10.1016/j.chaos.2005.10.015 [20] Serrano-Gotarredona, T.; Linares-Barranco, B.: Log-domain implementation of complex dynamics reaction – diffusion neural networks, IEEE trans. Neural networks 14, No. 5, 1337-1355 (2003) [21] Wang, L. S.; Xu, D. Y.: Global exponential stability of Hopfield reaction – diffusion neural networks with time-varying delays, Sci. China (Ser. F) 46, No. 6, 466-474 (2005) · Zbl 1186.82062 · doi:10.1360/02yf0146 [22] Chen, T. P.; Lu, W. L.; Chen, G. R.: Dynamical behaviors of a large class of general delayed neural networks, Neural comput. 17, 949-968 (2005) · Zbl 1080.68615 · doi:10.1162/0899766053429417 [23] Gopaldamy, K.; Sariyasa, S.: Time delays and stimulus-dependent pattern formation in periodic environments in isolated neurons, IEEE trans. Neural networks 13, No. 3, 551-563 (2002) [24] Zhou, J.; Liu, Z.; Chen, G.: Dynamics of periodic delayed neural networks, Neural networks 17, 87-101 (2004) · Zbl 1082.68101 · doi:10.1016/S0893-6080(03)00208-9 [25] Huang, T. W.; Cao, J. D.; Li, C. D.: Necessary and sufficient condition for the absolute exponential stability of a class of neural networks with finite delay, Phys. lett. A 352, No. 1 – 2, 94-98 (2006) · Zbl 1187.34100 · doi:10.1016/j.physleta.2005.11.038 [26] Gao, H.; Lam, J.; Wang, C.: Robust energy-to-peak filter design for stochastic time-delay systems, Syst. control lett. 55, No. 2, 101-111 (2006) · Zbl 1129.93538 · doi:10.1016/j.sysconle.2005.05.005 [27] Gao, H.; Lam, J.; Chen, G.: New criteria for synchronization stability of general complex dynamical networks with coupling delays, Phys. lett. A 360, No. 2, 263-273 (2006) [28] Gao, H.; Chen, T.; Lam, J.: A new delay system approach to network-based control, Automatica 44, No. 1, 39-52 (2008) · Zbl 1138.93375 · doi:10.1016/j.automatica.2007.04.020 [29] Roska, T.; Chua, L. O.: Cellular neural networks with nonlinear and delay-type template elements nonumiform grids, Int. J. Circ. theory appl. 20, 469-481 (1992) · Zbl 0775.92011 · doi:10.1002/cta.4490200504

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1
JEE Main 2021 (Online) 24th February Evening Slot
+4
-1
A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[Use $$g = {{GM} \over {{R^2}}}$$ = 9.8 ms$$-$$2 and radius of earth, R = 6400 km.]
A
49 N
B
49.83 N
C
48.83 N
D
49.17 N
2
JEE Main 2021 (Online) 24th February Morning Slot
+4
-1
Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :
A
$$\sqrt {{G \over 2}(1 + 2\sqrt 2 )}$$
B
$$\sqrt {{G \over 2}(2\sqrt 2 - 1)}$$
C
$$\sqrt {G(1 + 2\sqrt 2 )}$$
D
$${1\over2}\sqrt {G(1 + 2\sqrt 2 )}$$
3
JEE Main 2021 (Online) 24th February Morning Slot
+4
-1
Consider two satellites S1 and S2 with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is :
A
1 : 4
B
8 : 1
C
2 : 1
D
1 : 8
4
JEE Main 2021 (Online) 24th February Morning Slot
+4
-1
Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is :
A
$${1 \over {2\pi }}\sqrt {{{{d^3}} \over {3Gm}}}$$
B
$$2\pi \sqrt {{{3Gm} \over {{d^3}}}}$$
C
$${1 \over {2\pi }}\sqrt {{{3Gm} \over {{d^3}}}}$$
D
$$2\pi \sqrt {{{{d^3}} \over {3Gm}}}$$
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text:
7.2 Sum and difference identities  (Page 5/6)

 Page 5 / 6

Verify the identity: $\text{\hspace{0.17em}}\mathrm{tan}\left(\pi -\theta \right)=-\mathrm{tan}\text{\hspace{0.17em}}\theta .$

Using sum and difference formulas to solve an application problem

Let $\text{\hspace{0.17em}}{L}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{L}_{2}\text{\hspace{0.17em}}$ denote two non-vertical intersecting lines, and let $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ denote the acute angle between $\text{\hspace{0.17em}}{L}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{L}_{2}.\text{\hspace{0.17em}}$ See [link] . Show that

$\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{{m}_{2}-{m}_{1}}{1+{m}_{1}{m}_{2}}$

where $\text{\hspace{0.17em}}{m}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{m}_{2}\text{\hspace{0.17em}}$ are the slopes of $\text{\hspace{0.17em}}{L}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{L}_{2}\text{\hspace{0.17em}}$ respectively. ( Hint: Use the fact that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}{\theta }_{1}={m}_{1}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}{\theta }_{2}={m}_{2}.$ )

Using the difference formula for tangent, this problem does not seem as daunting as it might.

