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# Predicting Customer Churn for a Bank

Acknowledgment: This post uses data obtained from this Kaggle competition, Predicting churn for Bank Customers.

## Introduction

No business wants its valuable customers to churn. Retaining existing customers is often the most efficient and cost-effective way to bring in revenue. In this tutorial, we will show you how to build a machine learning model to help predict custom churn for a bank.

## Set-up a project

To begin using this platform, you would need to set up an Organization and a project

## Add Data to the platform

Once we have set up a Project to organize the work, we will supply the platform with data on which to build the machine learning models. In this case, we will upload a CSV containing the data. The platform supports many other data ingestion methods, including from traditional SQL databases like PostgreSQL and from NoSQL databases like MongoDB.

## Data Wrangling

Once the data has been uploaded, you will see the data wrangling user interface

and on the left panel you will see the iteration you are at and on the right the insights on the datasets. The first insight you will encounter is the schema inference

Click on the solution and then "Select Solution".

You can now press "Review" and "Commit Actions" before proceeding to the left panel which will show the progress bar on iteration 1.

Click continue once the progression bar is clear of animations to begin the next iteration, and you will be reminded to set the target column

The target column is Exited which you can set on the target column drop down

You may now proceed to the second iteration and continue to accept recommendations following the same procedure as with the schema confirmation. After a couple of iterations, the Engine may have no more recommendations for the dataset.

You can finalize the dataset by pressing "Finalize" to complete the data wrangling step

Note: The Finalization process also creates a recipe consisting of the committed actions thus far and is a reproducible data processing pipeline that can be applied to other any other dataset with a compliant schema.

## Data Visualization

Once the data wrangling process is complete, every column of data gets categorized as numeric, categorical, or other. The data is then visualized using the appropriate visualizations

## Setting up Apps

The concept of an App in central to the AI & Analytics Engine platform. Each app is tied to a dataset, and it can contain multiple machine learning models built on that dataset. One can have multiple apps under one project. In this case study, we will create an app for predicting the propensity of churn.

Each App takes care of details like partitioning the data into development and test samples so that we can scientifically assess the model's performance.

## Creating new models

You can now proceed to the Model Recommender. The Model Recommender will rank our expertly-crafted model templates by their Predictive Performance, Training Time, and Prediction Time metrics! For this use-case, we select the top two models according to predictive performance, which are XGBoost and LightGBM.

We will select our models based on model accuracy and here are the models we have selected

Rank Model AI & Analytics Engine Predicted Accuracy^ Actual Accuracy Actual ROC/AUC Actual Gini
1 XGBoost 0.74 0.75 0.856 0.712
2 LightGBM 0.73 0.74 0.857 0.714

## Feature Importance

For each app, you can take a look at the Feature Importance chart and see which features are the most useful at predicting churn. For this app and dataset, we see that Age and NumberOfProduts are the most important features for predicting churn.

## Automated creation of Key Diagnostics

The below diagnostic statistics are computed automatically. It automates away one of the most tedious albeit important tasks for you.

## Diagnostics

Let’s compare the model built by the AI & Analytics Engine with the models built by the top Kaggle kernels

As predicted by the AI & Analytics Engine, XGBoost Classifier is the best performing model and has an AUC of 0.856. This compares very favorably to the results from the top voted Kaggle results above. Also, the engine reports on a lot more statistics and measure for you.

If you want to look at a model's performance in-depth, you can go to the model page and get detailed metrics and diagnostics for each model. Below is a list of metrics and visualizations we automatically generate for you.

The page also creates a number of key visualizations for you including the ROC and PR curves. As an example, this is the visualization of the confusion matrix

## Deploying the model

With a few clicks of the button, the chosen model can be operationalized and deployed ready to do the grunt work of prediction! Deployment can be performed on the cloud or to an on-premise server of your choosing.

Once the deployed, the model can be accessed via an URL endpoint, and code examples for calling the endpoint are provided in Python, R, bash, Java, and NodeJS.

You can also test the endpoint using the API Test service directly in the browser.


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Coding directly in Python functions that will be executed on GPU may allow to remove bottlenecks while keeping the code short and simple. In this introduction, we show one way to use CUDA in Python, and explain some basic principles of CUDA programming.

We choose to use the Open Source package Numba. Numba is a just-in-time compiler for Python that allows in particular to write CUDA kernels. Numba is freely available at https://github.com/numba/numba.

Note that there are other packages, such as PyCUDA, that also allow to launch CUDA kernels in Python.

## Preliminary

Let’s define first some vocabulary:

• a CUDA kernel is a function that is executed on the GPU,
• the GPU and its memory are called the device,
• the CPU and its memory are called the host.

Now let’s check the versions used in this notebook:

import sys
import numba
import numpy

print("Python version:", sys.version)
print("Numba version:", numba.__version__)
print("Numpy version:", numpy.__version__)
Python version: 3.6.5 |Anaconda, Inc.| (default, Apr 29 2018, 16:14:56)
[GCC 7.2.0]
Numba version: 0.38.1+1.gc42707d0f.dirty
Numpy version: 1.14.5

!nvcc --version
nvcc: NVIDIA (R) Cuda compiler driver
Built on Fri_Nov__3_21:07:56_CDT_2017
Cuda compilation tools, release 9.1, V9.1.85

## Launching our first CUDA kernel

Numba’s cuda module interacts with Python through numpy arrays. Therefore we have to import both numpy as well as the cuda module:

from numba import cuda
import numpy as np

Let’s start by writing a function that adds 0.5 to each cell of an (1D) array. To tell Python that a function is a CUDA kernel, simply add @cuda.jit before the definition. Below is our first CUDA kernel!

@cuda.jit
def cudakernel0(array):
for i in range(array.size):
array[i] += 0.5

Note that inside the definition of a CUDA kernel, only a subset of the Python language is allowed. The whole list is available at: http://numba.pydata.org/numba-doc/latest/cuda/cudapysupported.html.

Now let’s launch our kernel!

array = np.array([0, 1], np.float32)
print('Initial array:', array)

print('Kernel launch: cudakernel0[1, 1](array)')
cudakernel0[1, 1](array)

print('Updated array:',array)
Initial array: [0. 1.]
Kernel launch: cudakernel0[1, 1](array)
Updated array: [0.5 1.5]

You can see that we simply launched the previous kernel using the command cudakernel0[1, 1](array). But what is the meaning of [1, 1] after the kernel name?

On the GPU, the computations are executed in separate blocks, and each of these blocks uses many threads.
The set of all the blocks is called a grid. Therefore the grid size is equal to the number of blocks used during a computation.
Furthermore, each block uses the same number of threads. We call this number the block size.

To launch a kernel over multiple threads, we use the following syntax:

kernelname[gridsize, blocksize](arguments)

This is equivalent to:

kernelname[number_of_blocks, number_of_threads_per_block](arguments)

The total number of threads used by such kernel launch is by definition: $number\_of\_threads\_per\_block \times number\_of\_blocks = blocksize \times gridsize$

Therefore in the previous example, [1, 1] indicates that we are using only one thread to launch the kernel. That’s not really smart since the purpose of using CUDA kernels is to launch computations over a very big number of threads.

Let’s test what happens if we launch this kernel with more threads:

array = np.array([0, 1], np.float32)
print('Initial array:', array)

gridsize = 1024
blocksize = 1024
print("Grid size: {}, Block size: {}".format(gridsize, blocksize))

print("Total number of threads:", gridsize * blocksize)

print('Kernel launch: cudakernel0[gridsize, blocksize](array)')
cudakernel0[gridsize, blocksize](array)

print('Updated array:',array)
Initial array: [0. 1.]
Grid size: 1024, Block size: 1024
Kernel launch: cudakernel0[gridsize, blocksize](array)
Updated array: [56.5 56. ]

The result when using so many threads is quite weird. Let’s execute the cell above a few more times… The result is not always the same!

The above kernel is launched with a big number of threads. But each thread tries to add 0.5 to exactly the same array. Since the threads are launched in parallel, it is impossible to know at what moment exactly the threads read or write a value in the array.
Let’s consider two threads $$t_0$$ and $$t_1$$ among these 1048576 threads. If $$t_0$$ and $$t_1$$ read the array at the same time, their output will be the same. However, if $$t_0$$ writes its output just before $$t_1$$ reads the array, the output of $$t_1$$ will be different.

That’s not a problem specific to CUDA programming but to parallel programming in general. We should modify our function to work correctly with multiple threads.

## Toward parallel programming

We are now going to write a kernel better adapted to parallel programming. A way to proceed is to assign each thread to update one array cell, and therefore use as many threads as the array size. For that, we will use the cuda.grid() function that returns the absolute position of the current thread inside the whole grid.

@cuda.jit
def cudakernel1(array):
array[thread_position] += 0.5

We used in this kernel the argument 1 in the cuda.grid() function. We will explain later, in Part 2, why we used this value.

To update an array of size 2, we use 2 threads as follow:

array = np.array([0, 1], np.float32)
print('Initial array:', array)

print('Kernel launch: cudakernel1[1, 2](array)')
cudakernel1[1, 2](array)

print('Updated array:',array)
Initial array: [0. 1.]
Kernel launch: cudakernel1[1, 2](array)
Updated array: [0.5 1.5]

Note that if we launch this kernel with a grid of only one thread, only the first cell of the array will be updated:

array = np.array([0, 1], np.float32)
print('Initial array:', array)

print('Kernel launch: cudakernel1[1, 1](array)')
cudakernel1[1, 1](array)

print('Updated array:',array)
Initial array: [0. 1.]
Kernel launch: cudakernel1[1, 1](array)
Updated array: [0.5 1. ]

Let’s test this kernel with an array of size $$2^{20}$$. We therefore need a total number of threads of $$2^{20} = 1024 \times 1024$$.

array = np.zeros(1024 * 1024, np.float32)
print('Initial array:', array)

print('Kernel launch: cudakernel1[1024, 1024](array)')
cudakernel1[1024, 1024](array)

print('Updated array:', array)

# Since it is a huge array, let's check that the result is correct:
print('The result is correct:', np.all(array == np.zeros(1024 * 1024, np.float32) + 0.5))
Initial array: [0. 0. 0. ... 0. 0. 0.]
Kernel launch: cudakernel1[1024, 1024](array)
Updated array: [0.5 0.5 0.5 ... 0.5 0.5 0.5]
The result is correct: True

Note that there are some limitations on the number of threads depending on the hardware. For recent GPUs, the maximum number of threads by block is usually equal to 1024.

