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Song* , Liu* , Tian** , and Fong**: Pointwise CNN for 3D Object Classification on Point Cloud

# Pointwise CNN for 3D Object Classification on Point Cloud

Abstract: Three-dimensional (3D) object classification tasks using point clouds are widely used in 3D modeling, face recognition, and robotic missions. However, processing raw point clouds directly is problematic for a traditional convolutional network due to the irregular data format of point clouds. This paper proposes a pointwise convolution neural network (CNN) structure that can process point cloud data directly without preprocessing. First, a 2D convolutional layer is introduced to percept coordinate information of each point. Then, multiple 2D convolutional layers and a global max pooling layer are applied to extract global features. Finally, based on the extracted features, fully connected layers predict the class labels of objects. We evaluated the proposed pointwise CNN structure on the ModelNet10 dataset. The proposed structure obtained higher accuracy compared to the existing methods. Experiments using the ModelNet10 dataset also prove that the difference in the point number of point clouds does not significantly influence on the proposed pointwise CNN structure.

Keywords: Point Clouds , Pointwise CNN , 3D Object Classification

## 1. Introduction

In recent years, using point clouds for three-dimensional (3D) object classification has become a popular method in various fields, such as face recognition [1], 3D modeling [2], intelligence surveillance [3], and robotic missions [4]. The characteristics of point clouds, for example, high precision, are vital for object classification. Compared to traditional two-dimensional (2D) images captured by a camera, point clouds have some significant advantages. For example, point clouds are invariant to lighting, rotation, and color. However, the unordered arrangement, inhomogeneous densities, and non-structural distributions of point clouds present challenges for object classification.

To address these problems, researchers use point cloud features as the basis for classification rather than processing raw point cloud data. Most methods, such as 3D shape context [5], point signature [6], and clustered viewpoint feature histogram [7], focus on analyzing and extracting point cloud features, such as geometric, shape, and structural attributes, or combinations of multiple attributes [8]. Traditionally, data size may increase unnecessarily when input to convolutional networks because convolutional networks require regular inputs. Thus, PointNet [9], a neural network structure that takes the original point clouds as input, has been developed. This paper proposes a lightweight pointwise convolution neural network (CNN) structure for 3D object classification. The proposed pointwise CNN requires fewer training epochs than the PointNet and achieves 87.07% classification accuracy on the ModelNet10 dataset.

The remainder of this paper is organized as follows. Section 2 introduces previous studies to 3D object classification in point clouds. Section 3 describes the proposed network structure. Section 4 describes the experiments and evaluates the classification results. Conclusions and suggestions for future work are provided in Section 5.

## 2. Related Work

Researchers have done considerable work in object classification using 3D point cloud data. Such studies primarily employed feature-based, graph matching-based, and machine learning-based object classification methods.

In general, feature-based object classification methods can be classified as global and local methods. Methods that combine two categories of features also occur in the face recognition domain [10]. In global feature-based methods, target objects must first be separated from the background. Then, their geometric features are matched to classify the objects. Drost et al. [11] introduced oriented point pair features and employed a fast voting scheme to match models locally. Their proposed method grouped similar features of a model when mapping point pair feature space to the model, which increased the speed of the algorithm when processing point cloud data. Rusu et al. [12] introduced the view feature histogram (VFH), which added viewpoint information to the extended fast point feature histogram (FPFH) information introduced in their previous work [13]. The VFH method could recognize both objects and their pose. In addition, it collected statistics of the relative angle between the surface normal and the direction of the central viewpoint. Therefore, the computational cost of the VFH was reduced, which allowed it to be used in real-time processing. Rusu et al. [14] also proposed global fast point feature histograms to capture the local geometric relationships in an object. In that study, a support vector machine was employed to label and recognize objects, and high classification accuracy was achieved. Marton et al. [15] introduced a global radius-based surface descriptor (RSD) that rasterized the point cloud and sorted the cambers first. Then, the RSD was used to calculate features for classification. Wohlkinger and Vincze [16] introduced an ensemble of shape functions (ESF) for real-time 3D object recognition. This descriptor combined angle, distance between two points, and shape functions, and significantly improved recognition rate. However, experimental results indicated that the ESF descriptor had difficulty identifying similarly shaped objects. Chen et al. [17] proposed the global Fourier histogram (GFH) descriptor, which could use the cylindrical angular coordinate and achieve rotational independence around the vertical axis. Their experimental results indicated that the GFH classified almost all objects correctly; however, it encountered the same problem as ESF, i.e., the descriptor might return incorrect results if two objects had a similar shape. In general, global feature-based methods failed to describe some details of the object and were likely influenced by noise and occlusion.

To address the limitations of global features, researchers focused on developing methods that used local features for classification, for example an object’s key points. Spin image (SI) is a classic method introduced by Johnson and Hebert [18]. In SI, 2D spin images were used to represent 3D point clouds. Although SI showed robustness against noise and occlusion, it required uniform distribution of point cloud data. Guo et al. [19] introduced the Tri-Spin-Image (TriSI) descriptor, which recognized 3D objects under clutter interference and occlusion conditions. Their system included a local reference frame and generated signatures. Then, these signatures were used to compress TriSI features, and a hierarchical feature matching method was employed for object recognition. TriSI showed robustness against noise and varying resolutions, which resolved the weakness of the classic SI method. Rusu et al. [20] introduced a point feature histogram (PFH) and a FPFH [13]. The PFH calculated features, such as the distance between two points and the angle between two points’ normal vectors, and then mapped those features to a histogram to obtain statistics. However, the PFH algorithm was highly complex. Thus, the FPFH was developed to improve the efficiency of PFH. FPFH used the relationship between a normal vector and its k-neighbor, which made FPFH more efficient than PFH. Salti et al. [21] introduced the signature of histograms of orientation (SHOT) method, which was robust against noise and rotational invariance. Prakhya et al. [22] improved the SHOT method [21] by introducing a “binary” 3D feature descriptor called binary signature of histograms of orientations (B-SHOT). They employed binary quantization to convert a real vector to a binary vector, which made B-SHOT run faster and require less memory than SHOT. He and Mei [23] proposed a new spin image-based registration (SIR) algorithm using 3D feature space that included the Tsallis entropy of the spin image and the laser sensor’s reflection intensity. The new SIR algorithm improved the robustness and computational speed of previous SIR algorithms.

With the successful application of deep learning systems in many fields, researchers also focused on applying deep learning systems to 3D data. Maimaitimin et al. [24] converted point clouds to a surfacecondition- feature map for feature extraction using an autoencoder. The geometric features extracted work well even with extremely noisy data. Su et al. [25] proposed a multi-view CNN that rendered point clouds into 2D images (views), and then combined information from different views for classification. A viewpooling layer was employed to pool the extracted view-based features from different views. The classification accuracy of their proposed method was significantly better than other 3D shape descriptors. Bobkov et al. [26] developed a point pair descriptor and a four-dimensional (4D) CNN for object classification. The proposed method showed robustness to noise and occlusion in point clouds; however, tuning network hyperparameters required optimization and the computational cost of a 4D convolution cloud was quite high. Ben-Shabat et al. [27] introduced a new representation of a 3D point cloud called 3D modified Fisher vectors. Employing continuous generalized Fisher vectors and a coarse discrete grid structure provided robustness against data loss. Li et al. [28] introduced a new deep neural network structure called a field probing neural network that employed field probing filters to extract features from volumetric space effectively; therefore, it ran much faster than traditional 3D CNNs. Zhou and Tuzel [29] proposed VoxelNet, a 3D detection network that puts a point cloud into an equidistant 3D voxel and then calculates a unified feature for each voxel using a voxel feature encoding layer. Engelcke et al. [30] proposed an efficient CNN to recognize 3D objects in point clouds. They introduced feature-centric voting scheme convolutional layers for recognition. Their model used fewer layers than previous models and simultaneously realized competitive processing time. Fang et al. [31] defined a deep shape descriptor (DeepSD) for a deep neural network. DeepSD demonstrated robustness against noise, incompleteness, and structural variations. Klokov and Lempitsky [32] proposed a deep learning network structure named Kd-network. Differing from traditional convolutional networks, Kd-networks did not require rasterization prior to processing point clouds, which eliminate the time required for rasterization and made the network run faster.

Researchers have also attempted to find ways to handle unordered input data because significant time is required to convert unordered data to ordered data. Vinyals et al. [33] proposed a framework with attention mechanisms that include read, process, and write blocks. Although this framework could handle unordered input sets, their method addressed language processing rather than 3D object classification, and employing that method for 3D object classification tasks was difficult. In this paper, we employed a CNN to classify 3D objects under point cloud data.

## 3. Pointwise CNN-Based 3D Object Classification Method

We propose a pointwise CNN for 3D object classification (Fig. 1). In our work, raw point cloud data is input to the network directly without any preprocessing. The developed pointwise CNN consists of four convolution layers (Conv1–Conv4), one max pooling layer, and four fully connected layers (FC1– FC4). The Conv1 layer applies 64 1×3 convolution kernels. The other three layers apply 128, 256, and 512 1×1 kernels, respectively. The max pooling layer employs a global max pooling method to extract 512 features from the convolution results of the Conv4 layer. Four fully connected layers, which contain 512, 256, 128, and k neurons in each layer, provide the classification results. A backward process is implemented for training to update the weight matrixes in the convolutional and fully connected layers, and the bias vectors in the fully connected layers. A dropout layer with a drop rate of 0.2 is implemented before FC4 to prevent overfitting. Variable N in Fig. 1 is the number of points.