Investigating a guy-wire problem

For a climbing wall, a guy-wire $\text{\hspace{0.17em}}R\text{\hspace{0.17em}}$ is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire $\text{\hspace{0.17em}}S\text{\hspace{0.17em}}$ attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ between the wires. See [link] .

Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta =\frac{47}{50},$ and $\text{\hspace{0.17em}}\mathrm{tan}\left(\beta -\alpha \right)=\frac{40}{50}=\frac{4}{5}.\text{\hspace{0.17em}}$ We can then use difference formula for tangent.

$\mathrm{tan}\left(\beta -\alpha \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\beta -\mathrm{tan}\text{\hspace{0.17em}}\alpha }{1+\mathrm{tan}\text{\hspace{0.17em}}\beta \mathrm{tan}\text{\hspace{0.17em}}\alpha }$

Now, substituting the values we know into the formula, we have

Use the distributive property, and then simplify the functions.

$\begin{array}{l}\text{\hspace{0.17em}}4\left(1\right)+4\left(\frac{47}{50}\right)\mathrm{tan}\text{\hspace{0.17em}}\alpha =5\left(\frac{47}{50}\right)-5\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4+3.76\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =4.7-5\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}5\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha +3.76\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =0.7\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8.76\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha =0.7\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\alpha \approx 0.07991\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(0.07991\right)\approx .079741\hfill \end{array}$

Now we can calculate the angle in degrees.

$\alpha \approx 0.079741\left(\frac{180}{\pi }\right)\approx {4.57}^{\circ }$

Access these online resources for additional instruction and practice with sum and difference identities.

Key equations

 Sum Formula for Cosine $\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \mathrm{sin}\text{\hspace{0.17em}}\beta$ Difference Formula for Cosine $\mathrm{cos}\left(\alpha -\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ Sum Formula for Sine $\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ Difference Formula for Sine $\mathrm{sin}\left(\alpha -\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$ Sum Formula for Tangent $\mathrm{tan}\left(\alpha +\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }$ Difference Formula for Tangent $\mathrm{tan}\left(\alpha -\beta \right)=\frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha -\mathrm{tan}\text{\hspace{0.17em}}\beta }{1+\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }$ Cofunction identities $\begin{array}{l}\mathrm{sin}\text{\hspace{0.17em}}\theta =\mathrm{cos}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{cos}\text{\hspace{0.17em}}\theta =\mathrm{sin}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{tan}\text{\hspace{0.17em}}\theta =\mathrm{cot}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{cot}\text{\hspace{0.17em}}\theta =\mathrm{tan}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{sec}\text{\hspace{0.17em}}\theta =\mathrm{csc}\left(\frac{\pi }{2}-\theta \right)\\ \mathrm{csc}\text{\hspace{0.17em}}\theta =\mathrm{sec}\left(\frac{\pi }{2}-\theta \right)\end{array}$

Key concepts

• The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.
• The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle. See [link] and [link] .
• The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle. See [link] .
• The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions. See [link] .
• The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles. See [link] .
• The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. See [link] .
• The cofunction identities apply to complementary angles and pairs of reciprocal functions. See [link] .
• Sum and difference formulas are useful in verifying identities. See [link] and [link] .
• Application problems are often easier to solve by using sum and difference formulas. See [link] and [link] .

Questions & Answers

what is f(x)=
Karim Reply
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
unknown Reply
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
Ef Reply
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
KARMEL Reply
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
Rima Reply
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
Jhon Reply
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
Diddy Reply
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply

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text:
2,169 views
What is time and space complexity to evaluate postfix expression ?

Algorithm

1. Read all the symbols from left to right of the postfix expression

2. If the reading symbol is  $operand$ then push into the stack

3. If the reading symbol is  $operator$, then perform two pop operations, do the arithmetic operation and again pushed the resultant into the stack

4. finally, perform a pop operation and print the popped value as the final result.