This is the end of the first part of this introduction, but you can learn more about CUDA programming in Python by reading Part 2!

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Evaluate $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$

A trigonometric function $\cos{\Big(\dfrac{\pi}{x}\Big)}$ and an algebraic function $x-2$ formed a fractional function by division. The limit of this algebraic trigonometric function has to evaluate as $x$ approaches $2$ in this trigonometric limit problem.

Let us try to evaluate the limit of the trigonometric function by the direct substitution method as $x$ approaches $2$.

$= \,\,\,$ $\dfrac{\cos{\Big(\dfrac{\pi}{2}\Big)}}{2-2}$

The cosine of angle $90$ degrees is zero as per trigonometry.

$= \,\,\,$ $\dfrac{0}{0}$

It is an indeterminate form. So, the direct substitution method is failed in calculating the limit of the given function as $x$ approaches $2$. So, we have to evaluate the limit of the function in another approach. The trigonometric limit problem can be done in two methods.

Method – 1

It is given that $x \,\to\, 2$, then $x-2 \,\to\, 2-2$. Therefore, $x-2 \,\to\, 0$. It clears that $x-2$ approaches zero when $x$ approaches $2$.

$= \,\,\,$ $\displaystyle \large \lim_{x-2 \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$

Convert the whole function in another variable

Now, take $y = x-2$, then $x = y+2$ mathematically. Finally, transform the entire function in terms of $y$ from $x$.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{y+2}\Big)}}{y}}$

You will get indeterminate form if you try to evaluate this trigonometric function as $y$ approaches zero by the direct substitution method. So, we have to take a different step here.

Express the cos function in terms of sine

In calculus, there is no trigonometric limit rule in cosine but we have a trigonometric limit rule in sine. So, we have to express the cosine function in terms of sine for moving ahead in simplification. It can be done by the cofunction identity of sine.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{y+2}\Big)}}{y}}$

Simplify the Algebra trigonometric function

The algebraic trigonometric function is almost same as the trigonometric limit rule of sine function. So, let us try to adjust the whole function for transforming the current function into required form.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi(y+2)-\pi(2)}{2(y+2)}\Big)}}{y}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y+2\pi-2\pi}{2(y+2)}\Big)}}{y}}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y+\cancel{2\pi}-\cancel{2\pi}}{2(y+2)}\Big)}}{y}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}}$

The denominator of the fractional function should be same as the angle in the sine function for applying the trigonometric limit rule of sine function in this problem. Hence, let us try to adjust the denominator of the function by taking some acceptable mathematical steps.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[1 \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\dfrac{\pi}{\pi} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\pi y}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{1 \times \pi y}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{2(y+2)}{2(y+2)} \times \pi y}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{1}{2(y+2)} \times \pi y}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{1 \times \pi y}{2(y+2)}}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{\pi y}{2(y+2)}}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{1 \times \sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{2(y+2) \times \dfrac{\pi y}{2(y+2)}}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\pi \times \dfrac{1}{2(y+2)} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\dfrac{\pi \times 1}{2(y+2)} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg[\dfrac{\pi}{2(y+2)} \times \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}\Bigg]}$

Evaluate the Limit of the function

A fractional function is split as the product of two fractional functions. The limit of the product of them can be evaluated by calculating the product of their limits. So, use the product rule of limits.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\pi}{2(y+2)}}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

The limit of first fractional function can be evaluated by the direct substitution method but do not disturb the limit of second fractional function.

$= \,\,\,$ $\dfrac{\pi}{2(0+2)}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

$= \,\,\,$ $\dfrac{\pi}{2(2)}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

$= \,\,\,$ $\dfrac{\pi}{2 \times 2}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

$= \,\,\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

The function in the denominator is same as the angle in the sine function. So, the trigonometric limit rule of sine function can be applied.

Now, take $z \,=\, \dfrac{\pi y}{2(y+2)}$. We have to express the input in terms of the $z$ completely in the function.

$(1) \,\,\,$ If $y \,\to\, 0$, then $\pi \times y \,\to\, \pi \times 0$. Therefore, $\pi y \,\to\, 0$.

$(2) \,\,\,$ If $\pi y \,\to\, 0$, then $\dfrac{\pi y}{2} \,\to\, \dfrac{0}{2}$. Therefore, $\dfrac{\pi y}{2} \,\to\, 0$.

$(3) \,\,\,$ If $y \,\to\, 0$, then $y+2 \,\to\, 0+2$. Therefore, $y+2 \,\to\, 2$. If $\dfrac{\pi y}{2} \times \dfrac{1}{y+2} \,\to\, 0 \times \dfrac{1}{y+2}$, then $\dfrac{\pi y}{2(y+2)} \,\to\, 0 \times \dfrac{1}{2}$. Therefore, $\dfrac{\pi y}{2(y+2)} \,\to\, 0$ but we have assumed that $z \,=\, \dfrac{\pi y}{2(y+2)}$.

Therefore, the three steps have cleared that if $y$ approaches zero, then the $z$ also approaches zero. Now, convert the function in terms of $z$ from $y$.

$\implies$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$ $\,=\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{\sin{z}}{z}}$

According to the trigonometric limit rule of sine function, the limit rule of sinx/x as x approaches 0 is equal to one. Therefore, the limit of $\dfrac{\sin{z}}{z}$ as $z$ approaches $0$ is also equal to one.

$=\,\,\,$ $\dfrac{\pi}{4} \times 1$

$=\,\,\,$ $\dfrac{\pi}{4}$

Method – 2

In this method, we do not use the transformation technique but we take some acceptable mathematical steps, which help us to evaluate the limit of the given function.

$\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$

Express the cos function in terms of sine

In the given function, the numerator is a trigonometric function and the denominator is algebraic function. In limits, there is no limit rule in cosine function but we have a trigonometric limit rule in sine function. Hence, we have to convert the cosine function into sine function and it is possible by the complementary angle identity of sine function.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x-2}}$

Adjust denominator same as Angle of sine

In order to use the trigonometric limit rule of sine function, we have to adjust the algebraic function in the denominator same as the angle in the sine function.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[1 \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x-2}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{\pi} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x-2}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\pi \times (x-2)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\pi \times x-\pi \times 2}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\pi x-2\pi}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{1 \times (\pi x-2\pi)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\dfrac{x}{x} \times (\pi x-2\pi)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \dfrac{1}{x} \times (\pi x-2\pi)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{1}{x} \times \pi x-\dfrac{1}{x} \times 2\pi\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{1 \times \pi x}{x}-\dfrac{1 \times 2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{\pi x}{x}-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\dfrac{\pi \cancel{x}}{\cancel{x}}-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{x \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\pi \times \dfrac{1}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi \times 1}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{1 \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\dfrac{2}{2} \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \dfrac{1}{2} \times \Big(\pi-\dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{1}{2} \times \pi-\dfrac{1}{2} \times \dfrac{2\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{1 \times \pi}{2}-\dfrac{1 \times 2\pi}{2 \times x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{2\pi}{2x}\Big)}\Bigg]}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{\cancel{2}\pi}{\cancel{2}x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{1 \times \sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{2 \times \Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{x} \times \dfrac{1}{2} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi \times 1}{x \times 2} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Bigg[\dfrac{\pi}{2x} \times \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}\Bigg]}$

Evaluate the Limits of the functions

Now, split the limit of the product of functions as product of their limits by the product rule of limits.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \Big(\dfrac{\pi}{2x}\Big)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$

Now, find the limit of the first fractional function by the direct substitution method as $x$ approaches $2$.

$=\,\,\,$ $\dfrac{\pi}{2(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$

$=\,\,\,$ $\dfrac{\pi}{2 \times 2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$

$=\,\,\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$

Now, let us evaluate the limit of the second trigonometric function. The algebraic function in the denominator is exactly same as the angle in the sine function. So, we can use the trigonometric limit rule of sine function but we have to set the input of the limit same as the denominator or angle in the sine function.

$(1) \,\,\,$ If $x \,\to\, 2$ then $\dfrac{1}{x} \,\to\, \dfrac{1}{2}$

$(2) \,\,\,$ If $\dfrac{1}{x} \,\to\, \dfrac{1}{2}$, then $\pi \times \dfrac{1}{x} \,\to\, \pi \times \dfrac{1}{2}$. Therefore, $\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}$

$(3) \,\,\,$ If $\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}$, then $-\dfrac{\pi}{x} \,\to\, -\dfrac{\pi}{2}$

$(4) \,\,\,$ If $-\dfrac{\pi}{x} \,\to\, -\dfrac{\pi}{2}$, then $\dfrac{\pi}{2}-\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}-\dfrac{\pi}{2}$. So, $\dfrac{\pi}{2}-\dfrac{\pi}{x} \,\to\, 0$

The four steps have cleared to us that $\dfrac{\pi}{2}-\dfrac{\pi}{x}$ approaches zero when $x$ approaches $2$.

Now, take $m = \dfrac{\pi}{2}-\dfrac{\pi}{x}$ and convert the function in terms of $m$ from $x$.

$\implies$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$ $\,=\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin{m}}{m}}$

According to the trigonometric limit rule of sine function, the limit of $\dfrac{\sin{m}}{{m}}$ as $m$ approaches zero is equal to one.