Fig. 1.

Proposed pointwise CNN for 3D object classification.
##### 3.1 Forward Process for Training and Testing

Two types of convolutional kernels are used in this study. In the first convolutional layer, we use 1×3 kernels to combine the coordinate information of each point in the point cloud. Fig. 2 and Eq. (1), respectively, illustrate and represents how these kernels work.

In Eq. (1), operator * represents convolution, variable [TeX:] $$p_{n}$$ is the coordinate information of a point where n ranges from 1 to N where N is the number of points, [TeX:] $$r_{u, n}$$ is the result of a convolution, [TeX:] $$k_{u}$$ is the [TeX:] $$u^{t h}$$ convolutional kernel where variable u ranges from 1 to m, and variable m is the number of convolution kernels.

##### (1)
[TeX:] $$r_{u, n}=p_{n} * k_{u}$$

The size of the output array is calculated in two steps. First, the width and height are calculated using equations [TeX:] $$O_{w}=I_{w}+1-k_{w} \text { and } O_{h}=I_{h}+1-k_{h},$$ respectively. Second, [TeX:] $$O_{w} \text { and } O_{h}$$ are multiplied to obtain the array size. Variables [TeX:] $$O_{w}, O_{h}, I_{w}, I_{h}, k_{w}, \text { and } k_{h}$$ represent the width and height of the output array, input array, and kernel, respectively.

Fig. 2.

Convolution computed by 1×3 kernels.

In the other three convolutional layers, 1×1 convolution kernels are used to combine the information from different channels. Each kernel convolutes the values at the same position from different channels in the input arrays. The convolutional results will be saved at the same position in the output array. The output arrays and the kernels are in a one-to-one correspondence. Fig. 3 illustrates how a 1×1 convolutional kernel works where variable d represents input and kernel channel.

Fig. 3.

Convolution computed by the 1×1 kernel.

Note that no biases are added to the output results in all convolutional layers prior to the results being sent to the activation function. The activation function for all four convolutional layers is the leaky ReLU defined in Eq. (2).

##### (2)
[TeX:] $$\sigma(m)=\left\{\begin{array}{cc} 0.01 m, & m \leq 0 \\ m, & m>0 \end{array}\right.$$

Here, m represents the values in output arrays.

To solve the problem of irregular input, we need to utilize a symmetry function as a pooling method. Thus, we employ the global max pooling function to extract 512 features from the last convolutional layer, as shown in Fig. 4. The pool size of this max pooling function equals the size of the output array in the last convolutional layer.

Fig. 4.

Global max pooling operation.

The features extracted from the last convolutional layer are then sent to the four fully connected layers. The operational principle of the fully connected layers is shown in Eqs. (3) and (4) where matrix W is a matrix that contains the weights of the layer, function [TeX:] $$\sigma(\cdot)$$ is the leaky ReLU function defined in Eq. (2), and variables [TeX:] $$b_{i} \text { and } x_{i}$$ represent the bias and input vectors in the [TeX:] $$i^{t h}$$ layer, respectively. The activation function for first three fully connected layers is the same as the activation function defined in Eq. (2).

##### (3)
[TeX:] $$z_{i}=W x_{i}+b_{i}$$

##### (4)
[TeX:] $$a_{i}=\sigma\left(z_{i}\right)$$

The results of the last layer are then sent to the softmax activation function shown in Eq. (5) where [TeX:] $$S_{i}$$ is the result of a softmax operation on a neuron, [TeX:] $$a_{i}$$ is calculated by Eq. (4), and variable c is the number of object categories.

##### (5)
[TeX:] $$s_{i}=\frac{e^{a_{i}}}{\sum_{j=1}^{c} e^{a_{j}}}$$

We select the max value from [TeX:] $$s_{i},$$ and then take the number i as the classification result. Note that testing is performed at this stage. However, training requires an updating process for weights and kernels in the network.

##### 3.2 Backward Process

The first step of the backward process used for training is calculating the loss rate q using a loss function. We employ the cross-entropy function shown in Eq. (6), where yi is an one-hot encoded label for the [TeX:] $$\dot{l^{t h}}$$ neuron, [TeX:] $$S_{i}$$ is the output of this neuron after the softmax operation, and c is the number of categories.

##### (6)
[TeX:] $$q=-\sum_{i=1}^{c} y_{i} \ln s_{i}$$

The gradient descent method is implemented to update the weight or the kernel in the network. Thus, we need to calculate the gradient for each layer and then update the weight or the kernel of each layer accordingly. The gradients of each layer are calculated by the chain derivation rule, which means that the gradient of the last fully connected layer is the most important for updating. Eq. (7) shows how to calculate the gradient of the [TeX:] $$i^{t h}$$ output result [TeX:] $$z_{i}$$ in the last fully connected layer. In Eq. (7), variables q and [TeX:] $$s_{j}$$ have the same meaning as in Eq. (6), [TeX:] $$z_{i}$$ is calculated by Eq. (3), and variable c is the number of categories.

##### (7)
[TeX:] $$\frac{\partial q}{\partial z_{i}}=\sum_{j=1}^{c} \frac{\partial q}{\partial s_{j}} \frac{\partial s_{j}}{\partial z_{i}}$$

Eqs. (8) and (9) provide the results of two partial differentials in Eq. (7).

##### (8)
[TeX:] $$\frac{\partial s_{j}}{\partial z_{i}}=\left\{\begin{array}{l} s_{i}-s_{i}^{2}, i=j \\ -s_{i} s_{j}, i \neq j \end{array}\right.$$

##### (9)
[TeX:] $$\frac{\partial q}{\partial s_{j}}=-\sum_{j=1}^{c} \frac{y_{j}}{s_{j}}$$

Combining the above results, Eq. (7) is equivalent to Eq. (10), where variable c has the same meaning as in Eq. (7) and variable [TeX:] $$y_{i}$$ has the same meaning as in Eq. (6). The one-hot encoder is employed in this study; therefore, the gradient of the last fully connected layer can be calculated using Eq. (11) where [TeX:] $$y_{i}$$ and [TeX:] $$S_{i}$$ have the same meaning as in Eq. (6), and [TeX:] $$z_{i}$$ is calculated by Eq. (3).

##### (10)
[TeX:] $$\frac{\partial q}{\partial z_{i}}=s_{i} \sum_{j=1}^{c} y_{j}-y_{i}$$

##### (11)
[TeX:] $$\frac{\partial q}{\partial z_{i}}=s_{i}-y_{i}$$

The gradients are then used to update the weights and biases in the fully connected layers. The error rates of the last layer and other fully connected layers can be calculated by Eq. (12). Eqs. (13) and (14) show how to calculate the new weight of a layer.

##### (12)
[TeX:] $$e_{i}=W_{i+1}^{T} e_{i+1} \circ \sigma^{\prime}\left(z_{i}\right)$$

##### (13)
[TeX:] $$W_{i}=W_{i}-e_{i} \cdot a_{i-1}^{T} l$$

##### (14)
[TeX:] $$b_{i}=b_{i}-l e_{i}$$

In Eqs. (12)–(14), operator denotes the Hadamard product, variable [TeX:] $$e_{i}$$ is the error rate of the [TeX:] $$i^{t h}$$ layer, [TeX:] $$W_{i}$$ is the weight matrix of the [TeX:] $$i^{t h}$$ layer, and variable l is the learning rate for fully connected layers. Variable [TeX:] $$b_{i}$$ is the bias vector of the [TeX:] $$i^{t h} \text { layer, } a_{i-1}$$ is the input of the [TeX:] $$i^{t h}$$ layer, and [TeX:] $$z_{i}$$ is the output of the [TeX:] $$i^{t h}$$ layer before the activation function operation. According to the leaky ReLU function defined Eq. (2), the values of function [TeX:] $$\sigma^{\prime}\left(z_{i}\right)$$ are shown in Eq. (15).

##### (15)
[TeX:] \sigma^{\prime}\left(z_{i}\right)=\left\{\begin{aligned} 0.01, z_{i} \leq 0 \\ 1, z_{i}>0 \end{aligned}\right.

The upsampling operation is done by the max pooling layer, which puts the error rates transmitted from the previous fully connected layer into the position where the maximum was in the forward process. The blank positions in the pool will be filled with zeroes.

There are two ways to obtain the error rate for convolutional layers in the backward process.

1. If the previous layer is the max pooling layer, the error rate of the previous layer is transmitted to current layer directly.

2. If the previous layer is a convolutional layer, the error rate of the current layer should be calculated based on the error rate of the previous layer. To calculate the error rate, convolution is implemented as shown in Eq. (16). In Eq. (16) [TeX:] $$e_{i}$$ is the error rate of the [TeX:] $$i^{t h} \text { layer, } k_{i}$$ are the kernels in the [TeX:] $$i^{t h}$$ layer, operator * means convolution, and function [TeX:] $$\operatorname{rot}(\cdot)$$ means to rotate the kernel 180. Note, function [TeX:] $$\sigma^{\prime}\left(z_{i}\right)$$ and operator are defined as in Eq. (12).