Time complexity of evaluation algorithm is $O(n)$ where n is the number of characters in input expression.

Can u explain how the time complexity is O(n) ?
there are n characters in input expression array and we  are scanning it from left to right, so the loop will run for n times

1
1,267 views


text:
$$\require{cancel}$$

# 11.5: Measuring Body Temperature

## Liquid Thermometers

We now know that an increase in temperature corresponds to an increase in the average kinetic energy of atoms and molecules. A result of that increased motion is that the average distance between atoms and molecules increases as the temperature increases. This phenomenon, known as thermal expansion is the basis for temperature measurement by liquid thermometer.

A clinical thermometer based on the thermal expansion of a confined liquid. Image Credit: Clinical Thermometer by Menchi via Wikimedia Commons

[1]

Common liquid thermometers use the thermal expansion of alcohol confined within a glass or plastic tube to measure temperature. Due to thermal expansion, the alcohol volume changes with temperature. The thermometer must be calibrated by marking the various fluid levels when the thermometer is placed in an environment with a known temperature, such as water boiling at sea level.

#### Reinforcement Exercise

You may have noticed that when you place a liquid thermometer into hot water, the liquid level actually drops momentarily, before beginning to climb as expected. Knowing what you do about thermal expansion, can you explain why this occurs?

## Bimetallic Strips

Different materials will thermally expand (or contract) by different amounts when heated (or cooled). Bimetallic strips rely on this phenomenon to measure temperature. When two different materials are stuck together, the resulting structure will bend as the temperature changes due to the different thermal expansion experienced by each material.

The curvature of a bimetallic strip depends on temperature. (a) The strip is straight at the starting temperature, where its two components have the same length. (b) At a higher temperature, this strip bends to the right, because the metal on the left has expanded more than the metal on the right. At a lower temperature, the strip would bend to the left. Image Credit: Openstax University Physics

[2]

## Linear Thermal Expansion

For most common materials the change in length () caused by a change in temperature () is proportional to the original length () and can be modeled using the linear thermal expansion coefficient () and the following equation:

$\Delta L=\alpha L_{0} \Delta T \label{1}$

The following table provides the linear thermal expansion coefficients for different solid materials. More expansive (ha!) tables can be found online.

Thermal Expansion Coefficients
Material Coefficient of Linear Expansion (1/°C)
Solids
Aluminum 25 × 10−6
Brass 19 × 10−6
Copper 17 × 10−6
Gold 14 × 10−6
Iron or steel 12 × 10−6
Invar (nickel-iron alloy) 0.9 × 10−6
Silver 18 × 10−6
Glass (ordinary) 9 × 10−6
Glass (Pyrex®) 3 × 10−6
Quartz 0.4 × 10−6
Concrete, brick ~12 × 10−6
Marble (average) 2.5 × 10−6

#### Everyday Example

The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from –15 °C to 40 °C. What is its change in length between these temperatures? Assume that the bridge is made entirely of steel.

We can use the equation for linear thermal expansion:

Substitute all of the known values into the equation, including the linear thermal expansion coefficient for steel and the initial and final temperatures:

Although not large compared to the length of the bridge, the change in length of nearly one meter is observable and important. Thermal expansion could causes bridges to buckle if not for the incorporation of gaps, known as expansion joints, into the design.

Expansion joint on the Golden Gate Bridge. Image Credit: Expansion Joint Golden Gate Bridge by Michiel1972 via Wikimedia Commons

[3]

#### Reinforcement Exercises

A deep-space satellite made primarily of aluminum is 12 m long when launched at 22 °C (72 °F).

If the satellite comes eventually come to thermal equilibrium with the cosmic microwave background temperature at -270.45 °C (-455 °F).[4]

How much will the satellite change length?

What will the new length be?

## Temperature Units

Thermometers measure temperature according to well-defined scales of measurement. The three most common temperature scales are Fahrenheit, Celsius, and Kelvin. On the Celsius scale, the freezing point of water is 0 °C and the boiling point is 100 °C. The unit of temperature on this scale is the degree Celsius (°C). The Fahrenheit scale (°F) has the freezing point of water at 32 °F and the boiling point 212 °F. You can see that 100 Celsius degrees span the same range as 180 Fahrenheit degrees. Thus, a temperature difference of one degree on the Celsius scale is 1.8 times as large as a difference of one degree on the Fahrenheit scale, as illustrated by the top two scales in the following diagram.

Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales are shown. The relative sizes of the scales are also shown. Image Credit: Temperature Scales diagram from OpenStax University Physics[/footnote]

### The Kelvin Scale

The definition of temperature in terms of molecular motion suggests that there should be a lowest possible temperature, where the average microscopic kinetic energy of molecules is zero (or the minimum allowed by the quantum nature of the particles). Experiments confirm the existence of such a temperature, called absolute zero. An absolute temperature scale is one whose zero point corresponds to absolute zero. Such scales are convenient in science because several physical quantities, such as the pressure in a gas, are directly related to absolute temperature. Additionally, absolute scales allow us to use ratios of temperature, which relative scales do not. For example, 200 K is twice the temperature of 100 K, but 200 °C is not twice the temperature of 100 °C.

#### Reinforcement Exercise

Convert 50 °F to °C and then to K.

### Temperature Measurement

In addition to thermal expansion, other temperature dependent physical properties can be used to measure temperature. Such properties include electrical resistance and optical properties such as reflection, emission and absorption of various colors. Light-based temperature measurement will come up again in the next chapter.

[5]

1. Clinical Thermometer by Menchi [GFDL (http://www.gnu.org/copyleft/fdl.html) or CC-BY-SA-3.0 (http://creativecommons.org/licenses/by-sa/3.0/)] via Wikimedia Commons
2. OpenStax University Physics, University Physics. OpenStax CNX. May 10, 2018 http://cnx.org/contents/74fd2873-157d-4392-bf01-2fccab830f2c@5.301.
3. Michiel1972 [CC BY-SA 3.0 (https://creativecommons.org/licenses/by-sa/3.0)], via Wikimedia Commons
4. "Web-based hypothermia information: a critical assessment of Internet resources and a comparison to peer-reviewed literature" by Dr. Eric Christian, Cosmicopia, NASA is in the Public Domain


text:
# Page layout design in KOMA script

This is follow up of the previous question, Document class for designing a book.

Majority of respondents suggested that I design the the page layout using the class I was using for writing the text, in my case scrbook.

I have, of course studied the KOMA script manual and I find page layout settings pretty cumbersome. The page is divided into n columns and n rows, of which outer two, inner one, bottom two and top one are margins. On the other hand, I have the following instructions from the book designer:

page dimensions: 210 x 245 mm2
inner margin: 25 mm
outer margin: 10 mm
upper margin: 10 mm
bottom margin: 25 mm
text block width: 120 mm

My calculation from data above:

marginal notes width: 55 mm

I also attach the sample of the required layout, created in non-LaTeX program.

Please, advice me how to do this layout in scrbook class. In this moment, I am interested in page layout only, as I have solved some of other design problems myself or with other people's help already. Thanks.

Using only KOMA-Script, the ratio between top and bottom margin will always be 1:2. Therefore, an additional package (e.g., geometry) is needed for your desired layout. Note that you have to split your calculated "marginal notes width" into marginparsep and marginparwidth and to add both parameters to outer. See section 3 of the geometry manual for details and additional modifications (e.g., regarding headers/footers).

\documentclass{scrbook}

\usepackage[
paperwidth=210mm,
paperheight=245mm,
top=10mm,
bottom=25mm,
inner=10mm,
outer=80mm,
marginparsep=7mm,
marginparwidth=48mm,
]{geometry}

\usepackage{lipsum}

\begin{document}

Some text.\marginpar{\lipsum[1]}

\lipsum[1-5]

\end{document}


text:
Loughborough University
Leicestershire, UK
LE11 3TU
+44 (0)1509 263171

# Loughborough University Institutional Repository

 Please use this identifier to cite or link to this item: https://dspace.lboro.ac.uk/2134/25482