$=\,\,\,$ $\dfrac{\pi}{4} \times 1$

$=\,\,\,$ $\dfrac{\pi}{4}$

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# Velocity-dependent potentials and the dissipation function

From this previous question Charge, velocity-dependent potentials and Lagrangian where the citation is shown at the page 22, §1.5 of the book Classical Mechanics of Goldstein, we read that

"an electric charge $$q$$ of mass $$m$$ moving at a velocity $$\mathbf{v}$$ in a region containing both electric field $$\mathbf E(t,x,y,z)$$ and magnetic field $$\mathbf B(t,x,y,z)$$ ($$\mathbf B$$ and $$\mathbf E$$ are derivable from a scalar potential $$\phi(t, x, y, z)$$ and a vector potential $$\mathbf A(t,x,y,z)$$), nowing that $$\mathbf E=- \nabla \phi - \frac{\partial {\mathbf A}} {\partial t}, \quad \mathbf B= \nabla \times {\mathbf A}. "\tag{1.61}$$

Why must be

$$\color{red}{\large \mathcal U=q \phi - q {\mathbf A} \cdot{\mathbf v} \quad ?}\tag{1.62}$$

Exist a physics proof of this equality?

Because that combination, after use of the Euler-Lagrange equation, gives you the Lorentz force.

Lagrangians can never really be "proved". Once you know the equation of motion (be it classical mechanics, quantum or electrodynamics), you can "come up" with a Lagrangian that gives the correct answer. Lagrangian are better for they are scalars, manifestly symmetric, and more compact.

• In the meantime, I would like to thank you very much for your reply. Could you do me a kindness, if it is possible, to have a simple proof...because that combination, after use of the Euler-Lagrange equation, gives you the Lorentz force. Thank you very much. Sep 6 '19 at 21:15
• phys.ufl.edu/~pjh/teaching/phy4605/notes/… Sep 6 '19 at 21:16
• Is it also the answer of Claudius' user? I have understood 50% the method of Claudius and 50% your notes. I'm not gonna make a mixture and just get confused, am I? Sep 6 '19 at 21:18
• ... I beg your pardon? Who's Claudius? Sep 6 '19 at 21:19
• Ah I see, on your other question. I think he does the beginning, whereas the link goes all the way. I'm happy to discuss steps of the proof in the chat you'd like. Sep 6 '19 at 21:20

If you use that definition for a potential then the generalised force Qi( see Goldstein equation 1.53) can be replaced by a velocity dependent potential( equation 1.58) that leads to a beautiful equation called Lagrange equation( equation 1.57)

• While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review Apr 7 '20 at 10:56
• I'm just a newbie here Apr 7 '20 at 11:07
• As it was already said in the review, it is good practice to be more explanatory (write corresponding formulas in the answer), and if the latter seems to make too much effort instead of posting an answer make rather a comment. The comment section offers quite a lot editing space (:-)). Apr 7 '20 at 12:22
• I cannot comment as it says that one should have at least 50 reputation. Apr 7 '20 at 13:35
• @YasirSadiq There is a reason for that. You shouldn't try to get around the site design by posting comments as answers. You should post answers as answers, and then when you get enough reputation it shows that you know what you are talking about and are a little more familiar with the site, and hence you will be able to leave comments Apr 7 '20 at 16:57


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Calculus Volume 3

# 7.4Series Solutions of Differential Equations

Calculus Volume 37.4 Series Solutions of Differential Equations

### Learning Objectives

• 7.4.1 Use power series to solve first-order and second-order differential equations.

In Introduction to Power Series, we studied how functions can be represented as power series, $y(x)=∑n=0∞anxn.y(x)=∑n=0∞anxn.$ We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives $y′(x)=∑n=1∞nanxn−1y′(x)=∑n=1∞nanxn−1$ and $y″(x)=∑n=2∞n(n−1)anxn−2.y″(x)=∑n=2∞n(n−1)anxn−2.$ In some cases, these power series representations can be used to find solutions to differential equations.

Be aware that this subject is given only a very brief treatment in this text. Most introductory differential equations textbooks include an entire chapter on power series solutions. This text has only a single section on the topic, so several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises in this section were chosen for which power solutions exist. However, it is not always the case that power solutions exist. Those of you interested in a more rigorous treatment of this topic should consult a differential equations text.

### Problem-Solving Strategy

#### Problem-Solving Strategy: Finding Power Series Solutions to Differential Equations

1. Assume the differential equation has a solution of the form $y(x)=∑n=0∞anxn.y(x)=∑n=0∞anxn.$
2. Differentiate the power series term by term to get $y′(x)=∑n=1∞nanxn−1y′(x)=∑n=1∞nanxn−1$ and $y″(x)=∑n=2∞n(n−1)anxn−2.y″(x)=∑n=2∞n(n−1)anxn−2.$
3. Substitute the power series expressions into the differential equation.
4. Re-index sums as necessary to combine terms and simplify the expression.
5. Equate coefficients of like powers of $xx$ to determine values for the coefficients $anan$ in the power series.
6. Substitute the coefficients back into the power series and write the solution.

### Example 7.25

#### Series Solutions to Differential Equations

Find a power series solution for the following differential equations.

1. $y″−y=0y″−y=0$
2. $(x2−1)y″+6xy′+4y=−4(x2−1)y″+6xy′+4y=−4$
Checkpoint 7.22

Find a power series solution for the following differential equations.

1. $y′+2xy=0y′+2xy=0$
2. $(x+1)y′=3y(x+1)y′=3y$

We close this section with a brief introduction to Bessel functions. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order n is given by

$x2y″+xy′+(x2−n2)y=0.x2y″+xy′+(x2−n2)y=0.$

This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.

### Example 7.26

#### Power Series Solution to the Bessel Equation

Find a power series solution to the Bessel equation of order 0 and graph the solution.

Checkpoint 7.23

Verify that the expression found in Example 7.26 is a solution to the Bessel equation of order 0.

### Section 7.4 Exercises

Find a power series solution for the following differential equations.

104.

$y″+6y′=0y″+6y′=0$

105.

$5y″+y′=05y″+y′=0$

106.

$y″+25y=0y″+25y=0$

107.

$y″−y=0y″−y=0$

108.

$2y′+y=02y′+y=0$

109.

$y′−2xy=0y′−2xy=0$

110.

$(x−7)y′+2y=0(x−7)y′+2y=0$

111.

$y″−xy′−y=0y″−xy′−y=0$

112.

$(1+x2)y″−4xy′+6y=0(1+x2)y″−4xy′+6y=0$

113.

$x2y″−xy′−3y=0x2y″−xy′−3y=0$

114.

$y″−8y′=0,y(0)=−2,y′(0)=10y″−8y′=0,y(0)=−2,y′(0)=10$

115.

$y″−2xy=0,y(0)=1,y′(0)=−3y″−2xy=0,y(0)=1,y′(0)=−3$

116.

The differential equation $x2y″+xy′+(x2−1)y=0x2y″+xy′+(x2−1)y=0$ is a Bessel equation of order 1. Use a power series of the form $y=∑n=0∞anxny=∑n=0∞anxn$ to find the solution.

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text:
chrono::particlefactory::ChRandomParticlePositionRectangleOutlet Class Reference

## Description

Class for generator of random particle positions scattered over a rectangle outlet in 3D space.

#include <ChRandomParticlePosition.h>

Inheritance diagram for chrono::particlefactory::ChRandomParticlePositionRectangleOutlet:
Collaboration diagram for chrono::particlefactory::ChRandomParticlePositionRectangleOutlet:

## Public Member Functions

virtual ChVector RandomPosition () override
Function that creates a random position each time it is called. More...

ChCoordsysOutlet ()
Access the coordinate system of the rectangular outlet. More...

double & OutletWidth ()
Access the width of the rectangular outlet, that is on the X axis of the coordinate.

double & OutletHeight ()
Access the height of the rectangular outlet, that is on the Y axis of the coordinate.

## Member Function Documentation

 ChCoordsys& chrono::particlefactory::ChRandomParticlePositionRectangleOutlet::Outlet ( )

Access the coordinate system of the rectangular outlet.

The outlet width is on the X direction of this csys, and the outled height is on the Y direction of this csys.

 virtual ChVector chrono::particlefactory::ChRandomParticlePositionRectangleOutlet::RandomPosition ( )
overridevirtual

Function that creates a random position each time it is called.

Reimplemented from chrono::particlefactory::ChRandomParticlePosition.


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Prove by performing the required integration that the total forceon a small mass located
inside a spherical shell is zero. The density of the sphericalshell is constant, 3, the inside
radius is Ro and has a thickness, t, and the sphere is center atthe origin. The small mass,
m is located on the x axis at a distance, x, from the origin.

the link has a picture copy and paste

http://learn.vccs.edu/courses/1/NV280.PHY.231.002A.FA08/content/_5812703_1/gravi.pdf?bsession=56138852&bsession_str=session_id=56138852,user_id_pk1=174741,user_id_sos_id_pk2=1,one_time_token=ECE3F22DC518D23E8FEC7437D8414DED


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# ReStructuredText Sample¶

ReStructuredText is an extensible markup language used by LSST.

ReStructuredText provides basic italic, bold and monospaced typesetting. There is also the concept of roles that provide sophisticated typesetting, such as $$\mu = -2.5 \log_{10}(\mathrm{DN} / A) + m_0$$, and referencing.

## Sectioning¶

Sections are formed with underlining the headline text. We use a conventional sequence of underline symbols to indicate different levels of hierarchy.

## Directives¶

Besides roles that are used for inline markup, reStructuredText has the concept of directives to markup blocks of content. One example is the is the code-block directive:

print('hello world!')


text:
# Real Analysis/Riemann integration

 Real Analysis Riemann integration

## Definition

Riemann integration is the formulation of integration most people think of if they ever think about integration. It is the only type of integration considered in most calculus classes; many other forms of integration, notably Lebesgue integrals, are extensions of Riemann integrals to larger classes of functions. The Riemann integral was developed by Bernhard Riemann in 1854 and was, when invented, the first rigorous definition of integration applicable to not necessarily continuous functions.

We will first define some preliminary ideas.

## Partitions

### Definition

Let $a,b\in\mathbb{R}$

A Partition $\mathcal{P}$ is defined as the ordered n-tuple of real numbers $\mathcal{P}=(x_0,x_1,x_2,\ldots ,x_n)$ such that $a

### Norm of a Partition

Let $\mathcal{P}$ be a partition given by $\mathcal{P}=(x_0,x_1,x_2,\ldots ,x_n)$

Then, the Norm (or the "mesh") of $\mathcal{P}$ is defined as $\|\mathcal{P}\| =\sup \{x_{k+1}-x_k|0\leq k\leq n-1\}$

### Tagged Partition

Let $\mathcal{P}=(x_0,x_1,x_2,\ldots ,x_n)$ be a partition

A Tagged Partition $\mathcal{\dot{P}}$ is defined as the set of ordered pairs $\mathcal{\dot{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n$ such that $x_{i-1}\leq t_i\leq x_i$. The points $t_i$ are called Tags.