##### (16)
[TeX:] $$e_{i}=e_{i+1} * \operatorname{rot}\left(k_{i+1}\right) \circ \sigma^{\prime}\left(z_{i}\right)$$

Eq. (17) shows how the kernels are updated. Here variable [TeX:] $$k_{i}$$ is the kernel in the [TeX:] $$i^{t h}$$ convolutional layer, [TeX:] $$e_{i}, a_{i-1}, \text { and } l$$ are the error rate, input data, and the learning rate for this layer, respectively. Operator * denotes convolution.

##### (17)
[TeX:] $$k_{i}=k_{i}-e_{i} * a_{i-1} l$$

## 4. Object Recognition Experiments

The proposed network was tested on the ModelNet10 dataset [34]. The dataset contains 4,899 objects from 10 categories, i.e., bathtub, bed, chair, desk, dresser, monitor, nightstand, sofa, table, and toilet, as shown in Fig. 5. Note that the objects in this dataset are aligned in orientation.

Fig. 5.

Examples of categories in the ModelNet10 dataset.

In our experiments, we employed balanced and unbalanced datasets. The partition in the unbalanced set was the same as that of the original ModelNet10 dataset. For the balanced set, we selected 106 samples from the original training set of 10 categories. The remainder of the samples in the original training set were added to the test set. Table 1 shows the partition of training and test sets for each category in the balanced and unbalanced datasets. The learning rate for all fully connected layers and all convolutional layers was set to 0.0001. The experiments were performed on a computer with an Intel Xeon CPU E5- 2670 v3 @ 2.30 GHz, 56 GB RAM, using the Windows 10 operating system.

Table 1.

Partition of training sets and test sets
Category Balanced set Unbalanced set
Training set Test set Total Training set Test set Total
Bathtub 106 50 156 106 50 156
Bed 106 509 615 515 100 615
Chair 106 883 989 889 100 989
Desk 106 180 286 200 86 286
Desk 106 180 286 200 86 286
Monitor 106 459 565 465 100 565
Nightstand 106 180 286 200 86 286
Sofa 106 674 780 680 100 780
Table 106 386 492 392 100 492
Toilet 106 338 444 344 100 444
Total sample 1,060 3,839 4,899 3,991 908 4,899

Table 2.

Classification accuracy of each category on test datasets
Category Accuracy (%)
Balanced dataset Unbalanced dataset
Bathtub 74.00 80.00
Bed 87.22 90.00
Chair 93.54 99.00
Desk 78.33 60.46
Dresser 86.11 87.20
Monitor 96.29 98.00
Nightstand 81.66 86.04
Sofa 88.27 95.00
Table 73.31 79.00
Toilet 94.67 96.00
Average 85.34 87.07

Fig. 6.

Confusion matrix of classification results on balanced set.

Fig. 7.

Confusion matrix of classification results on unbalanced set.

We extracted x, y, z coordinate information from the original ModelNet10 Object File Format files and saved the information in individual files that included the coordinate information of a 3D object. In both the training and testing processes, we input the files one at a time. Table 2 shows the classification accuracy of each category on the balanced and unbalanced datasets. With the proposed network, the average accuracy reached 87.07% on the unbalanced dataset after 90 iterations on the training dataset and 85.34% on the balanced set after 140 iterations on the training dataset. The proposed network had a better performance on the unbalance dataset because more samples were used for training.

Figs. 6 and 7 show the classification result confusion matrixes for the balanced and unbalanced datasets, respectively. Note that the values in the matrixes are the number of examples, not percentages. As shown in Fig. 6, the network confused samples in the table and desk classes on the balanced dataset. We thought this occurred because the objects had a similar shape. However, there were some exceptions, e.g., the object in Fig. 5(i) is a round table. As mentioned in Section 2, global features have difficulty classifying objects with similar shape. However, on the unbalanced set, the network misclassified most samples in the desk category as sofa and nightstand. We considered that this result occurred because there are a greater number of samples in the sofa category in the unbalanced training dataset. Another reason for the misclassification was information loss in the point cloud data. The information loss made objects in the desk category share more similarities with samples in the sofa and nightstand categories. This also explained why sofa and nightstand were misclassified as desk (second and third place) on the balanced set.

We compared our results on the unbalanced set against PointNet structure [9] and 3D ShapeNets developed by Wu et al. [34], as shown in Table 3. Compared with these previous studies, the classification accuracy in our work increased by 9.47% and 3.53%, respectively. Compared to the PointNet structure [9], the iteration times on the training set also decreased from 200 to 90.

Table 3.

Classification accuracy on test set
Method Average accuracy (%) Training iteration
PointNet [9] 77.60 200
3D ShapeNet [34] 83.54 -
Proposed 87.07 90

Fig. 8 is a line chart of the average accuracies for the test set under different training iterations. Here, the horizontal axis is iteration times, and the vertical axis is average accuracy. We found that our proposed network outperformed the PointNet structure [9] after 20 iterations (81.91%). The average accuracy of the proposed method is almost the same as that of the method developed by Wu et al. [34] after 30 iterations.

Fig. 8.

Testing accuracy under different iterations.

## 5. Conclusion

This paper proposed a pointwise CNN for 3D point cloud classification that consumes raw point cloud data directly. The proposed system employs four convolutional layers and a max pooling layer to extract features from origin point cloud data. Then, four fully connected layers are employed for classification. The system was tested on the ModelNet10 dataset and achieved average accuracy of 87.07% on the unbalanced training set after 90 iterations and 85.34% on the balanced training set after 140 iterations. Average accuracy improved 9.47% compared to the existing PointNet structure with fewer iterations. However, the proposed network confused objects with similar shape, particularly objects in the desk, nightstand, and table categories. We believe this occurred because we employed global features for object classification and due to information loss in the point cloud data. In future, to improve accuracy, we will consider implementing local features for classification.

## Acknowledgement

This research was funded by the MSIT (Ministry of Science, ICT), Korea, under the High-Potential Individuals Global Training Program (No. 2020-0-01576) supervised by the IITP (Institute for Information Communications Technology Planning Evaluation), National Nature Science Foundation of China (No. 61503005), the Great Wall Scholar Program (No. CITTCD20190305), and NCUT funding (No. 110052972027/008).

## Biography

##### Wei Song
https://orcid.org/0000-0002-5909-9661

He received his B.Eng. degree in Software Engineering from Northeastern University, Shenyang, China, in 2005, and his M.Eng. and Dr.Eng. segrees in the Department of Multimedia from Dongguk University, Seoul, Korea, in 2008 and 2013, respectively. Since September 2013, he has been an associate professor at the Department of Digital Media Technology of North China University of Technology. His current research interests are focused on parallel computation, 3D modeling, object recognition, virtual reality, and Internet of Things.

## Biography

##### Zishu Liu
https://orcid.org/0000-0002-3444-5919

He is a postgraduate student at the School of Information of North China University of Technology, Beijing, China, since September 2018. His current research interests are focus on machine learning, 3D object recognition and deep learning.

## Biography

##### Yifei Tian
https://orcid.org/0000-0001-8884-2170

She is a postgraduate student at the Computer and Information Science Department of the University of Macau, Macau, China, since September 2018. Her current research interests are focused on IoT, big data mining, computer graphics, and 3D reconstruction.

## Biography

##### Simon Fong
https://orcid.org/0000-0002-1848-7246

He graduated from La Trobe University, Australia, with a 1st Class Honours B.Eng. Computer Systems degree and a Ph.D. Computer Science degree in 1993 and 1998 respectively. Simon is now working as an Associate Professor at the Computer and Information Science Department of the University of Macau. He is also one of the founding members of the Data Analytics and Collaborative Computing Research Group in the Faculty of Science and Technology. Prior to his academic career, Simon took up various managerial and technical posts, such as systems engineer, IT consultant and e-bommerce director in Australia and Asia.

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text:
Seminar

# Cubes, side lengths and centres and their Furstenberg type siblings

#### Han Yu - University of St Andrews

We will discuss Bourgain-Wolff-Schlag circle maximal operators to see what happens if we replace circles by cubes or polygonal shapes. In recent work by (Keleti, Nagy, Shmerkin) and (Chang, Csörnyei, Héra and Keleti) (rotated)cubes with large centre set were considered. In this talk we talk about cubes with large side length set and similar result holds in that case for example, in dimension n, a set which contains surfaces of cubes with all side lengths has Hausdorff dimension at least n-1 and box/Assouad dimension n-0.5, this bound is sharp. This is not the end of the story. We show such sets must be large in general in a precise sense that will be formulated in the talk. As a by-product we will encounter a Kakeya type construction which we shall call 'grass sets', a corresponding 'grass conjecture' will be formulated. Everything above has Furtenberg type generalization which we will briefly discuss. As a by-product, we revisit Wolff's circle maximal inequality and a $\alpha,\beta$-Wolff problem will be formulated as a philosophical analogy of $\alpha,\beta$ Furstenberg problem. To be more related with Furtenberg sets, we discuss a certain dynamical defined set which has self-similar centre set and 'self-similar' way of assigning radius we shall see that under some rational independence such set must hit the expected large Assouad dimension. This can be seen as the Furtenberg times 2 times 3 set conjecture which was proved by Shmerkin and Wu independently. We conjectured that all other dimensions (Hausdorff for example) must hit the large expected value. Subject to time limit, only a subset of the above results which has 0 dimension can be presented
Organizers
Kenneth J. Falconer
University of St Andrews
Maarit Järvenpää
University of Oulu
Antti Kupiainen
University of Helsinki
Francois Ledrappier
University of Notre Dame
Pertti Mattila
University of Helsinki

# ProgramContact

Maarit Järvenpää

maarit.jarvenpaa@oulu.fi

# Otherinformation

For practical matters at the Institute, send an e-mail to administration@mittag-leffler.se

format: plain


text:
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### Learning Objectives

1. Solve equations involving square roots.
2. Solve equations involving cube roots.

A radical equationAny equation that contains one or more radicals with a variable in the radicand. is any equation that contains one or more radicals with a variable in the radicand. Following are some examples of radical equations, all of which will be solved in this section:

 $2x−1=3$ $4x2+73−2=0$ $x+2−x=1$

We begin with the squaring property of equalityGiven real numbers a and b, where $a=b$, then $a2=b2.$; given real numbers a and b, we have the following:

$If a=b, then a2=b2.$

In other words, equality is retained if we square both sides of an equation.