 Title: Non-linear model fitting for the measurement of thin films and surface topography Authors: Yoshino, Hirokazu Keywords: Coherence scanning interferometryThin filmRefractive indexInterfacial surface topographyFreeformAreal stylus profilometryZernike polynomials Issue Date: 2017 Publisher: © Hirokazu Yoshino Abstract: Inspection of optical components is essential to assure the quality and performance of optical systems. Evaluation of optical components includes metrology measurements of surface topography. It also requires optical measurements including refractive index, thin film thickness, reflectivity and transmission. The dispersion characteristics of optical constants including refractive index are also required. Hence, various instruments are used to make these measurements in research laboratories and for quality assurance. Clearly, it would be a significant advantage and cost saving if a technique was developed that could combine surface metrology with optical measurements. {Coherence Scanning Interferometry} (CSI) (also referred to as {Scanning White Light Interferometry} (SWLI)) has been used widely to measure surface topography with sub-nanometre vertical resolution. One of the benefits of the CSI is that the technique is non-contacting and hence non-destructive. Thus the test surfaces are not affected by the measurement using a CSI instrument whereas damage to the surfaces can occur when using traditional contact methods such as stylus profilometry. However use of CSI is geometrically limited to small areas ($\lesssim 10 \times 10$ mm) with gentle slopes ($\lesssim \ang{40}$) because of the numerical aperture of objective lens whereas stylus profilometry works well with larger areas and higher slopes due to the range of motion of the gauge and the traverse unit. Since the CSI technique is optical and involves light reflection and interference it is possible to extend the technique for the measurement of the thickness of transparent films, the roughness of surfaces buried beneath thin films or interfacial surfaces. It may also be used to determine spectral complex refractive index. This thesis provides an analytical framework of new methods to obtain complex refractive index in a visible light domain and interfacial surface roughness (ISR). It also provides experimental verification of these new capabilities using actual thin film model systems. The original Helical Complex Field (HCF) function theory is presented followed by its existing extensions that enable determination of complex refractive index and interfacial surface roughness. Further theoretical extensions of the HCF theory are also provided: A novel theory to determine the refractive index of a (semi-)transparent film is developed to address the constraint of the current HCF theory that restricted its use to opaque materials; Another novel theory is provided to measure ISR with noise compensation, which avoids erroneous surface roughness caused by the numerical optimisation affected by the existence of noise. The effectiveness of the ISR measurement with noise compensation has been verified using a number of computer simulations. Stylus profilometry is a well established method to provide a profile and has been used extensively as a 'reference' for other techniques. It normally provides a profile on which the roughness and the waviness are computed. Extension of the stylus profilometry technique to areal measurement of asymmetrical surfaces, namely raster scan measurement, requires a system to include error compensation between each traverse. The system errors and the random errors need to be separately understood particular when the measurement of a surface with nanometre-order accuracy is required. In this thesis a mathematical model to locate a stylus tip considering five mechanical errors occurring in a common raster scan profilometer is provided. Based on the model, the simulator which provides an areal measurement of a sphere was developed. The simulator clarified the relationship between the Zernike coefficients obtained from the form residual and the size of the errors in the form of partial derivatives of Zernike coefficients with respect to the errors. This provides theoretical support to the empirical knowledge of the relationship between the coefficients and the errors. Furthermore, a method to determine the size of errors directly from Zernike coefficients is proposed supported by simulations. Some of the error parameters were accurately determined avoiding iterative computation with this method whereas the errors are currently being determined by iterative computation. Description: A Doctoral Thesis. Submitted in partial fulfilment of the requirements for the award of Doctor of Philosophy of Loughborough University. Sponsor: Taylor Hobson Ltd. URI: https://dspace.lboro.ac.uk/2134/25482 Appears in Collections: PhD Theses (Mechanical, Electrical and Manufacturing Engineering)

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text:
Physics

posted by on .

A carnival ferris wheel completes five turns about its horizontal axis every minute. What is the acceleration of a passenger at his/ her lowest point on the ride?

I understand that 1/f= T which is how you find the period and that minutes must be converted to sec, but I'm getting really strange answers for the centrip. acceleration. Is the question asking for the linear acceleration or both? Any help would be really appreciated!

• Physics - ,

we assume that the speed is constant so there is no linear acceleration, only centripetal.

We HAVE to know the radius.

T = 60 sec /5 turns = 12 seconds/turn

v = 2 pi r/T
and
Ac = v^2/r


text:
# Proving $\limsup\frac 1 {a_n}=\frac 1 {\liminf a_n}$ and $\limsup a_n\cdot \limsup \frac 1 {a_n} \ge 1$

Let $a_n$ be a sequence such that $\forall n\in \mathbb n: 0<a\le a_n\le b <\infty.$

Prove:

1. $\displaystyle\limsup_{n\to\infty}\frac 1 {a_n}=\frac 1 {\displaystyle\liminf_{n\to\infty}a_n}$

2. $\displaystyle\limsup_{n\to\infty}a_n\cdot \limsup_{n\to\infty}\frac 1 {a_n} \ge 1$ and there's an equality iff $a_n$ is converging.