## Riemann

Riemann sum of a function

### Riemann Sums

Let $f:[a,b]\to\mathbb{R}$

Let $\mathcal{\dot{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n$ be a tagged partition of $[a,b]$

The Riemann Sum of $f$ over $[a,b]$ with respect to $\mathcal{\dot{P}}$ is given by

$S(f,\mathcal{\dot{P}})=\displaystyle\sum_{i=1}^n f(t_i)(x_i-x_{i-1})$

### Riemann Integral

Let $f:[a,b]\to\mathbb{R}$

Let $L\in\mathbb{R}$

We say that $f$ is Integrable on $[a,b]$ if and only if for every $\varepsilon>0$, there exists $\delta>0$ such that for every partition $\mathcal{\dot{P}}$ satisfying $\|\mathcal{\dot{P}}\|<\delta$, we have that $|S(f,\mathcal{\dot{P}})-L|<\varepsilon$

$L$ is said to be the integral of $f$ over $[a,b]$, and is written as

$L=\int_a^b f(x)dx$ or as $L=\int_a^b f$

## Properties

### Theorem (Uniqueness)

Let $f:[a,b]\to\mathbb{R}$ be integrable on $[a,b]$

Then the integral $L$ of $f$ is unique

#### Proof

Assume, if possible that $L_1\neq L_2$ are both integrals of $f$ over $[a,b]$. Consider $\varepsilon=\tfrac{|L_1-L_2|}{2}$

As $L_1,L_2$ are integrals, there exist $\delta_1,\delta_2>0$ such that $|S(f,\mathcal{\dot{P}})-L_1|<\varepsilon$ for all $\mathcal{\dot{P}}$ that satisfy $\|\mathcal{\dot{P}}\|<\delta_1$ and $|S(f,\mathcal{\dot{P}})-L_2|<\varepsilon$ for all $\mathcal{\dot{P}}$ that satisfy $\|\mathcal{\dot{P}}\|<\delta_2$

Let $\delta=\inf\{\delta_1,\delta_2\}$. Hence, if $\mathcal{\dot{P}}$ is a partition satisfying $\|\mathcal{\dot{P}}\|<\delta$, then we have $|S(f,\mathcal{\dot{P}})-L_1|<\varepsilon$ and that $|S(f,\mathcal{\dot{P}})-L_2|<\varepsilon$

That is, $|L_1-L_2|<2\varepsilon =|L_1-L_2|$, which is an obvious contradiction. Hence the integral $L$ of $f$ is unique.

We now state (without proof) two seemingly obvious properties of the integral.

### Theorem

Let $f,g:[a,b]\to\mathbb{R}$ be integrable and let $c\in (a,b)$

Then:

(i)$\int_a^b f+\int_a^b g=\int_a^b f+g$

(ii)$\int_a^c f+\int_c^b f=\int_a^b f$

### Theorem (Boundedness Theorem)

Let $f:[a,b]\to\mathbb{R}$ be Riemann integrable. Then, $f$ is bounded over $[a,b]$

#### Proof

Assume if possible that $f$ is unbounded. For every $n\in\mathbb{N}$ divide the interval $[a,b]$ into $n$ parts. Hence, for every $n\in\mathbb{N}$, $f$ is unbounded on at least one of these $n$ parts. Call it $I_n$.

Now, let $\varepsilon>0$ be given. Consider an arbitrary $\delta>0$. Let $\mathcal{\dot{P}}$ be a tagged partition such that $\|\mathcal{\dot{P}}\|<\delta$ and $(I_n,t_n)\in\mathcal{\dot{P}}$, where $t_n$ is taken so as to satisfy $|f(t_n)|>n\varepsilon$.

Thus we have that $|S(f,\mathcal{\dot{P}})-L|>\varepsilon$. But as $\delta>0$ is arbitrary, we have a contradiction to the fact that $f$ is Riemann integrable.

Hence, $f$ is bounded.

## Integrability

We now study classes of Riemann integrable functions. The first "constraint" on Riemann integrable functions is provided by the Cauchy Integrability Criterion.

### Theorem (Cauchy Criterion)

Let $f:[a,b]\to\mathbb{R}$

Then,

(i)$f$ is Riemann integrable on $[a,b]$ if and only if

(ii) For every $\varepsilon>0$, there exists $\delta>0$ such that if $\mathcal{\dot{P}},\mathcal{\dot{Q}}$ are two partitions satisfying $\|\mathcal{\dot{P}}\|,\|\mathcal{\dot{Q}}\|<\delta$ then $|S(f,\mathcal{\dot{P}})-S(f,\mathcal{\dot{Q}})|<\varepsilon$

#### Proof

($\Rightarrow$)Let $\int_a^b f=L$ and let $\varepsilon>0$ be given.

Then, there exists $\delta>0$ such that for every partition $\mathcal{\dot{P}}$ satisfying $\|\mathcal{\dot{P}}\|<\delta$,we have $|S(f,\mathcal{\dot{P}})-L|<\tfrac{\varepsilon}{2}$

Now, let partitions $\mathcal{\dot{P}},\mathcal{\dot{Q}}$ be such that $\|\mathcal{\dot{P}}\|,\|\mathcal{\dot{Q}}\|<\delta$.

Thus we have that $|S(f,\mathcal{\dot{P}})-L|,|S(f,\mathcal{\dot{Q}})-L|<\tfrac{\varepsilon}{2}$, that is $|S(f,\mathcal{\dot{P}})-S(f,\mathcal{\dot{Q}})|<\varepsilon$

($\Leftarrow$) For every $n\in\mathbb{N}$, consider $\delta_n>0$ such that for all partitions $\mathcal{\dot{P}},\mathcal{\dot{Q}}$ satisfying $\|\mathcal{\dot{P}}\|,\|\mathcal{\dot{Q}}\|<\delta_n$, we have $|S(f,\mathcal{\dot{P}})-S(f,\mathcal{\dot{Q}})|<\tfrac{1}{n}$.

Without loss of generality, we can assume that $\delta_m>\delta_n$ when $m. For every $\delta_n$, let $\mathcal{\dot{P}}_n$ be a partition such that $\|\mathcal{\dot{P}}_n\|<\delta_n$

The sequence $a_n=S(f,\mathcal{\dot{P}}_n)$ is a Cauchy sequence, and hence it has a limit $L\in\mathbb{R}$.

Now, for every $\varepsilon>0$, we have a $\delta>0$ such that $\|\mathcal{\dot{P}}\|<\delta$ implies $|S(f,\mathcal{\dot{P}})-L|<\varepsilon$.

Thus $\int_a^b f=L$

### Theorem (Squeeze Theorem)

Let $f:[a,b]\to\mathbb{R}$

Then,

(i) $f$ is Riemann integrable on $[a,b]$ if and only if

(ii) For every $\varepsilon>0$, there exist Riemann integrable functions $\alpha_{\varepsilon},\omega_{\varepsilon}:[a,b]\to\mathbb{R}$ such that

$\alpha_{\varepsilon}(x)\leq f(x)\leq\omega_{\varepsilon}(x)$ for all $x\in[a,b]$ and

$\int_a^b (\omega_{\varepsilon}-\alpha_{\varepsilon})<\varepsilon$

#### Proof

($\Rightarrow$)Take $\alpha_{\varepsilon}(x)=f(x)=\omega_{\varepsilon}(x)$. It is easy to see that $\int_a^b (\omega_{\varepsilon}-\alpha_{\varepsilon})<\varepsilon$

($\Leftarrow$)Let $n\in\mathbb{N}$. Then, there exist functions $\alpha_n,\omega_n$ such that $\int_a^b (\omega_n-\alpha_n)<\tfrac{1}{n}$. Further, if $\int_a^b \alpha_n=A_n$ and $\int_a^b \omega_n=Z_n$, then there exist $\delta_1,\delta_2>0$ such that if a partition $\mathcal{\dot{P}}$ satisfies $\|\mathcal{\dot{P}}\|<\delta_1$ then $|S(\alpha_n,\mathcal{\dot{P}})-A_n|<\tfrac{1}{n}$ and $\|\mathcal{\dot{P}}\|<\delta_2$ then $|S(\omega_n,\mathcal{\dot{P}})-Z_n|<\tfrac{1}{n}$

Now let $\mathcal{\dot{P}}_n$ be an interval satisfying $\|\mathcal{\dot{P}}_n\|<\inf \{\delta_1,\delta_2\}$.

Now, we can easily see that $|S(f,\mathcal{\dot{P}}_n)-S(f,\mathcal{\dot{P}}_{n-1})|<\tfrac{1}{n}$. Hence, $S(f,\mathcal{\dot{P}}_n)$ is a Cauchy sequence, with a limit $L\in\mathbb{R}$, and as in the previous proof, we can show that $\int_a^b f=L$


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## Lines and points

The equation of a straight line is given by:

where a is the slope of the line. If a = 0, the line is horizontal. b indicate the intersection of the line with the y-axis.

## Finding the equation of a line, when two points are given

The equation of a straight line, two points are given

It’s possible to find the equation of a straight line, when two of its points are given:

(x1,y1) and (x2, y2).

First you have to determine a and b of the equation of a straight line:

a and b can then be substituted into the equation of a straight line.

## The equation of a straight line when two points are given

 Point 1 x1: y1: Point 2 x2: y2:

## Finding the equation of a line, when a point and the line’s slope is given

The equation of a straight line passing through the point (x0, y0) and with the slope a is given by:

If the angle θ between the x-axis and the line is given instead of the slope a, it’s possible to calculate a in the following way:


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# Rational canonical form of the matrix $A$

Let the matrix

$$A=\begin{bmatrix} 2 & 1 & 2 \\ -2 & -1 & -4 \\\ 1 & 1 & 3 \end{bmatrix}.$$

So far I found the characteristic polynomial $C_A(x)=(2-x)(x-1)^2$, the minimal polynomial $m_A(x)=(2-x)(x-1)$ and the Jordan matrix form is

$$J=\begin{bmatrix} 2 & * & *\\ * & 1 & * \\\ * & * & 1 \end{bmatrix}.$$

Does someone could tell me from what I did how it possible to find the rational canonical form of $A$?