$−3=−3⇒(−3)2=(−3)29=9 ✓$

The converse, on the other hand, is not necessarily true,

$9=9(−3)2=(3)2⇒−3≠3 ✗$

This is important because we will use this property to solve radical equations. Consider a very simple radical equation that can be solved by inspection,

$x=5$

Here we can see that $x=25$ is a solution. To solve this equation algebraically, make use of the squaring property of equality and the fact that $(a)2=a2=a$ when a is nonnegative. Eliminate the square root by squaring both sides of the equation as follows:

$(x)2=(5)2x=25$

As a check, we can see that $25=5$ as expected. Because the converse of the squaring property of equality is not necessarily true, solutions to the squared equation may not be solutions to the original. Hence squaring both sides of an equation introduces the possibility of extraneous solutionsA properly found solution that does not solve the original equation., which are solutions that do not solve the original equation. For example,

$x=−5$

This equation clearly does not have a real number solution. However, squaring both sides gives us a solution:

$(x)2=(−5)2x=25$

As a check, we can see that $25≠−5.$ For this reason, we must check the answers that result from squaring both sides of an equation.

### Example 1

Solve: $3x+1=4.$

Solution:

We can eliminate the square root by applying the squaring property of equality.

$3x+1=4(3x+1)2=(4)2Square both sides.3x+1=16Solve.3x=15x=5$

Next, we must check.

$3(5)+1=415+1=416=44=4 ✓$

There is a geometric interpretation to the previous example. Graph the function defined by $f(x)=3x+1$ and determine where it intersects the graph defined by $g(x)=4.$

As illustrated, $f(x)=g(x)$ where $x=5.$

### Example 2

Solve: $x−3=x−5.$

Solution:

Begin by squaring both sides of the equation.

$x−3=x−5(x−3)2=(x−5)2Square both sides.x−3=x2−10x+25$

The resulting quadratic equation can be solved by factoring.

$x−3=x2−10x+250=x2−11x+280=(x−4)(x−7)x−4=0or x−7=0x=4x=7$

Checking the solutions after squaring both sides of an equation is not optional. Use the original equation when performing the check.

$Check x=4$

$Check x=7$

$x−3=x−54−3=4−51=−11=−1 ✗$

$x−3=x−57−3=7−54=22=2 ✓$

After checking, you can see that $x=4$ is an extraneous solution; it does not solve the original radical equation. Disregard that answer. This leaves $x=7$ as the only solution.

Geometrically we can see that $f(x)=x+3$ is equal to $g(x)=x−5$ where $x=7.$

In the previous two examples, notice that the radical is isolated on one side of the equation. Typically, this is not the case. The steps for solving radical equations involving square roots are outlined in the following example.

### Example 3

Solve: $2x−1+2=x.$

Solution:

Step 1: Isolate the square root. Begin by subtracting 2 from both sides of the equation.

$2x−1+2=x2x−1=x−2$

Step 2: Square both sides. Squaring both sides eliminates the square root.

$(2x−1)2=(x−2)22x−1=x2−4x+4$

Step 3: Solve the resulting equation. Here we are left with a quadratic equation that can be solved by factoring.

$2x−1=x2−4x+40=x2−6x+50=(x−1)(x−5)x−1=0or x−5=0x=1x=5$

Step 4: Check the solutions in the original equation. Squaring both sides introduces the possibility of extraneous solutions; hence the check is required.

$Check x=1$

$Check x=5$

$2x−1+2=x2(1)−1+2=11+2=11+2=13=1 ✗$

$2x−1+2=x2(5)−1+2=59+2=53+2=55=5 ✓$

After checking, we can see that $x=1$ is an extraneous solution; it does not solve the original radical equation. This leaves $x=5$ as the only solution.

Sometimes there is more than one solution to a radical equation.

### Example 4

Solve: $22x+5−x=4.$

Solution:

Begin by isolating the term with the radical.

$22x+5−x=4Add x to both sides.22x+5=x+4$

Despite the fact that the term on the left side has a coefficient, we still consider it to be isolated. Recall that terms are separated by addition or subtraction operators.

$22x+5=x+4(22x+5)2=(x+4)2Square both sides.4(2x+5)=x2+8x+16$

$4(2x+5)=x2+8x+168x+20=x2+8x+160=x2−40=(x+2)(x−2)x+2=0or x−2=0x=−2x=2$

Since we squared both sides, we must check our solutions.

$Check x=−2$

$Check x=2$

$22x+5−x=422(−2)+5−(−2)=42−4+5+2=421+2=42+2=44=4 ✓$

$22x+5−x=422(2)+5−(2)=424+5−2=429−2=46−2=44=4 ✓$

After checking, we can see that both are solutions to the original equation.

Sometimes both of the possible solutions are extraneous.

### Example 5

Solve: $4−11x−x+2=0.$

Solution:

$4−11x−x+2=0Isolate the radical.4−11x=x−2(4−11x)2=(x−2)2Square both sides.4−11x=x2−4x+4Solve.0=x2+7x0=x(x+7)$

$x=0orx+7=0x=−7$

Since we squared both sides, we must check our solutions.

$Check x=0$

$Check x=−7$

$4−11x−x+2=04−11(0)−0+2=04+2=02+2=04=0 ✗$

$4−11x−x+2=04−11(−7)−(−7)+2=04+77+7+2=081+9=09+9=018=0 ✗$

Since both possible solutions are extraneous, the equation has no solution.

The squaring property of equality extends to any positive integer power n. Given real numbers a and b, we have the following:

$If a=b, then an=bn.$

This is often referred to as the power property of equalityGiven any positive integer n and real numbers a and b where $a=b$, then $an=bn.$. Use this property, along with the fact that $(an)n=ann=a$, when a is nonnegative, to solve radical equations with indices greater than 2.

### Example 6

Solve: $4x2+73−2=0.$

Solution:

Isolate the radical, and then cube both sides of the equation.

$4x2+73−2=0Isolate the radical.4x2+73=2(4x2+73)3=(2)3Cube both sides.4x2+7=8Solve.4x2−1=0(2x+1)(2x−1)=02x+1=0or2x−1=02x=−12x=1x=−12x=12$

Check.

$Check x=−12$

$Check x=12$

$4x2+73−2=04(−12)2+73−2=04⋅14+73−2=01+73−2=083−2=02−2=00=0 ✓$

$4x2+73−2=04(12)2+73−2=04⋅14+73−2=01+73−2=083−2=02−2=00=0 ✓$

Answer: The solutions are $±12$.

Try this! $x−33x+1=3$

It may be the case that the equation has more than one term that consists of radical expressions.

### Example 7

Solve: $5x−3=4x−1.$

Solution:

Both radicals are considered isolated on separate sides of the equation.

$5x−3=4x−1(5x−3)2=(4x−1)2Square both sides.5x−3=4x−1Solve.x=2$

Check $x=2.$

$5x−3=4x−15(2)−3=4(2)−110−3=8−17=7 ✓$

### Example 8

Solve: $x2+x−143=x+503.$

Solution:

Eliminate the radicals by cubing both sides.

$x2+x−143=x+503(x2+x−143)3=(x+503)3Cube both sides.x2+x−14=x+50Solve.x2−64=0(x+8)(x−8)=0x+8=0or x−8=0x=−8x=8$

Check.

$Check x=−8$

$Check x=8$

$x2+x−143=x+503(−8)2+(−8)−143=(−8)+50364−8−143=423423=423 ✓$

$x2+x−143=x+503(8)2+(8)−143=(8)+50364+8−143=583583=583 ✓$

Answer: The solutions are $±8.$

It may not be possible to isolate a radical on both sides of the equation. When this is the case, isolate the radicals, one at a time, and apply the squaring property of equality multiple times until only a polynomial remains.

### Example 9

Solve: $x+2−x=1$

Solution:

Begin by isolating one of the radicals. In this case, add $x$ to both sides of the equation.

$x+2−x=1x+2=x+1$

Next, square both sides. Take care to apply the distributive property to the right side.

$(x+2)2=(x+1)2x+2=(x+1)(x+1)x+2=x2+x+x+1x+2=x+2x+1$

At this point we have one term that contains a radical. Isolate it and square both sides again.

$x+2=x+2x+11=2x(1)2=(2x)21=4x14=x$

Check to see if $x=14$ satisfies the original equation $x+2−x=1.$

$14+2−14=194−12=132−12=122=11=1 ✓$

Answer: The solution is $14.$

Note: Because $(A+B)2≠A2+B2$, we cannot simply square each term. For example, it is incorrect to square each term as follows.