1. Suppose there are two subsequences: $a_{n_l}, \ a_{n_k}$ such that $\lim a_{n_k} = k, \ \lim a_{n_l}=l$ and suppose $l\le k$, so $\lim \frac 1 {a_{n_k}}=\frac 1 k , \ \lim \frac 1 {a_{n_l}}=\frac 1 l$ so clearly: $\frac 1 k\le \frac 1 l\le l\le k$ so it's easy to see once the largest limit (supermum) is 'inverted' it has to become the smallest limit (infimum).

I realize this doesn't show equality, I don't know how to do the other way and I'm not even sure if what I did is good.

2. If $a_n$ converges, suppose to $L$ as its limit then we have: $L\cdot \frac 1 L=1$.

If it does not converge then $a_n$ may tend to infinity or won't have a limit. From 1 we can change it to $\displaystyle\limsup_{n\to\infty}a_n\cdot \frac 1 {\displaystyle\liminf_{n\to\infty}a_n} \ge 1$ and from BW, every sequence has a converging subsequence, and since for converging subsequences: $\liminf a_n\le \limsup a_n$ we have $\frac {\limsup a_n} {\liminf a_n}=\limsup a_n\cdot \limsup\frac 1 {a_n} \ge 1$.

This should probably be in absolute value since one of those subsequnce limits can be negative, but it isn't in absolute value in the question.

1. This is easy. Assume that $m>0$ is an eventual lower bound for $a_n$ (i.e. $a_n \geq m$ for $n \gg 1$). Then $1/a_n \leq 1/m$, so that $\limsup_{n \to +\infty} \frac{1}{a_n} \leq \frac{1}{m}$. By definition of $\liminf$, this means that $\limsup_{n \to +\infty} \frac{1}{a_n} \leq \liminf_{n \to +\infty} a_n$. Now take an eventual upper bound $M$ for $1/a_n$, and deduce similarly that $\liminf_{n \to +\infty} a_n \leq \limsup_{n \to +\infty} \frac{1}{a_n}$.
2. This is proved here.

By your assumption, all the quantities are actually positive.

I think the following lemmas are particularly useful (In fact, some take these as the definitions of limsup's and liminf's).

1) $\limsup a_n \le a$ if and only if for every $\epsilon>0$, $a_n\le a+\epsilon$ for all but a finite number of $a_n$'s.

2) $\limsup a_n \ge a$ if and only if for every $\epsilon>0$, $a_n\ge a-\epsilon$ for infinitely many $a_n$'s.

Similar statements hold for the liminf.

Assuming $$\limsup \frac{1}{a_n}=a$$ we wish to show that $$\liminf a_n=\frac{1}{a}.$$ Let $\epsilon>0$ and choose $\delta>0$ so that $\frac{1}{a+\delta}=\frac{1}{a}+\epsilon$. Then by (1), $\frac{1}{a_n}\le a+\delta$ for all but a finite number of $a_n$'s, hence, $a_n\ge \frac{1}{a+\delta}=\frac{1}{a}+\epsilon$ for all but a finite number of $a_n$'s. Likewise by (2), $\frac{1}{a_n}\ge a+\delta$ for infinitely many $a_n$'s, hence, $a_n\le \frac{1}{a+\delta}=\frac{1}{a}+\epsilon$ for infinitely many $a_n$'s. We have just shown that, $$\liminf a_n = \frac{1}{a}.$$ This is what we wanted to show.

Similar arguments hold for the other cases.

To get the first one, you just need to notice that $$\sup\{1/{a_k}; k\ge n\} = \frac1{\inf\{a_k; k\ge n\}}$$ and take limit for $n\to\infty$ to get $$\lim_{n\to\infty}\sup\{1/a_k; k\ge n\} = \lim_{n\to\infty}\frac1{\inf\{a_k; k\ge n\}} = \frac1{\lim_{n\to\infty}\inf\{a_k; k\ge n\}}.$$ (Although you should also check the cases when the LHS is $+\infty$ and when you have zero in the denominator.)

Once you have shown the first part, you get $$\limsup_{n\to\infty} \frac1{a_n} = \frac1{\liminf_{n\to\infty} a_n} \ge \frac1{\limsup_{n\to\infty} a_n}.$$ Then you can simply multiply this by $\limsup\limits_{n\to\infty} a_n$.

Again, you should check separately the cases where some of the above values is $+\infty$ or when you divide by zero.