There are two "rational canonical forms", one obtained using elementary divisors and another obtained by invariant factors.

The invariant factors form a list where each term divides the next, their product is the characteristic polynomial, and the minimal polynomial is the largest one.

Therefore in this problem, $(x-2)(x-1)$ is the largest invariant factor. The other invariant factor is $(x-1)$ simply because there is nothing left over.

The rational canonical form is obtained by putting the companion matrices corresponding to the invariant factors on the main diagonal.

So the rational canonical form of $A$ given by the invariant factors is: $$\begin{pmatrix} 0 & -2 & 0\\ 1 & 3 & 0\\ 0 & 0 & 1 \end{pmatrix}_.$$

The matrix $$\begin{pmatrix} 0 & -2 \\ 1 & 3 \end{pmatrix}_.$$

Is the companion matrix of $(x-2)(x-1)$.

https://en.wikipedia.org/wiki/Companion_matrix

• It is easy to obtain the invariant factors from the elementary divisors, and there is a great algorithm for computing the elementary divisors of a matrix in the book "Linear Algebra" by Friedberg, Insel, and Spence. – Ken Duna Apr 29 '16 at 14:38
• Is it necessary to use this algorithm to obtain the elementary divisor? – user332681 Apr 29 '16 at 14:45
• No. This is just a convenient way, and the way I like to do it. There are methods to find the invariant factors, such as the Smith-Normal Form. However, I find that very cumbersome. math.stackexchange.com/questions/340352/… Here is a link to a question about computing rational canonical forms that has an answer (by the guy who wrote the book the questioner was studying from) on how to computing the invariant factors. – Ken Duna Apr 29 '16 at 14:48
• So far, no one has said anything mathematically incorrect. We just have a disagreement on terminology. If you do not want to refer to the companion matrix of elementary divisors decomposition as a rational canonical form, that is fine and I don't think I will be able to change your mind. Let's just move on and be friends :) – Ken Duna Apr 29 '16 at 21:59
• Fair enough, I had never seen such a definition before. Thanks for the reference! – André 3000 Apr 30 '16 at 2:06


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# A triangle has sides A,B, and C. If the angle between sides A and B is (3pi)/4, the angle between sides B and C is pi/12, and the length of B is 9, what is the area of the triangle?

May 24, 2018

Area of triangle color(maroon)(A_t = 14.83 sq units

#### Explanation:

Given : $\hat{A} = \frac{\pi}{12} , \hat{C} = \frac{3 \pi}{4} , b = 9$

$\hat{B} = \pi - \frac{\pi}{12} - \frac{3 \pi}{4} = \frac{\pi}{6}$

$\frac{a}{\sin} A = \frac{b}{\sin} B$

$a = \frac{b \sin A}{\sin} B = \frac{9 \cdot \sin \left(\frac{\pi}{12}\right)}{\sin} \left(\frac{\pi}{6}\right)$

$a = 4.66$

Area of triangle ${A}_{t} = \left(\frac{1}{2}\right) a b \sin C$

${A}_{t} = \left(\frac{1}{2}\right) \cdot 4.66 \cdot 9 \cdot \sin \left(\frac{3 \pi}{4}\right)$

${A}_{t} = 14.83$ sq units


text:
An ant walks in a straight line on a plane leaving behind a trace of its movement. The initial position of the ant is at point $\text{P}$ facing east.

The ant first turns $72^{\circ}$ anticlockwise at $\text{P},$ and then does the following two steps in sequence exactly FIVE times before halting.

1. moves forward for $10 \; \text{cm.}$
2. turns $144^{\circ}$ clockwise.

The pattern made by the trace left behind by the ant is

A.

B.

C.

D.

Here, ${\color{Blue}{\text{PS = SQ = QT = TR = RP = 10 cm}}}$
Correct Answer $:\text{C}$


text:
# ctable cell vertical align

I have a table like this:

\documentclass{article}
\usepackage{ctable}
\begin{document}
\ctable
[caption=Sample]
{|p{2cm}|c|}
{}
{
\hline
some multiline (2 or 3 fields) field & need to vertically center this \\ \hline
one more field & and this \\ \hline
}
\end{document}

I want cells at the second column to be vertically center aligned. Which is the simplest and best method?

-

Use an m{2in} column type rather than p{2in} (which centres its contents vertically). This is provided by the array package, already loaded by ctable:

\ctable
[caption=Sample]
{|m{2cm}|c|}

Alternatively, the multirow can also do this:

\documentclass{article}
\usepackage{ctable}% http://ctan.org/pkg/ctable
\usepackage{multirow}% http://ctan.org/pkg/multirow
\begin{document}
\ctable
[caption=Sample]
{|p{2cm}|c|}
{}
{
\hline
some multiline (2 or 3 fields) field & \multirow{3}*{need to vertically center this} \\ \hline
one more field & \multirow{2}*{and this} \\ \hline
}
\end{document}

-
Thanks, m{2cm} works perfect! –  Drakmail May 31 '12 at 17:00


text:
Perpetual Contracts
Perpetual contracts are one of the most prominent types of derivatives within the market. Buying (going long) or selling (going short) perpetual contracts, allows traders to speculate on the future price of an asset, without having to specify an expiry date. This implies that the trader, buyer and seller, can hold the position for as long as they want. Additionally, there are no added costs apart from funding costs. As a result of this flexibility, perpetual contract trading is ideal. Traders are not locked into a losing trade even in a lousy market or on the weekends.
Within the fxdx protocol, the contract price is linked to the cost of the underlying asset by a dynamic interest rate, this is similar to current perpetual contracts. For liquidation, an on-chain pricing oracle is used (and secondarily for interest payments). The market's order book is off-chain, allowing for faster price changes and more liquidity.
Each perpetual contract has its own set of contract specifications.
• Funding Rate
• Leverage
• Margin Requirement
• Perpetual Liquidations

## Funding Rate

Perpetual contract prices frequently diverge from those of spot markets, dependening on the market sentiment. If most traders believe the underlying asset's value will rise over time, the price of the perpetual contract will almost certainly outperform the current spot price.
Short positions pay funding to long positions if the perpetual is trading at a discount to the underlying index. As a result of this, traders are incentivized to drive the future rate closer to the index price.
• Perpetual futures have funding payments every hour, fxdx protocol follows FTX's method to compute the funding rate.

## Leverage

Leverage allows traders to enter a position that is worth a lot more, by investing a small amount of money. As a consequence of this, the profit, as well as losses, can be dramatically amplified.
• Fxdx protocol allows traders to use leverage by backing a position with margin collateral that is worth less than the total position size. Traders are able to open positions with leverage up to 10x.

## Margin Requirement

The minimal margin percentage that an account must maintain to prevent liquidation is known as the margin requirement. The minimum collateral required to maintain an open position (maintenance margin) and all open orders (initial margin) are calculated periodically.
Traders on derivative exchanges must constantly maintain a healthy ratio between the value of the margin used to open a position, in addition to the notional value of that position. This is known as the margin ratio and can be calculated by the formula given below.
$(Margin + Unrealized PnL)/Notional Value of the Position$
When evaluating liquidation conditions, PnL is calculated using both index Price and 15-minute TWAP of Mark Price; the highest of the two values is then utilized.
Unlike centralized markets, margin calls will be performed by client software, algorithms, or third-party services and not issued by the network.

## Perpetual Liquidations

If the account's margin percentage falls below the maintenance margin minimum (7.5 percent), liquidation may be required. If a trader's position drops and if the loss begins to approach the value of his margin(initial collateral), these sudden price changes can make the position worth less than the collateral. If the value of assets is dangerously close to the value of the trader's collateral, the exchange will liquidate the trader's position to avoid losses. Partially liquidating an account is also possible. In this scenario, the liquidator will assume proportional amounts of the account's margin.
The liquidation takes place on-chain, and any account can act as a liquidator, as long as its final amount matches the maintenance margin criteria.
Incentives
Because a liquidator may take on the whole balance, the temptation to liquidate grows as the target account gets closer to the liquidation threshold. Other elements that influence the liquidation reward include:
• Avoidance of taking on additional price exposure, thus closing costs for the liquidated position
• Gas costs to win liquidation
• Price direction and time to liquidation
Alternatively explained, the more leverage a trader employs, the easier it is to liquidate the position. Due to the possibility that the margin ratio may quickly go below the maintenance margin during volatile market conditions. In order to free up the margin and maintain the position, fxdx will cancel all open orders on the current contract if liquidation is triggered. Orders on other contracts will continue to be accepted.


text:
# What does (x+sin(x))/(x+cos(x)) equal as limit x-> "infinity"?? Thank you!!!

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
bp Share
Feb 11, 2018

1

#### Explanation:

Divide the numerator and denominator by x, so that given function becomes

$f \left(x\right) = \frac{1 + \frac{\sin x}{x}}{1 + \frac{\cos x}{x}}$.

Now as $x \to \infty$. $\frac{\sin x}{x} \to 0$, because sin x would oscillate between +1 and -1, which in either case divided by $\infty$ would be 0. Thus the limit of the numerator would be 1. Like wise the limit of the denominator would also be 1.
Thus limit as a whole would be 1

• 6 minutes ago
• 7 minutes ago
• 7 minutes ago
• 8 minutes ago
• 5 seconds ago
• 2 minutes ago
• 4 minutes ago
• 5 minutes ago
• 5 minutes ago
• 6 minutes ago
• 6 minutes ago
• 7 minutes ago
• 7 minutes ago
• 8 minutes ago


text:
# An optimization problem, non complete bipartite graph and hungarian algorithm

I have two tables at my disposal, one work dataset and one reference dataset. Each dataset has got two columns, lets say these are fields A and B. I would like the rows in reference dataset with the rows in work dataset that are 'closest' w.r.t. some distance. I have more rows in work dataset than in reference dataset, and i will match all rows in reference dataset to some rows in work dataset.