$(x+2)2−(x)2=(1)2Incorrect!$

This is a common mistake and leads to an incorrect result. When squaring both sides of an equation with multiple terms, we must take care to apply the distributive property.

### Example 10

Solve: $2x+10−x+6=1$

Solution:

Begin by isolating one of the radicals. In this case, add $x+6$ to both sides of the equation.

$2x+10−x+6=12x+10=x+6+1$

Next, square both sides. Take care to apply the distributive property to the right side.

$(2x+10)2=(x+6+1)22x+10=x+6+2x+6+12x+10=x+7+2x+6$

At this point we have one term that contains a radical. Isolate it and square both sides again.

$2x+10=x+7+2x+6x+3=2x+6(x+3)2=(2x+6)2x2+6x+9=4(x+6)x2+6x+9=4x+24x2+2x−15=0(x−3)(x+5)=0x−3=0or x+5=0x=3x=−5$

Check.

$Check x=3$

$Check x=−5$

$2x+10−x+6=12(3)+10−3+6=116−9=14−3=11=1 ✓$

$2x+10−x+6=12(−5)+10−−5+6=10−1=10−1=1−1=1 ✗$

Try this! Solve: $4x+21−2x+22=1$

### Key Takeaways

• Solve equations involving square roots by first isolating the radical and then squaring both sides. Squaring a square root eliminates the radical, leaving us with an equation that can be solved using the techniques learned earlier in our study of algebra.
• Squaring both sides of an equation introduces the possibility of extraneous solutions. For this reason, you must check your solutions in the original equation.
• Solve equations involving nth roots by first isolating the radical and then raise both sides to the nth power. This eliminates the radical and results in an equation that may be solved with techniques you have already mastered.
• When more than one radical term is present in an equation, isolate them one at a time, and apply the power property of equality multiple times until only a polynomial remains.

### Part A: Solving Radical Equations

Solve

1. $x=7$

2. $x=4$

3. $x+8=9$

4. $x−4=5$

5. $x+7=4$

6. $x+3=1$

7. $5x−1=0$

8. $3x−2=0$

9. $3x+1=2$

10. $5x−4=4$

11. $7x+4+6=11$

12. $3x−5+9=14$

13. $2x−1−3=0$

14. $3x+1−2=0$

15. $x+1=x+1$

16. $2x−1=2x−1$

17. $4x−1=2x−1$

18. $4x−11=2x−1$

19. $x+8=x−4$

20. $25x−1=5x+1$

21. $x3=3$

22. $x3=−4$

23. $2x+93=3$

24. $4x−113=1$

25. $5x+73+3=1$

26. $3x−63+5=2$

27. $4−2x+23=0$

28. $6−32x−33=0$

29. $3(x+10)5=2$

30. $4x+35+5=4$

31. $8x+11=3x+1$

32. $23x−4=2(3x+1)$

33. $2(x+10)=7x−15$

34. $5(x−4)=x+4$

35. $5x−23=4x3$

36. $9(x−1)3=3(x+7)3$

37. $3x+13=2(x−1)3$

38. $9x3=3(x−6)3$

39. $3x−55=2x+85$

40. $x+35=2x+55$

41. $4x+21=x$

42. $8x+9=x$

43. $4(2x−3)=x$

44. $3(4x−9)=x$

45. $2x−1=x$

46. $32x−9=x$

47. $9x+9=x+1$

48. $3x+10=x+4$

49. $x−1=x−3$

50. $2x−5=x−4$

51. $16−3x=x−6$

52. $7−3x=x−3$

53. $32x+10=x+9$

54. $22x+5=x+4$

55. $3x−1−1=x$

56. $22x+2−1=x$

57. $10x+41−5=x$

58. $6(x+3)−3=x$

59. $8x2−4x+1=2x$

60. $18x2−6x+1=3x$

61. $5x+2=x+8$

62. $42(x+1)=x+7$

63. $x2−25=x$

64. $x2+9=x$

65. $3+6x−11=x$

66. $2+9x−8=x$

67. $4x+25−x=7$

68. $8x+73−x=10$

69. $24x+3−3=2x$

70. $26x+3−3=3x$

71. $2x−4=14−10x$

72. $3x−6=33−24x$

73. $x2−243=1$

74. $x2−543=3$

75. $x2+6x3+1=4$

76. $x2+2x3+5=7$

77. $25x2−10x−73=−2$

78. $9x2−12x−233=−3$

79. $4x2−13−2=0$

80. $4x23−1=0$

81. $x(2x+1)5−1=0$

82. $3x2−20x5−2=0$

83. $2x2−15x+25=(x+5)(x−5)$

84. $x2−4x+4=x(5−x)$

85. $2(x2+3x−20)3=(x+3)23$

86. $3x2+3x+403=(x−5)23$

87. $2x−5+2x=5$

88. $4x+13−2x=3$

89. $8x+17−22−x=3$

90. $3x−6−2x−3=1$

91. $2(x−2)−x−1=1$

92. $2x+5−x+3=2$

93. $2(x+1)−3x+4−1=0$

94. $6−5x+3−3x−1=0$

95. $x−2−1=2(x−3)$

96. $14−11x+7−9x=1$

97. $x+1=3−2−x$

98. $2x+9−x+1=2$

99. $x1/2−10=0$

100. $x1/2−6=0$

101. $x1/3+2=0$

102. $x1/3+4=0$

103. $(x−1)1/2−3=0$

104. $(x+2)1/2−6=0$

105. $(2x−1)1/3+3=0$

106. $(3x−1)1/3−2=0$

107. $(4x+15)1/2−2x=0$

108. $(3x+2)1/2−3x=0$

109. $(2x+12)1/2−x=6$

110. $(4x+36)1/2−x=9$

111. $2(5x+26)1/2=x+10$

112. $3(x−1)1/2=x+1$

113. $x1/2+(3x−2)1/2=2$

114. $(6x+1)1/2−(3x)1/2=1$

115. $(3x+7)1/2+(x+3)1/2−2=0$

116. $(3x)1/2+(x+1)1/2−5=0$

Determine the roots of the given functions. Recall that a root is a value in the domain that results in zero. In other words, find $x$ where $f(x)=0.$

1. $f(x)=x+5−2$

2. $f(x)=2x−3−1$

3. $f(x)=2x+2−8$

4. $f(x)=3x−7−6$

5. $f(x)=x+13+2$

6. $f(x)=2x−13+6$

Solve for the indicated variable.

1. Solve for P: $r=P−1$

2. Solve for x: $y=x−h+k$

3. Solve for s: $t=2sg$

4. Solve for L: $T=2πL32$

5. Solve for R: $I=PR$

6. Solve for h: $r=3Vπh$

7. Solve for V: $r=3V4π3$

8. Solve for c: $a=b2π2c3$

9. The square root of 1 less than twice a number is equal to 2 less than the number. Find the number.

10. The square root of 4 less than twice a number is equal to 6 less than the number. Find the number.

11. The square root of twice a number is equal to one-half of that number. Find the number.

12. The square root of twice a number is equal to one-third of that number. Find the number.

13. The distance d in miles a person can see an object on the horizon is given by the formula $d=6h2$ where h represents the height in feet of the person’s eyes above sea level. How high must a person’s eyes be to see an object 5 miles away?

14. The current I measured in amperes is given by the formula $I=PR$ where P is the power usage measured in watts and R is the resistance measured in ohms. If a light bulb requires 1/2 amperes of current and uses 60 watts of power, then what is the resistance through the bulb?

The period of a pendulum T in seconds is given by the formula $T=2πL32$ where L represents the length in feet. Calculate the length of a pendulum given the period. Give the exact value and the approximate value rounded to the nearest tenth of a foot.

1. 1 second

2. 2 seconds

3. $12$ second

4. $13$ second

The time t in seconds, an object is in free fall is given by the formula $t=s4$ where s represents the distance it has fallen, in feet. Calculate the distance an object will fall given the amount of time.

1. 1 second

2. 2 seconds

3. $12$ second

4. $14$ second

### Part B: Discussion Board

1. Discuss reasons why we sometimes obtain extraneous solutions when solving radical equations. Are there ever any conditions where we do not need to check for extraneous solutions? Why or why not?

2. If an equation has multiple terms, explain why squaring all of them is incorrect. Provide an example.

1. 49

2. 1

3. Ø

4. $125$

5. 1

6. 3

7. $134$

8. 0

9. $14$

10. Ø

11. 27

12. 9

13. −3

14. 6

15. $23$

16. 2

17. 7

18. 2

19. −3

20. 13

21. 7

22. 2, 6

23. 2

24. −1, 8

25. 5

26. Ø

27. −3, 3

28. 2, 5

29. −4, 4

30. $12$

31. 2, 7

32. Ø

33. 10

34. −6, −4

35. $−12,32$

36. Ø

37. −5, 5

38. −9, 3

39. $15$

40. $−32,32$

41. −1, 1/2

42. 5, 10

43. −7, 7

44. $92$

45. 1

46. 10

47. Ø

48. 3

49. −1, 2

50. 100

51. −8

52. 10

53. −13

54. $52$

55. −6, −4

56. −2, 2

57. 1

58. −2

59. −1

60. 14

61. −9

62. $P=(r+1)2$

63. $s=gt22$

64. $R=PI2$

65. $V=4πr33$

66. 5

67. 0, 8

68. $1623$ feet

69. $8π2$ feet; 0.8 feet

70. $2π2$ feet; 0.2 feet

71. 16 feet

72. 4 feet


text:
# Calculating covariances

Currently I am preparing for my exam. Therefore, we have a sample exam with the answers included. Unfortunately, even with the answers it not all makes sense to me. This is the question with the answer provided:


text:
# Mass conservation for pulsating sphere

1. May 15, 2013

### enc08

Hi,

I'm trying to understand the mass conservation equation for a pulsating sphere which has thickness dr. Please refer to the attached solution.

$$\rho = \rho_{0} (ambient density) + \rho' (small deviation)$$

There are two things I don't follow.