• You show $\limsup a_k = 1/\liminf a_k$. But then you say, using this, we have $\limsup (1/a_n) = 1/\liminf a_n$, but surely now there's a fraction which changes things? – Numbersandsoon Jun 29 '14 at 9:09
• Thanks for pointing out my mistake @BoSchmidt. I have edited $\limsup a_k$ (which was incorrect) to $\limsup 1/a_k$. – Martin Sleziak Jun 29 '14 at 9:36


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# 28th IAEA Fusion Energy Conference (FEC 2020)

May 10 – 15, 2021
Virtual Event
Europe/Vienna timezone
The Conference will be held virtually from 10-15 May 2021

## Edge Coherent Mode Providing Continuous Transport during ELM Mitigation with n=1 RMP in HL-2A H-mode plasma

May 12, 2021, 8:30 AM
4h
Virtual Event

#### Virtual Event

Regular Poster Magnetic Fusion Experiments

yi liu

### Description

Due to concerns of erosion and damage to divertors caused by Type-I edge-localized modes (ELMs) in ITER tokamak$[1]$, resonant magnetic perturbation (RMP) has been proposed as an efficient technique for ELM control and has been extensively studied in the past few decades. Complete suppressed of type-I ELMs by RMPs was first demonstrated in DIII-D $[2]$. Since then, suppression or mitigation of ELMs has been achieved in several present day devices. It has been speculated that application of RMP may modify the turbulence behavior of the plasma, though there has so far been no conclusive evidence that enhanced turbulence by RMP is responsible for ELM control. Here, we report the first observation of an edge coherent mode (ECM) in the HL-2A H-mode plasmas with ELM mitigation by RMP, with a frequency range of 2–25 kHz. It was found that the mode is located at the edge pedestal region with steep pressure gradient, being excited by the three-wave interaction of turbulence enhanced by the application of RMP. The ECM drives a significant outflow of particles and heat flux, thus providing a channel for continuous extra particle and heat transport across the pedestal during the ELM mitigation.

Mitigation of Type-I ELMs, utilizing the n = 1 RMP field, is observed in HL-2A as shown in Fig.1.The D$_{\alpha}$ signal shows a reduction in amplitude and an about two-third increase in frequency of the ELM bursts [Fig. 1(a)]. A strong reduction in the amplitude of the peak heat flux deposited onto the divertor targets, due to ELM bursts, was also observed [Fig. 1(b)]. Figure 1(c) shows reduced heat flux and a broader flux distribution during the ELM mitigation phase. The MARS-F toroidal modeling for this type of HL-2A discharges has suggested a strong edge-peeling response as the main reason leading to the ELM mitigation $[3]$. A key finding during this experiment is the occurrence of small 2kHz coherent oscillations (ECM), appearing between the reduced Type-I ELM events or riding on them as shown in Fig.2.

The impact of RMP on the plasma turbulence is examined by a new BES system covering the density fluctuation in the pedestal region. A pronounced increase in the density fluctuation level is found in the power spectrum after RMP is switched on. To gain a deeper insight into the RMP impact on the plasma turbulence and the ECM, we have investigated the possible nonlinear interaction or energy cascading among different turbulent components. We found that the nonlinear coupling between different frequency components was substantially enhanced during RMP as shown in Fig.3. The bi-spectrum exhibits strong coupling during RMP as compared with that before RMP, suggesting that the ECM is excited by the three-wave interaction of turbulences enhanced by RMP.

Furthermore, clear evidence of particle exhaust by the ECM has been obtained, via a close examination of the measurement of particle flux in the plasma edge region as shown in Fig.4.The envelope of the saturate particle flow exhibits an oscillation at about 2 kHz, suggesting that the ECM contributes to the outward particle transport. This significant correlation indicates that the ECM can indeed induce a continuous outward particle flux, leading to extra particle transport during the application of the n=1 RMP in HL-2A.

This work was supported by NSFC under Contracts No. 11875018.

References
$[1]$ A. Loarte et al, Nucl. Fusion 54, 033007 (2014)
$[2]$ Evans T.E. et al, Phys. Rev. Lett. 92, 235003 (2004)
$[3]$ Liu et al, Phys. Plasmas 24, 056111 (2017)

Country or International Organization China Southwestern Institute of Physics

### Primary authors

Tengfei Sun (Southwestern Institute of Physics) Xiaoquan JI (Southwestern Institute of Physics, Chengdu 610041 China) Yueqiang Liu (General Atomics, PO Box 85608, San Diego, CA 92186-5608, USA) Rui Ke (SWIP, China) Dr Jinming Gao (Southwestern Institute of Physics, Chengdu 610041, People’s Republic of China) Wei Deng (Southwestern Institute of Physics) Dr Na Wu (Southwestern Institute of Physics, Chengdu 610041, People’s Republic of China) Dr Min Xu (Southwestern Institute of Physics, Chengdu 610041, People’s Republic of China) Xuru Duan (Southwestern Institute of Physics)


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# Reward

Janko and Peter divided the reward from the brigade so that Peter got 5/8 of the reward. What percentage of Janko's reward got?