I can define distance between two rows in reference and wrk dataset like this:

$d_{i,j} = (A_{w}(i) - A_{r}(j))^{2} + (B_{w}(i) - B_{r}(j))^{2}$

with $A_{w}, A_{r}, B_{w}, B_{r}$ the A and B columns work and reference datasets with obvious notation.

In other word, i want to minimize over all permutations of rows from the work dataset (with the same number of rows as in reference dataset) the sum of distances between one row in reference dataset and one row in wrk dataset.

I do not know how to proceed if not examining all permutations.

This is combinatorial problem. What about stochastic methods: genetic algorithm, swarm...

Could you hint at some idea for a start ?

I had a look at linear assignment problem. However, my problem is not a complete bipartatite graph representation, it is not complete since there can be more peaks in work signal than there are in reference signal.

Thanks !

-

There's a standard trick to convert the min cost matching problem on a balanced bipartite graph to one on an unbalanced bipartite graph. Let $G = (X \cup Y, E, w)$ be the bipartite graph where $E \subset X \times Y$ and $|X| \le |Y|$.
Now create a copy of $G$ and add it "reversed", so that in the new graph both sides have exactly $|X| + |Y|$ vertices. All old edges have their same weight: edges between vertices in $X$ and their copies have weight $\infty$, and edges between vertices in $Y$ and their copies have weight zero.


text:
# Surfaces

Surfaces are features of volumes and act as their limits - like the square surface of the cube.


text:
Algebra II : Multiplying and Dividing Logarithms

Example Questions

Example Question #1 : Multiplying And Dividing Logarithms

Rewrite the following logarithmic expression in expanded form (i.e. as a sum and/or difference):

Explanation:

By logarithmic properties:

;

Combining these three terms gives the correct answer:

Example Question #2 : Simplifying Logarithms

Which of the following is equivalent to

Explanation:

Recall that log implies base  if not indicated.Then, we break up . Thus, we have .

Our log rules indicate that

.

So we are really interested in,

.

Since we are interested in log base , we can solve  without a calculator.

We know that , and thus by the definition of log we have that .

Therefore, we have

Example Question #3 : Simplifying Logarithms

Find the value of the Logarithmic Expression.

Explanation:

Use the change of base formula to solve this equation.

Example Question #2 : Multiplying And Dividing Logarithms

Many textbooks use the following convention for logarithms:

What is ?

Explanation:

Remembering the rules for logarithms, we know that .

This tells us that .

This becomes , which is .

Example Question #3 : Multiplying And Dividing Logarithms

What is another way of expressing the following?

Explanation:

Use the rule

Example Question #2 : Multiplying And Dividing Logarithms

Expand this logarithm:

Explanation:

In order to solve this problem you must understand the product property of logarithms  and the power property of logarithms . Note that these apply to logs of all bases not just base 10.

log of multiple terms is the log of each individual one:

now use the power property to move the exponent over:

Example Question #3 : Multiplying And Dividing Logarithms

Which of the following is equivalent to ?

Explanation:

We can rewrite the terms of the inner quantity.  Change the negative exponent into a fraction.

This means that:

Split up these logarithms by addition.

According to the log rules, the powers can be transferred in front of the logs as coefficients.


text:
# How can I draw this diagram?

How can I draw this diagram in LaTeX?

• Yes. (As an answer this would be about as helpful as the question) – Chris H Feb 14 '17 at 13:26
• Possible duplicate of TikZ - flowchart nodes alignment. And many other TikZ questions about flowcharts and similar – Chris H Feb 14 '17 at 13:27
• In its current form, your question might not receive many answers. Please take a look at the How to Ask-page and try to improve your question according to the guidance found there. This may require you to show some effort on your part in terms of attempting a solution. If you have questions about what to do or if you don't quite understand what this means, please ask for clarification using the add comment function. – Martin Schröder Feb 14 '17 at 13:55
• Why would you want to, though? – MissMonicaE Feb 14 '17 at 14:01

## 2 Answers

By way of stacks.

\documentclass{article}
\usepackage[usestackEOL]{stackengine}
\begin{document}
\sffamily
{\footnotesize\stackanchor{Self-adjusted}{Market}}
$\leftarrow$
\Centerstack{1840s -- 1870s\\$\downarrow$\\Adam Smith \& David Ricardo\\
$\downarrow$\\ \small Restriction on Economic Competition}
$\rightarrow$
{\footnotesize\stackanchor{Comparitive}{Advantage}}
\end{document}

• Stackengine to the rescue! ;-) – JPi Feb 14 '17 at 13:32
• @JPi Takes a whacking and keeps on stacking! – Steven B. Segletes Feb 14 '17 at 13:33

   \documentclass{article}

\usepackage{tikz}
\usetikzlibrary{positioning}

\begin{document}

\begin{tikzpicture}[every text node part/.style={align=center}]
\node (a) at (0,0) {Adam Smith \& David Ricardo};
\node[below=of a] (b) {Restriction on Economic Competition};
\node[above=of a] (c) {1840s--1870s};
\node[left=of a] (d) {Self--adjusted\\Market};
\node[right=of a] (e) {Comparative\\Advantage};
\draw[->] (c.south)--(a.north);
\draw[->] (a.west)--(d.east);
\draw[->] (a.east)--(e.west);
\draw[->] (a.south)--(b.north);
\end{tikzpicture}
\end{document}


text:
A Analytical results for relaxation times and conductivities

# Relaxation times and charge conductivity of silicene

## Abstract

We investigate the transport and single particle relaxation times of silicene in the presence of neutral and charged impurities. The static charge conductivity is studied using the semiclassical Boltzmann formalism when the spin-orbit interaction is taken into account. The screening is modeled within Thomas-Fermi and random phase approximations. We show that the transport relaxation time is always longer than the single particle one. Easy electrical controllability of both carrier density and band gap in this buckled two-dimensional structure makes it a suitable candidate for several electronic and optoelectronic applications. In particular, we observe that the dc charge conductivity could be easily controlled through an external electric field, a very promising feature for applications as electrical switches and transistors. Our findings would be qualitatively valid for other buckled honeycomb lattices of the same family, such as germanine and stanine.

###### pacs:
68.65.Pq, 72.10.-d, 71.10.Ca, 72.80.Vp

## I Introduction

Silicene, a single layer of silicon atoms arranged in a honeycomb structure was foreseen theoretically (1). Although silicene was predicted to be stable in its freestanding fashion, so far its existence has been evidenced only on supporting templates on metallic substrates both as a sheet (2); (3) and as ribbons (4); (5) and also on semiconductor substrates (6). Silicon atoms in silicene tend to adopt the tetrahedral hybridization over the hybridization of carbon atoms in graphene. This results in a slightly buckled structure of silicene (1); (7); (8) where atoms in sublattices and are displaced from each other in the out-of-plane direction with a distance Å (see, Fig. 1). The buckled nature of silicene that originates from the mixed hybridization (9), together with its large atomic volume, leads to a relatively large spin-orbit coupling (SOC) which is roughly 1000 times larger than graphene (10); (11), and it is expected to open a visible band gap to be measured at room temperature. This large SOC is the origin of the envisioned quantum spin-Hall insulator (QSHI) phase, a state with dissipationless spin currents along the edges of the sample, in silicene (12). Moreover, a Rashba spin-orbit coupling term also naturally appears in the effective models for silicene, however, this term has a negligible effect on its energy dispersion and could be neglected for most purposes (13); (14).

The magnitude of the energy gap in silicene is controllable in different ways, such as photo-irradiation (15), anti-ferromagnetic exchange interactions (16), and by surface adsorption (17). Also, as the structure of silicene is buckled, application of a perpendicular electric field breaks its sublattice symmetry, and, as a result, a controllable band gap is induced in its energy dispersion (19); (18). This perpendicular electric field makes it possible to control the mass term in the Dirac equation independently in valleys and . The possibility of controlling the gap term in different valleys, in addition to the strong SOC in silicene, makes valleytronics and spin-valleytronics (20); (14); (16) applications feasible in this material. The external perpendicular electric field could also spin-polarize the states of each valley. This could be useful in silicene based spin-filtering (21). Remarkably, a transition from a two-dimensional topological insulator state to a band insulator is also predicated through increasing the strength of the electric field (19). The large spin-orbit coupling in silicene in comparison with graphene, together with the tunability of its band dispersion in particular, makes it a very good candidate to be used in the field of spintronics.

In this paper, we use the semiclassical Boltzmann formalism, within the relaxation time approximation (RTA), to study the charge conductivity of a silicene sheet, subjected to a perpendicular electric field. The effects of both short-range and charged scatterers on the conductivity are investigated. The Rashba spin-orbit coupling in intrinsic silicene and germane is very tiny and therefore, it is usually neglected in literature. However, one can think of enhancing this term by adding heavy adatoms (22); (23); (24), or by placing the layer on special substrates such as Ni(111) (26); (25); (27). For this purpose, we keep the Rashba SOC throughout our calculations. Charged impurities are naturally screened by the medium and we take care of this within two different schemes, namely, the Thomas-Fermi approximation (TF) and random phase approximation (RPA). We calculate the static density-density response function (28) in the presence of the SOC term numerically and illustrate how Rashba SOC affects it. We also explore two different momentum relaxation times, namely, the transport relaxation time and the single particle relaxation time. The former determines the charge conductivity and other transport properties, while the latter shows the quantum-level broadening (29). The effects of both short-range and screened charged impurities are investigated. Finally, we use the semiclassical Boltzmann formalism to find the charge conductivity of silicene from the transport relaxation time.

The organization of our paper is as follows. In Sec. II, we briefly describe our model Hamiltonian, together with its band dispersion and eigenfunctions in the presence of Rashba spin-orbit coupling. Then, we describe the density-density response function both at zero and finite Rashba spin-orbit coupling limit. This will be the basis to derive the many-body screening of charged impurities. Also in this section, we describe the single particle as well as transport relaxation times, and charge conductivity within the Boltzmann formalism. Section III is devoted to our analytic and numerical results for relaxation times and conductivity of silicene, in the presence of charged long-range as well as neutral short-range scatterers. We summarize and conclude our main findings in Sec. IV.