First, is that to obtain the mass, the area of the sphere is multiplied by the speed (and density) at which it is pulsating. Although the units work out, I don't see the physical meaning behind taking the product with speed .

Secondly, on the LHS, the ρ turns into a ρ'. I can't see how this substitution was made.

#### Attached Files:

• ###### Capture.JPG
File size:
16.4 KB
Views:
76
2. May 15, 2013

### DEvens

It's not clear what you mean by "pulsating sphere." What changes and what stays the same? It looks like the density is changing. The drawing makes it look like the sphere is a shell with inside radius r and outside radius r + dr.

Can you make things a bit more clear?

3. May 15, 2013

### physwizard

The equation 1) mentioned in your figure doesn't look correct. It may be tentatively correct if the time derivative on the left hand side (L.H.S.) is really a total time derivative and not a partial time derivative(as indicated in the figure) and the deviations of the density are assumed to be sufficiently small.
Also apparently spherical symmetry appears to have been assumed for the density and the velocity.
If you want to figure things out for yourself you can start out with the equation of continuity:
$\nabla.(\rho\bar{v}) = -\frac{\partial\rho}{\partial t}$


text:
# zbMATH — the first resource for mathematics

Degeneration of $$l$$-adic weight spectral sequences. (English) Zbl 1033.14012
From the introduction: Let $$A$$ be a henselian discrete valuation ring and let $$l$$ be a prime number invertible in $$A$$. We prove:
Theorem 1. For a proper semistable family over $$A$$ whose special fiber has smooth irreducible components, the Steenbrink-Rapoport-Zink $$l$$-adic weight spectral sequence degenerates in $$E_2$$-terms.
The degeneracy was proved in cases where
(i) $$A=\mathbb{C}\{t\}$$ by J. H. M. Steenbrink [Invent. Math. 31, 229–257 (1976; Zbl 0303.14002)]; and
(ii) The residue field of $$A$$ is finite by a result by M. Rapoport and Th. Zink [Invent. Math. 68, 21-101 (1982; Zbl 0498.14010)].
The second case was proved by the Weil conjecture on Frobenius weights. L. Illusie [Astérisque 223, 9-57 (1994; Zbl 0837.14013)] conjectured that the degeneracy holds over an arbitrary A. Theorem 1 gives an affirmative answer to it.

##### MSC:
 14G20 Local ground fields in algebraic geometry 14F30 $$p$$-adic cohomology, crystalline cohomology 14D06 Fibrations, degenerations in algebraic geometry
Full Text:


text:
# Achieving tikz shift right/left with old version on arXiv

I have the following diagram using the tikz-cd package:

\begin{tikzcd}[]
A  \rar[equal]   & B \\
\uar[hookrightarrow,shift left=.60ex]
\end{tikzcd}

This compiles perfectly well on my own computer, but when submitting to the arXiv produces a number of compile errors.

• ! Package pgfkeys Error: I do not know the key '/tikz/equal' and I am going to ignore it. Perhaps you misspelled it.
• ! Package pgfkeys Error: I do not know the key '/tikz/shift right' and I am going to ignore it. Perhaps you misspelled it.

It seems the source of the problem is that arXiv uses older versions of tikz-cd and pgf, which do not support these keys.

The first error can be worked around by replacing [equal] with [-, double equal sign distance] (see e.g. here), however I don't know how to achieve the results of the shifts right and left with other commands. Does anybody know another way to achieve the same results, preferably with minimal extra work (i.e. I'd rather avoid translating all my diagrams to xy-pic if possible)? Including an updated tikzlibrarycd.code.tex with the submission (as suggested here) doesn't work for the reasons specified in that answer.

• Does it work if you copy (only!) the definition of the keys into the preamble ? – percusse Feb 18 '15 at 6:29
• Have you talked to the maintainers of arXiv? I have suggested that on several post around here, but never got an answer. Maybe they are willing to finally update to something newer. – LaRiFaRi Feb 18 '15 at 8:29
• @percusse: I'm in the process of trying this, but it's leading down a rabbit hole of defining the definitions in the definitions, and so on. I'm not sure if it will terminate before I've copied over the entire library. – Stephen Morgan Feb 18 '15 at 8:58
• @LaRiFaRi: I haven't talked to the maintainers (though I have spoken with a moderator), but I'll try this. I'm not sure how much hope there is that a random person with this issue will convince them to update their system in sufficient time to make a difference for this posting. – Stephen Morgan Feb 18 '15 at 9:00
• @StephenMorgan I know what you mean, but maybe everybody thinks like this. If you search this site for arXiv, you will find several post complaining about the missing updates of TeX on arXiv. You are not alone... Would be nice, if you get some maintainer informed. – LaRiFaRi Feb 18 '15 at 9:13

It seems that the same effect can be achieved in a way the arXiv accepts by substituting xshift=.60ex for shift right=.60ex and xshift=-.60ex for shift left=.60ex. This should be instead replaced with yshift if the arrows are horizontal. And of course if the arrows are diagonal you need to manually pick your xshift and yshift values to work well.


text:
# Reactive Power Apparent Power Issue

by azaharak
Tags: apparent power, reactive power
 P: 152 This is not a HW question. Assume you have a sinusoidally varying voltage, such that current lags by some phase angle phi Then you can write the power as a function of time as p(t)=i(t)v(t)=2*Vrms*Irms*cos(omega*t)*cos(omega*t-phi) You can break this up via trig identity into the following Eq1: p(t)=2*Vrms*Irms*( [1+cos(2*omega*t)]*cos(phi) + sin(2*omega*t)*sin(phi) ) If you average over 1 period, then you obtain that the average REAL power is P=Vrms*Irms*cos(phi) This is average real power The peak real power would be twice this value since (take t=0 in Eq1:) real power is associated with cos(phi) term. The average reactive power is zero, it bounces back and forth between inductive and capacitive loads (B and E fields). But the peak reactive power is defined as Q=Irms*Vrm*sin(phi). HERE IS THE QUESTion. I see often that the Apparent power S = P + i*Q. Unfortunately it is never throuroughly defined. Why would would be adding the Average value of real power and Peak value of Reactice power and call it the apparent power. Am I misunderstanding this? Shouldn't we be adding the PEAK real power and PEAK reactive power and call this the apparent power. Also, other relations are P=(I^2)R Q=(I^2)X S=(I^2)Z THANK YOU!! Am i correct in saying that the reason why it is defined this way is so that "you" could say that the phase angle is given by Phi = arctan(Q/P). I'm just a little weired out by combining the Average Real power with Peak Reactive power. Thanks
HW Helper
Thanks
P: 26,160
hi azaharak!
 Quote by azaharak HERE IS THE QUESTion. I see often that the Apparent power S = P + i*Q.
no, that's the complex power

the apparent power is VrmsIrms … it's voltage times current
 Unfortunately it is never throuroughly defined. Why would would be adding the Average value of real power and Peak value of Reactice power and call it the apparent power.
so far as i know, neither complex power nor apparent power have any precise physical significance

(but you need to know them because they come up in exam questions! )

from the pf library on impedance
Power:

Power = work per time = voltage times charge per time = voltage times current:

$$P = VI =\ V_{max}I_{max}\cos(\omega t + \phi/2)\cos(\omega t - \phi/2)$$
$$=\ V_{max}I_{max}(\cos\phi + \cos2\omega t)/2$$
$$=\ V_{rms}I_{rms}(\cos\phi + \cos2\omega t)$$
$$=\ (V_{rms}^2/|Z|)(\cos\phi + \cos2\omega t)$$

So (instantaneous) power is the constant part, $P_{av} = V_{rms}I_{rms}\cos\phi$ (the average power), plus a component varying with double the circuit frequency, $V_{rms}I_{rms}\cos2\omega t$ (so a graph of the whole power is a sine wave shifted by a ratio $\cos\phi$ above the x-axis).

Apparent power, reactive power, and complex power, are convenient mathematical definitions with no precise physical significance: apparent power is the (constant) product of the r.m.s. voltage and current, $S=V_{r.m.s.}I_{r.m.s.}$: it is what we would expect the average power to be if we knew nothing about reactance!

Similarly reactive power is defined as $Q=S\sin\phi=P_{av}\tan\phi$, and complex power is defined as $Se^{j\phi}=P_{av}+jQ$.

 Related Discussions Electrical Engineering 8 Electrical Engineering 1 Electrical Engineering 0 Engineering, Comp Sci, & Technology Homework 5 Electrical Engineering 6


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# How to obtain the final values of a DecisionTreeRegressor in Scikit-Learn?

This page shows the paths in the decision trees in scikit-learn. After reaching the leaf nodes of the decision tree, where do we obtain the final resultant value?

• Check in the value attribute i.e. regressor.tree_.value – 10xAI Dec 25 '20 at 10:18
• Alright. Is there any documentation for this? – Chong Lip Phang Dec 25 '20 at 10:37

For documentation, check the Scikit-Learn code at Github i.e. line#535 [Link]

value : array of double, shape [node_count, n_outputs, max_n_classes]
Contains the constant prediction value of each node.