Correct result:

p =  37.5 %

#### Solution:

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# The Unapologetic Mathematician

## Properties of Irreducible Root Systems III

Today we conclude with our series of lemmas on irreducible root systems.

If $\Phi$ is irreducible, then roots in $\Phi$ have at most two different lengths. Here I mean actual geometric lengths, as measured by the inner product, not the “length” of a Weyl group element. Further, any two roots of the same length can be sent to each other by the action of the Weyl group.

Let $\alpha$ and $\beta$ be two roots. We just saw that the $\mathcal{W}$-orbit of $\alpha$ spans $V$, and so not all the $\sigma(\alpha)$ can be perpendicular to $\beta$. From what we discovered about pairs of roots, we know that if $\langle\alpha,\beta\rangle\neq0$, then the possible ratios of squared lengths $\frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}$ are limited. Indeed, this ratio must be one of $\frac{1}{3}$, $\frac{1}{2}$, ${1}$, ${2}$, or ${3}$.

If there are three distinct root-lengths, let $\alpha$, $\beta$, and $\gamma$ be samples of each length in increasing order. We must then have $\frac{\lVert\beta\rVert^2}{\lVert\alpha\rVert^2}=2$ and $\frac{\lVert\gamma\rVert^2}{\lVert\alpha\rVert^2}=3$, and so $\frac{\lVert\gamma\rVert^2}{\lVert\beta\rVert^2}=\frac{3}{2}$, which clearly violates our conditions. Thus there can be at most two root lengths, as asserted. We call those of the smaller length “short roots”, and the others “long roots”. If there is only one length, we call all the roots long, by convention.

Now let $\alpha$ and $\beta$ have the same length. By using the Weyl group as above, we may assume that these roots are non-orthogonal. We may also assume that they’re distinct, or else we’re already done! By the same data as before, we conclude that $\alpha\rtimes\beta=\beta\rtimes\alpha=\pm1$. We can replace one root by its negative, if need be, and assume that $\alpha\rtimes\beta=1$. Then we may calculate:

$\displaystyle[\sigma_\alpha\sigma_\beta\sigma_\alpha](\beta)=[\sigma_\alpha\sigma_\beta](\beta-\alpha)=\sigma_\alpha(-\beta-\alpha+\beta)=\alpha$.

We may note, in passing, that the unique maximal root $\beta$ is long. Indeed, it suffices to show that $\langle\beta,\beta\rangle\geq\langle\alpha,\alpha\rangle$ for all $\alpha\in\Phi$. We may, without loss of generality, assume $\alpha$ is in the fundamental doman $\overline{\mathfrak{C}(\Delta)}$. Since $\beta-\alpha\succ0$, we must have $\langle\gamma,\beta-\alpha\rangle\geq0$ for any other $\gamma\in\overline{\mathfrak{C}(\Delta)}$. In particular, we have $\langle\beta,\beta-\alpha\rangle\geq0$ and $\langle\alpha,\beta-\alpha\rangle\geq0$. Putting these together, we conclude

$\displaystyle\langle\beta,\beta\rangle\geq\langle\alpha,\beta\rangle\geq\langle\alpha,\alpha\rangle$

and so $\beta$ must be a long root.

February 12, 2010 Posted by | Geometry, Root Systems | 4 Comments


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# nLab model structure on spectra

model category

## Model structures

for ∞-groupoids

### for $(\infty,1)$-sheaves / $\infty$-stacks

#### Stable homotopy theory

stable homotopy theory

# Contents

## Idea

A model category structure on a category of spectra presents a stable (∞,1)-category of spectrum objects.

Many such model structures model spectrum objects as symmetric spectra. See there for more details.

Typically and naturally, a model structure on spectra forms a stable model category.

## Examples

Among the model structures with a symmetric monoidal smash product of spectra (“highly structured spectra”) are

## References

A discussion of model structures on spectra in general ambient model categories (general spectrum objects) is in

Revised on November 15, 2013 05:23:12 by Urs Schreiber (82.169.114.243)