## Ii Model Hamiltonian and Formalism

The effective low-energy Hamiltonian of the buckled honeycomb lattice of two-dimensional silicene, in the presence of a perpendicular electric field, can be written as (30); (14); (31)

 Hη=ℏvF(kx^τx−ηky^τy)−ηΔSO^τz^σz+ηa0λR^τz(kx^σy−ky^σx)+Δz^τz , (1)

where is the Fermi velocity and is the equilibrium lattice constant of silicene. and are ,respectively, the and components of the momentum, and (with ), are the Pauli matrices in the sublattice and spin spaces, respectively, and refers to two different valleys and . and in Eq. (1) are the intrinsic and Rashba spin-orbit couplings, respectively. Owing to the buckled structure of the system, atoms of two sublattices and are located in two different planes, separated by . Therefore, application of a perpendicular electric field to the sample plane, breaks the sublattice symmetry and splits their on-site energies by . This leads to the splitting of the band dispersion into four subbands in each valley.

The parameters of the Hamiltonian (1) are Å, Å, ms, meV and meV. Note that the Hamiltonian (1) could be used to model other buckled honeycomb structures such as germanine and stanene, with larger energy gaps of and meV, respectively. Material specific parameters of model Hamiltonian (1) are listed in Ref. [(12)].

The eigenvalues of Hamiltonian (1) in valley are

 εηβ,s(k)=β√(ℏvFk)2+(√Δ2SO+(a0λRk)2+sηΔz)2 , (2)

where refers to the conduction and valence bands, and distinguishes between the light and heavy spin subbands of the conduction and valence bands. At zero electric field, two subbands are degenerate and a small direct energy gap equal to opens up between the valence and conduction bands at . Applying a finite gate voltage, splits the two bands into four subbands with the energy gaps of . At a critical value of the electric field , the smaller gap vanishes, and the light subbands become massless with graphene-like linear dispersions at small wave-vectors. Increasing the gate field further, the smaller gap opens up once again and increases with . Note that Hamiltonian (1) describes a two-dimensional topological insulator for , and a band insulator for (30)

In typical intrinsic buckled honeycomb lattices such as silicene, germanine, and stanene, the strength of Rashba spin-orbit coupling is much smaller than the intrinsic one, , and therefore the effects of the Rashba term on the energy dispersion are negligible, even at , which is much beyond the low-energy (i.e., small ) validity region of Hamiltonian (1). Although one can safely discard the Rashba term for most practical purposes and work with much simpler dispersions of , as it is possible to enhance the strength of this term by adding heavy adatoms or using special substrates (32), unless otherwise stated, we will retain the Rashba SOC term across our calculations.

The low-energy dispersions (2) of silicene are presented in Fig. 2. Here the effects of Rashba SOC are exaggerated, enhancing , 1000 times. Note that band dispersions with the intrinsic value of would be essentially indistinguishable from the zero Rashba ones.

The eigenvectors of Hamiltonian (1), in the basis, could be written in a very compact form as

 ψβ,s(k,r)=eik⋅r√2A⎛⎜ ⎜ ⎜ ⎜⎝βζβ(θs)ζs(γ)e−iηϕkζ−β(θs)ζs(γ)isβeiηϕkζβ(θs)ζ−s(γ)isζ−β(θs)ζ−s(γ)⎞⎟ ⎟ ⎟ ⎟⎠, (3)

where is the sample area, , and and are short-hand notes for the trigonometric functions sine and cosine, respectively. Moreover, we have introduced new angles

 γ=12arctan(a0λRkΔSO) , (4)

and

 θ±=12arctan(ℏvFkδ±k) , (5)

with .

### ii.1 Density-density response function and dielectric function

In order to obtain the relaxation times and the electrical conductivity in a system with charged impurities, one needs to know how these impurities are screened in the medium. In this section, we discuss the noninteracting density-density response function of silicene, which will be used to account for the screening of the electron-charged impurity scattering within the RPA. The noninteracting density-density response function could be obtained from (28)

 χ0(q,ω)=2A∑j,j′,kn[εj(k)]−n[εj′(k′)]ℏω+εj(k)−εj′(k′)+i0+Fj,j′(k,k′) , (6)

where , is the Fermi-Dirac distribution function, the prefactor comes from the valley degeneracy, and we use collective indexes: and for bands and spin subbands. The form-factor is the overlap between two wave-functions

 Fj,j′(k,k′)=|<ψj(k)|ψj′(k′)>|2=14(1+ββ′Gj,j′)(1+ss′Hj,j′) , (7)

where

 Gβ,s;β′,s′(k,k′)≡ℏ2v2Fk⋅k′+δskδs′k′|εβ,s(k)εβ′,s′(k′)| , (8)

and

 Hβ,s;β′,s′(k,k′)≡Δ2SO+a20λ2Rk⋅k′|εβ,s(k)εβ′,s′(k′)| . (9)

Using Eqs. (7)-(9) in Eq. (6), one can numerically calculate the polarizability of silicene with the Rashba spin-orbit coupling, and in the presence of a perpendicular electric field.

Density-density response function at the zero Rashba limit. As we have already pointed out, the intrinsic Rashba spin-orbit coupling is very small, and it is often reasonable to neglect it for simplicity (33); (34). Taking , the Hamiltonian (1) becomes diagonal in the spin space,

 Hη,s(k)=ℏvF(kx^τx−ηky^τy)+(Δz+sηΔSO)^τz , (10)

which is identical to the Hamiltonian of a gapped graphene with spin dependent gap values for each spin component (35). In other words, the -component of the spin is conserved, and the external electric field simply splits two spin sub-bands. Note that as the time reversal symmetry is still preserved, the direction of spin splitting should be reversed between two valleys.

The noninteracting density-density response of this simplified model becomes , where is the noninteracting response function of a gapped graphene with gap . This quantity is analytically known, both along the real (35) and imaginary (36) frequency axes, and in the static (i.e., ) limit reads

 χg0(q,Δ)=−νg(μ)⎡⎢ ⎢⎣1−Θ(q−2kF)⎛⎜ ⎜⎝√q2−4k2F2q−(ℏvFq)2−4Δ24ℏvFqμarctan(√(ℏvFq)2−4(ℏvFkF)22μ))] , (11)

for , and

 χg0(q,Δ)=−1πℏ2v2F[Δ+(ℏvFq)2−4Δ22ℏvFqarctan(ℏvFq2Δ)] , (12)

for . In Eqs. (11) and (12), is the chemical potential, , and is the density of states (per unit area) of the gapped graphene, where is the Heaviside step function. Note that the spin degeneracy of graphene is absent here.

The noninteracting density-density response function of silicene is presented in Fig. 3. The analytic expressions of Eqs. (11) and (12) are compared with the full numerical evaluation of Eq. (6) at non-zero Rashba SOC and for two different values of the external electric field. Similar to Fig. 2, we enhance the value of , 1000 times to illustrate its effect on the polarizability. In both cases, is constant and equal to at low wave-vectors, where is the density of states (per unit area) of silicene at its Fermi level. At larger values of the wave-vector, on the other hand, the absolute value of the response function increases linearly with .

Once the noninteracting density-density response function is known, one can obtain the screened interaction between electrons and charged impurities , where is the dielectric function, and within the RPA reads , where is the Fourier transform of the electron-electron interaction in two dimensions (), with being the vacuum permittivity, and the dielectric constant of the environment, which we take to be equal to 1 throughout this work. Replacing in the RPA dielectric function with its long-wavelength limit results in the well-known Thomas-Fermi approximation for screening .

### ii.2 Relaxation times and conductivity

Within the relaxation time approximation, the transport scattering time of an electron in the -th band and with wave-vector , could be written as (37)

 1τjtr(k)=1A∑j′,k′Wj,j′(k,k′)[1−vj′(k′)vj(k)cos(θk,k′)] , (13)

where the summation is over all possible final states with band and wave-vector , is the group velocity of band , and is the transition probability between states and , which , by using the Fermi’s golden rule for elastic scatterings, could be written as

 Wj,j′(k,k′)=2πℏ∣∣Tj,j′(k,k′)∣∣2δ(εj′,k′−εj,k) . (14)

Here is the -matrix (38), and within the first Born approximation for dilute and randomly distributed impurities becomes proportional to the matrix elements of the scattering potential (39)

 ∣∣Tj,j′(k,k′)∣∣2≈nimpFj,j′(k,k′)V2e−imp(k−k′) , (15)

where is the density of impurities and is the Fourier transformation of the interaction potential between an electron and a single impurity, and depends on the nature of impurities. In the case of short-ranged non-magnetic impurities (e.g., defects or neutral adatoms) it is quite safe to approximate it with a zero-range hard-core potential . On the other hand, scattering from charged impurities will be long-ranged Coulombic, . This potential is screened by other electrons of the system. The screening could be treated within the e.g., Thomas-Fermi or random-phase approximations. (28)

Energy conservation in an elastic scattering prevents any transition between valence and conduction bands. In the zero Rashba limit, commutes with Hamiltonian (1), and spin is a good quantum number. Therefore, with spin-independent scatterers such as the ones we consider in this work, the transition between two spin subbands of the same (valance or conduction) band will be prohibited too. The diagonal components of the form-factors in this case will read

 Fβ,s;β,s(k,k′)=12(1+ℏ2v2Fk⋅k′+Δ2sεk,sεk′,s) , (16)

where denotes the gap between -spin subbands, and is the dispersion of massive Dirac fermions with a mass term of . But the situation is different in the presence of a finite Rashba term. As the -component of the spin of an electron is no longer conserved, inter-subband transitions would be allowed even in the absence of magnetic impurities. However, as we will show below, the rate of this spin-flip transition would be much smaller than the intra-subband transitions, for the intrinsic values of .