You can check the same using this sample code.

• Identify the leaf nodes
• Slice the value attributes for leaf nodes
from sklearn.datasets import load_diabetes
import numpy as np, matplotlib.pyplot as plt
from sklearn.tree import DecisionTreeRegressor

regressor = DecisionTreeRegressor(max_depth=5)
regressor.fit(X, y)

children_left = regressor.tree_.children_left
children_right = regressor.tree_.children_right
leaf_nodes = []
for i in range(n_nodes):
if children_left[i] == children_right[i]:
leaf_nodes.append(i)

y_pred =  regressor.predict(X)
all_val = regressor.tree_.value[leaf_nodes,0,0] # Sliced on Leaf nodes

# Check if all the y_pred is from these values
set(y_pred) - set(all_val) # Or, [elem for elem in y_pred if elem not in all_val]

The decision depends on the loss function used. If you are using the Mean Square Error you use the within-leaf mean. The fastest way is to use the .apply() method and estimate the average of observations that fall in the same leaf.

• Could you show any relevant documentation? – Chong Lip Phang Dec 25 '20 at 11:02


text:
# Isolated point explained

In mathematics, a point x is called an isolated point of a subset S (in a topological space X) if x is an element of S and there exists a neighborhood of x which does not contain any other points of S. This is equivalent to saying that the singleton is an open set in the topological space S (considered as a subspace of X). Another equivalent formulation is: an element x of S is an isolated point of S if and only if it is not a limit point of S.

If the space X is a Euclidean space (or any other metric space), then an element x of S is an isolated point of S if there exists an open ball around x which contains only finitely many elements of S.

## Related notions

A set that is made up only of isolated points is called a discrete set (see also discrete space). Any discrete subset S of Euclidean space must be countable, since the isolation of each of its points together with the fact that rationals are dense in the reals means that the points of S may be mapped into a set of points with rational coordinates, of which there are only countably many. However, not every countable set is discrete, of which the rational numbers under the usual Euclidean metric are the canonical example.

A set with no isolated point is said to be dense-in-itself (every neighbourhood of a point contains infinitely many other points of the set). A closed set with no isolated point is called a perfect set (it contains all its limit points and no isolated points).

X

and

Y

are homeomorphic, the number of isolated points in each is equal.

## Examples

### Standard examples

Topological spaces in the following three examples are considered as subspaces of the real line with the standard topology.

• For the set

S=\{0\}\cup[1,2]

, the point 0 is an isolated point.
• For the set

S=\{0\}\cup\{1,1/2,1/3,...\}

, each of the points 1/k is an isolated point, but 0 is not an isolated point because there are other points in S as close to 0 as desired.
• The set

{N}=\{0,1,2,\ldots\}

of natural numbers is a discrete set.In the topological space

X=\{a,b\}

with topology

\tau=\{\emptyset,\{a\},X\}

, the element

a

is an isolated point, even though

b

belongs to the closure of

\{a\}

(and is therefore, in some sense, "close" to

a

). Such a situation is not possible in a Hausdorff space.

The Morse lemma states that non-degenerate critical points of certain functions are isolated.

### Two counter-intuitive examples

Consider the set

F

of points

x

in the real interval

(0,1)

such that every digit

xi

of their binary representation fulfills the following conditions:
• Either

xi=0

or

xi=1

.

xi=1

only for finitely many indices

i

.
• If

m

denotes the largest index such that

xm=1

, then

xm-1=0

.
• If

xi=1

and

i<m

, then exactly one of the following two conditions holds:

xi-1=1

or

xi+1=1

. Informally, these conditions means that every digit of the binary representation of

x

which equals 1 belongs to a pair ...0110..., except for ...010... at the very end.

Now,

F

is an explicit set consisting entirely of isolated points which has the counter-intuitive property that its closure is an uncountable set.

Another set

F

with the same properties can be obtained as follows. Let

C

be the middle-thirds Cantor set, let

I1,I2,I3,\ldots

be the component intervals of

[0,1]-C

, and let

F

be a set consisting of one point from each

Ik

. Since each

Ik

contains only one point from

F

, every point of

F

is an isolated point. However, if

p

is any point in the Cantor set, then every neighborhood of

p

contains at least one

Ik

, and hence at least one point of

F

. It follows that each point of the Cantor set lies in the closure of

F

, and therefore

F

has uncountable closure.


text:
# How GoTo Telescopes Work

The latest issue of The College Mathematics Journal contains a good article by Donald Teets, titled “Push-To Telescope Mathematics” (see references below), describing the alignment process used by computerized telescopes to determine their orientation, and thus how to point to selected stars.  This is a beautiful problem, in part because of just how little information is needed for its solution.  Just power up the telescope, point it at any two known stars, and after some simple computation, the telescope knows how to find any other star in its database.

But some care must be taken; I think the solution in the article overlooks an important practical consideration, which I’ll get to shortly.  Fortunately, there is an equally beautiful solution to a natural generalization of this problem, that I suspect is probably closer to what is actually implemented in the software of these computerized telescopes.

The basic problem is this: when the telescope is first powered up, it does not know its orientation relative to the stars.  That is, it does not know the relationship between its “body frame” azimuth and altitude (which it can sense and/or control with motors) and the corresponding “reference frame” right ascension and declination.

We can point the telescope at a known star $S_1$, and tell the telescope that it is pointing at that star.  This gives some information about the telescope’s orientation: a particular direction in the body frame corresponds to a known direction in the reference frame.  From now on, no matter where the telescope slews, it always knows how to return to $S_1$. [1]

However, knowing the direction to just one star is not enough to completely determine orientation.  Intuitively, this makes sense: for example, when the telescope is pointed at $S_1$, it may still be “rolled” through any angle about its axis without changing its direction.  Slewing the telescope again to point at one more known star $S_2$ will do the trick.

(Actually, this problem is interesting in that one measurement is not enough, but two is too many.  Three parameters are needed to specify the telescope’s orientation– think yaw, pitch, and roll– and each star direction measurement provides two parameters.  So with two stars, the telescope’s orientation is over-determined.  We’ll come back to this shortly.)

At this point, we have the following mathematical problem: given 3-D unit (column) vectors $\mathbf{r}_i$ and $\mathbf{s}_i$ indicating the position of star $S_i$ in the inertial reference frame and sensor body frame, respectively, what is the 3×3 rotation matrix $A$ such that $A \mathbf{s}_i = \mathbf{r}_i$?

The CMJ article solves the problem as follows: let $R$ be the 3×3 matrix with columns given by

$R = (\mathbf{r}_1, \mathbf{r}_2, \mathbf{r}_1 \times \mathbf{r}_2)$

and $S$ be the 3×3 matrix

$S = (\mathbf{s}_1, \mathbf{s}_2, \mathbf{s}_1 \times \mathbf{s}_2)$

Then $A S = R$; solving, we have $A = R S^{-1}$.  Simple, right?

This works fine… as long as our measurements of direction in the body and/or reference frames are exact.  If there are any errors– and there almost certainly are– then in general the matrix $A$ may not even represent a proper rotation (i.e., it may not be orthogonal with determinant 1).

Fortunately, there are two well-documented incremental improvements to this approach.  The first is the TRIAD algorithm, whose only difference is essentially to ensure that the input matrices are themselves rotations, even if the measurements are inaccurate:

$R = (\mathbf{r}_1, \frac{\mathbf{r}_1 \times \mathbf{r}_2}{|\mathbf{r}_1 \times \mathbf{r}_2|}, \mathbf{r}_1 \times (\frac{\mathbf{r}_1 \times \mathbf{r}_2}{|\mathbf{r}_1 \times \mathbf{r}_2|}))$

$S = (\mathbf{s}_1, \frac{\mathbf{s}_1 \times \mathbf{s}_2}{|\mathbf{s}_1 \times \mathbf{s}_2|}, \mathbf{s}_1 \times (\frac{\mathbf{s}_1 \times \mathbf{s}_2}{|\mathbf{s}_1 \times \mathbf{s}_2|}))$

$A = R S^T$

Note that the TRIAD algorithm effectively treats the first measurement as “perfect” (i.e., $A \mathbf{s}_1 = \mathbf{r}_1$ exactly), and accounts for error in the second measurement.

A more realistic approach would be to assume that all of the measurements contain errors, and to try to determine the “best” rotation that minimizes the effect of those errors in a least-squares sense.  It turns out that we can do just that, with a very elegant algorithm described by Markley (see references below) that looks only slightly more sophisticated than what we have already seen.  As a bonus, this algorithm can accept more than just two measurements:

Given n measurements of unit vector pairs in the reference and body frames, let $R$ be the 3xn matrix with columns corresponding to the reference frame vectors, and $S$ be the 3xn matrix of body frame vectors.  Then the rotation matrix $A$ is computed using the singular value decomposition of $R S^T$:

$U \Sigma V^T = R S^T$

$A = U diag(1, 1, \det(U) \det(V)) V^T$

Finally, it is worth emphasizing that this is really just one of several algorithms that solve the same least-squares problem.  Shuster provides a good history and survey of these algorithms in the paper referenced below, including his quaternion-based QUEST algorithm.

References:

1. Markley, F. L., Attitude Determination using Vector Observations and the Singular Value Decomposition, The Journal of the Astronautical Sciences, 36(3) (July-September 1988): 245-258. [PDF]
2. Shuster, M. D., The Quest for Better Attitudes, The Journal of the Astronautical Sciences, 54(3-4) (July-December 2006): 657–683. [PDF]
3. Teets, D., Push-To Telescope Mathematics. The College Mathematics Journal43(3) (May 2012): 227-231. [JSTOR]

[1] This is not quite true.  I am glossing over the fact that the telescope “body” frame is rotating with the Earth.  So after some time has passed, the telescope does not even have enough information to know how to rotate back to the star $S_1$.  But as will be shown, we can handle this problem, with a second star… and an onboard clock to measure the time difference between the two star measurements.

# The Hunger Games: Follow-up

This is a follow-up to the previous post, which presented a situation in the second book of the Hunger Games trilogy as a probability problem.  The problem may be re-stated as follows: when $n=59$ balls are randomly distributed into $d=24$ initially empty urns, what is the probability that every urn gets at least one ball?  (In the book, living past Hunger Games victors are the balls, and the districts/genders of the victors are the urns.  The problem asks how likely is it that a male and female from each of 12 districts is represented.)

I found that students really enjoyed tackling this problem, and it can lead in many interesting directions.  It makes for a very accessible programming exercise to estimate the probability by simulation.  But there is also a nice analytic solution, particularly when each urn is equally likely.  In that case, we can use inclusion-exclusion to count distributions of the balls leaving $k$ of the urns empty, yielding the probability:

$\sum_{k=0}^d (-1)^k {d \choose k} (\frac{d-k}{d})^n$

The following plot shows this probability as a function of $n$, the number of balls (i.e., past victors).  For $n=59$, the probability is only about 0.099, or less than 1 in 10.

Probability that all 12 districts have a living male and female past victor, vs. the number of living victors.

Finally, this problem is essentially a slight variant of the coupon collector problem, where in the latter we are asked for the expected number of balls needed so that every urn gets at least one ball.  There is plenty of interesting literature on the problem, including the case where the probabilities for each urn are not equal.

Which brings us back to the original episode in the book, where those probabilities are indeed not equal; 3 of the 12 districts are described as favorites for producing victors.  This makes for another nice simulation exercise, where it is necessary to quantify what “favorite” might mean.  Another interesting challenge is to show that this always makes the scenario even less likely.  That is, the probability of “hitting” all of the urns is maximized when the probabilities for the urns are equal.


text:
#### Vol. 14, No. 10, 2020

 Recent Issues
 The Journal About the Journal Editorial Board Editors’ Interests Subscriptions Submission Guidelines Submission Form Policies for Authors Ethics Statement ISSN: 1944-7833 (e-only) ISSN: 1937-0652 (print) Author Index To Appear Other MSP Journals
Motivic multiple zeta values relative to $\mu_2$

### Zhongyu Jin and Jiangtao Li

Vol. 14 (2020), No. 10, 2685–2712
##### Abstract

We establish a short exact sequence about depth-graded motivic double zeta values of even weight relative to ${\mu }_{2}$. We find a basis for the depth-graded motivic double zeta values relative to ${\mu }_{2}$ of even weight and a basis for the depth-graded motivic triple zeta values relative to ${\mu }_{2}$ of odd weight. As an application of our main results, we prove Kaneko and Tasaka’s conjectures about the sum odd double zeta values and the classical double zeta values. We also prove an analogue of Kaneko and Tasaka’s conjecture in depth three. Finally, we formulate a conjecture which is related to sum odd multiple zeta values in higher depth.

However, your active subscription may be available on Project Euclid at
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text:
# e

 A page for e, i, π, and φ
Approximation of ex with first four Maclaurin series polynomials

The role that the exponential function (ex) plays is extremely significant to an understanding of the laws of nature and harmonic organisation. Mathematically (ex) can be expressed several ways. A common form is the Maclaurin series expansion of;

$e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots$

Note that;

$e^1 = \sum_{n=0}^{\infty}{1^n \over n!} = 1 + 1 + {1^2 \over 2!} + {1^3 \over 3!} + {1^4 \over 4!} + \cdots$

and

$\frac{1}{e} = \sum_{n=0}^{\infty} (-1)^n\frac{1}{n!} = 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots$

where

$e^2 = e^{(1+1)} = e^1 \times e^1$.

or

$\sum_{n=0}^{\infty}{2^n \over n!} = \sum_{n=0}^{\infty}{1^n \over n!} \times \sum_{m=0}^{\infty}{1^m \over m!}$.

Expansion of the right hand side terms is an outer product, taking the summation of these terms equals e2.

This formula needs to be checked and converted to LaTeX

As an infinite sum ex is;

$e^x = \sum_{n = 0}^{\infty} {x^n \over n!} = 1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots$

We can look at the construction of this relation more easily by describing it by an algorithm. We can remove the power and factorial operations by basing the current numerator and denominator values on the previous values:

# Returns e^x from e( x, iterations )
sub e {
$x = shift;$e = $top =$bot = 1;
for $n ( 1 .. shift ) {$top = $top *$x;
$bot =$bot * $n;$e = $e +$top / $bot; } return$e;
}

But a better conceptual representation is to restructure it such that the division occurs outside the main loop since its also a higher-level operation. So each iteration is now maintaining the value of a numerator and denominator which represent the entire preceeding series as a single fraction which can be reduced afterwards.

# Returns e^x from e( x, iterations )
sub e {
$x =$y = shift;
$top =$bot = 1;
for $n ( 1 .. shift ) {$y = $y +$x;
$top =$top * $n +$y;
$bot =$bot * $n; } return$top / \$bot;
}

Here's a snipit in C which uses the taylor series of cosine to make Φ from π

// Calculate Phi from pi
double Phi = 0;
int n, i = -1;
for (n = 0; n < iterations; n += 2) {
Phi += pow(PI/5,n) / factorial(n) * (i = -i);
}
Phi *= 2;

## The unit circle

The nth roots of unity, or de Moivre numbers, are all the complex numbers which yield 1 when raised to a given power n. It can be shown that they are located on the unit circle of the complex plane and that in that plane they form the vertices of a n-sided regular polygon with one vertex on 1.

## Factorial & the Gamma Function

The Gamma function extends the factorial function to complex and non-integer numbers (it is already defined on the naturals, and has simple poles at the negative integers). The Gamma function "fills in" the factorial function for non-integer and complex values of n. If z is a real variable, then for natural number values only, we have

$\Gamma(z+1)= z \times \Gamma(z) = z!\,$

but for non-natural values of z, the above equation does not apply, since the factorial function is not defined.

If the real part of the complex number z is positive, then the following integral converges absolutely:

$\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,dt$

An interesting property is the Gamma functions link to π;

$\Gamma(\frac{1}{2}) = \sqrt\pi$

## Taylor series expansion

Taylor's theorem expresses a function f(x) as a power series in x, basically because the nth derivative of xn is n!.

Why is it that when we reduce the function down to degree-1 through differentiation it ends up as n!? --Nad 17:16, 26 Oct 2006 (NZDT)
Because each derivative multiplies by the power and reduces the power by one. --Nad 12:12, 12 July 2007 (NZST)
So maybe the n! denominator in talor-series of ex should be thought of as the nth derivative rather than as combinatorial. --Nad 12:16, 12 July 2007 (NZST)

Actually, the Taylor series article explains it - the theorem in general for a function f(a) is,

$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^{n}\,,$

where f (n)(a) denotes the nth derivative of f at the point a; the zeroth derivative of f is defined to be f itself and $(x-a)^0$ is defined to be 1. But all the derivatives are the still ex for the exponential function, and we're using a=0, so the top term here is always equal to 1, leaving only xn on the top. --Nad 18:22, 12 July 2007 (NZST)

## Spheres

The volume of an n-dimensional hypersphere can be expressed as:

$V_n={\pi^{n/2}R^n\over (n/2)!}.$

Note that the Gamma function is required for odd dimensions and that its value cancels out the apparent fractional power of π in those cases.

## Gaussian Function

$f(x) = a e^{- { \frac{(x-b)^2 }{ 2 c^2} } }$

for some real constants a, b, c > 0


text:
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# Homework Assignments

## Homework 3

1. Suppose and are the indicator functions of events and , respectively. Find , and show that and are independent if and only if .
2. Suppose is a ch.f. Show that is also a ch.f.
3. Suppose and are jointly Gaussian. Use ch.f.s to show that .
4. Suppose and are jointly continuous. (a) Show that

and thus that

(b) Suppose . Show that
5. Suppose and are independent continuous r.v.s with c.d.f.s and , respectively. Suppose further that for all . Show that
6. Prove Jensen's inequality for the case of simple-function r.v.'s
7. Prove the Schwartz inequality.

Problems from Grimmet & Stirzaker :

1. Prob 2.7.4
2. Prob 2.7.7. Hint: binomial distribution
3. Prob 2.7.9. Hint:

4. Ex 3.3.1. Hint: Let , and .
5. Ex 3.4.1. Let be the indicator function of the event that the outcome of the st toss is different from the outcome of the th toss. The number of distinct runs is . Observe that and are independent if . Show that

Show that
6. Ex. 3.4.2. Hint: Let , where is the number on the th ball. Show that: . show that . Hint:

7. Ex 3.5.2. Hint: .
8. Ex 3.6.5. Hint: , with equality if and only if .


text:
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