In addition to the transport relaxation times, one can define single particle relaxation times, which are related to the imaginary part of the single particle self-energies, and define the broadening of the single particle levels owing to the impurity scatterings(38); (29):

 1τjsp(k)=1A∑j′,k′Wj,j′(k,k′) . (17)

Having calculated the transport relaxation times, the charge conductivity of the system could be obtained using the semi-classical Boltzmann formalism (38); (40)

 σ=e2ℏ2v2F2∑j∫dενj(ε)τjtr(ε)(−∂n(ε)∂ε) , (18)

where is the density of states of the -th band. At zero temperature, is a step function, and we arrive at the renowned expression

 σ=e2v2F2∑sνs(μ)τstr(kF,s) . (19)

Naturally, only the bands which cross the Fermi energy contribute to the conductivity at low temperatures. Note that this semiclassical theory is valid only in the regime of dilute impurity concentration i.e., when the density of impurities is much smaller than the density of charge carriers.

## Iii Results and Discussions

In this section, we turn to the presentation of our analytical and numerical results for single particle and transport relaxation times as well as the charge conductivity of silicene in the presence of short-range and long-range charged scatterers.

### iii.1 Relaxation times

Relaxation times are generally functions of the incoming electrons’ wave-vector, and at low temperatures only states around the Fermi level will contribute to transport and single particle properties. In the following we will concentrate only on the behavior of relaxation times at the Fermi energy. The transport and single particle relaxation times could be conveniently obtained using Eqs. (13) and (17), respectively. In the case of short-range neutral impurities, assuming vanishing Rashba SOC, the corresponding summations over final states could be performed analytically and the final results read

 1τstr(εF)=nimpV204ℏ3v2Fμ(μ2+3Δ2s)Θ(μ−Δs) , (20)

and

 1τssp(εF)=nimpV202ℏ3v2Fμ(μ2+Δ2s)Θ(μ−Δs) . (21)

For charged impurities, as we already discussed in the previous section, the inclusion of many-body screening is essential. Within the Thomas-Fermi approximation the screened potential can be written as

 VTFe−imp(q)=Ve−imp(q)ϵTF(q)=e22ϵ0(q+qTF) , (22)

where is the bare Coulomb interaction of an electron with an impurity of charge , and is the TF wave-vector. At the zero Rashba SOC limit, again the relaxation times could be obtained analytically. The final expressions for single particle and transport relaxation times are quite lengthy and we have presented them in the appendix. A more accurate screening of the charged impurities would be achieved within the RPA. In this case we present our results for the relaxation times numerically, retaining the intrinsic value of the Rashba spin-orbit coupling in our calculations. In the following we will use for the impurity concentration of both short range and long range scatterers. This guarantees that the diluteness criteria will be satisfied for a wide range of chemical potentials , and perpendicular electric fields in the following results for relaxation times and charge conductivities. Note that in an intrinsic silicene, and in the absence of any perpendicular electric field, even a chemical potential of , roughly corresponds to the carrier density of , which is already much larger than .

Our results for the transport relaxation times of both conduction spin subbands at the Fermi levels, and in the presence of short-range (SR) and long-range (LR) charged impurities, are presented as functions of the chemical potential and the perpendicular electric field in Figs. 4 and 5, respectively. In the case of charged impurities, the electron-impurity interaction is screened within the RPA. Note that the relaxation time of a band at the Fermi energy is not defined when the Fermi energy does not intersect that band. In the case of charged impurities a cusp like feature in the relaxation time of the lower () band as a function of the chemical potential or perpendicular electric field appears when the chemical potential reaches the bottom of the upper () band. The origin of this behavior lies in the transition between metallic and insulating contributions of the upper bands to the screening of the electron impurity interaction. As the short range interaction is not screened, such a behavior is absent there.

As we already discussed, within the TF approximation for screening, one simply replaces the full wave-vector dependent polarizability , with its limit. In 2D systems is constant at long wavelengths (see, Fig. 3) therefore one would expect a good agreement between RPA and TF. In Fig. 6, we compare our results obtained for the transport relaxation times within these two different screening approximations of the long-range potential. The only region where there is a visible difference between two different schemes, is the relaxation time of the lower subband at small chemical potentials. This is precisely the regime where only the lower conduction band is occupied. The RPA would take into account the contribution of the upper band in screening, using Eq. (12) for its density-density response function, while the TF approximation will simply ignore it, as this response function vanishes at .

Now, we turn to the presentation of our results for single particle relaxation times, as introduced in Eq. (17). Figure 7 shows the ratio between transport and single particle relaxation times, as a function of the chemical potential. The transport relaxation time is always larger than the single particle one. For the upper conduction band, the ratio increases with chemical potential, while for the lower one this ratio quickly saturates to . In the case of long-range impurities, abrupt changes in the ratio between transport and single particle relaxation times for lower band again originates from the transition between insulating and metallic screenings of the upper band.

In the last figure of this section, we assume a finite value for Rashba spin-orbit coupling meV. This off-diagonal term in the spin space permits interband transitions between two conduction spin subbands. As the intrinsic value of the Rashba spin-orbit coupling is very small in silicene, it still makes sense to label two conduction subbands as up and down spin subbands. In this sense, the interband transition could be interpreted as spin relaxation or spin-flip. In Fig. 8 we present the inverse of the interband or spin-flip single particle relaxation time,

 1τ+−sp(k)=∑k′W+,+;+,−(k,k′) , (23)

for the intrinsic value of the Rashba spin-orbit coupling of silicene (12). This interband relaxation time is much (almost times) longer than the intraband relaxation times, presented in our previous figures.

### iii.2 Charge conductivity

Now, having calculated the transport relaxation times, we can calculate the charge conductivity using Eq. (19).

Figure 9 illustrates our results for the charge conductivity of silicene in the presence of long-range and short-range scatterers as a function of the chemical potential for two different values of the perpendicular electric field, i.e., , for which both bands are gapped, and for which makes the lower band gapless. In the case of long-range impurities, the comparison is made between RPA and TF screenings of the electron-impurity interaction, too. The conductivity is an increasing function of , and therefore of the carrier density. In the case of short-range impurities, one can see that at large chemical potentials it saturates to a constant value determined by the scattering strength and impurity concentration [see also Eq. (25)]. As one would expect from the behavior of relaxation times, conductivity jumps when the chemical potential reaches the bottom of the upper band. This makes silicene suitable for making charge switches. Note that, the conductivity becomes zero when the chemical potential lies inside the gap of both bands. Interestingly, in the case of short-range scatterers, and for where the lower band is massless, the conductivity remains constant, as long as has not reached the upper band. This is in agreement with what has been observed in graphene (40).

Finally, in Fig. 10 the dependence of charge conductivity on the perpendicular electric field is illustrated for long-range and short-range scatterings. A similar behavior to previous figure is also observed here. Increasing the perpendicular electric field, a sudden drop in the conductivity appears in the transition from the double band to the single band regime. Increasing further, conductivity drops to zero, when the lower band is also pushed above the chemical potential.

## Iv Summary and conclusions

In summary, we have investigated the transport properties of a silicene sheet in the presence of both intrinsic and Rashba spin-orbit couplings. Although the intrinsic value of the Rashba spin-orbit coupling is small, it can be enhanced using heavy adatoms or appropriate substrates. Also, we have considered a perpendicular electric field applied to silicene sheet which can be used to modify the band gap and therefore the transport properties of silicene.

Electron-impurity scatterings in two dimensional systems can originate from charged impurities, that mainly come from the substrate or from the local defects. In the latter case, the scattering potential is short range, while in the former it is long-ranged coulombic. Using an effective low-energy Hamiltonian for silicene in the presence of a perpendicular electric field, we have calculated the transport relaxation time for both types of scatterers. For charged impurities, we have employed Thomas-Fermi and random phase approximations for screening.

We have also compared the single particles relaxation time, which measures the quantum level broadening of the system, with the transport relaxation time. We have found that the transport relaxation time is always larger than the single particle one. Moreover, we have shown that the interband relaxation time is much longer than the interband relaxation time. Having calculated the transport relaxation times, and using the Boltzmann approach for conductivity, we calculated the charge conductivity in the presence of short-range and long-range impurities. As silicene is a material with a tunable band dispersion we have observed that the conductivity can possess abrupt changes with respect to the perpendicular electric field. This makes silicene a good candidate for electronic device applications where the switching process is a need.

*

## Appendix A Analytical results for relaxation times and conductivities

In the limit of vanishing Rashba SOC, i.e. , it is possible to obtain analytical expressions for relaxation times and conductivities in the presence of short-range impurities and in the presence of long-range impurities when the TF approximation is used for screening.

Short-range impurities- In the case of short-range impurities, the summation over in Eq. (13) could be performed analytically,

 1τstr(εF)=nimpV204πμℏ3v2FΘ(μ−Δs)×∫2π0dθ(1−cosθ)[μ2+Δ2s+(μ2−Δ2s)cosθ]=nimpV204ℏ3v2Fμ(μ2+3Δ2s)Θ(μ−Δs) , (24)

where . Using the above expression in Eq (19), one can find the charge conductivity at zero temperature, which in the units of reads

 σ [e2h]=4ℏ2v2Fμ2nimpV20∑sΘ(μ−Δs)μ2+3Δ2s . (25)

Long-range impurities with TF screening- The simplest approximation to include the static screening of charged impurities is the Thomas-Fermi approximation, in which one uses expression (22) for the screened electron-impurity interaction. Using Eq. (22) in Eq. (13), we can obtain an analytic expression for the transport relaxation time

 1τstr(εF)=nimpe432πε20ℏk2Fsμ2Θ(μ−Δs){{16k4Fsπbs−40k3FsqTF+πq2TF(3q2TF−2μ2)(bs−1)+4qTFkFs(3q2TF−2μ2)+2Δ[4πk2Fs(2bs−1)−πq2TF(bs−1)−4qTFkFs]+2πk2Fs[3q2TF(2−3bs)+2μ2(2bs−1)]}−2bs[16k4Fs+q2TF(3q2TF−2μ2)+2k2Fs(8μ2−9q2TF)+2Δ2(8k2Fs−q2TF)]arctan(2kFsas)} , (26)

where and , with being the Fermi wave-vector of the subband. Finally, the analytic form of the conductivity could be obtained from

 σ [e2h]=μℏ∑sτstr(kFs)Θ(μ−Δs) . (27)

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