text:
# Celia的博客

## 揭晓秘密

来不及等到开学了，我先在就说说之前那篇博文里提到的那个什么鬼秘密吧。

我在博文里说，秘密就在 Navigation Bar 导航栏里，其实如果和之前的Celia的博客对比，你会发现，原来的English Blog变成了CN Blog。这到底是什么一会事儿呢？

## 圣诞长假，在冰火两重天中揭开序幕

已经挺久没有在这个博客上发表文章了。为什么呢？原因在这个博客的 Navagation Bar 上。如果没懂没关系，答案会在开学后揭晓。

这才是假期的第一周，我就干了两件大事（还是在假期的第一天）：

## Secret

Mum, wanna see what's on ur blog?

If I see it's fixed tmr u probably didn't sleep! 😏

Relax, check it out on Dec 26 and it will be normal again.

## 中文戏剧社Activity申请成功！

在上课时收到这么一封邮件，真是又惊喜又害怕，感觉简直要从座位上蹦起来。

中文戏剧社是我与若谷、周祎一起计划创办的，针对UWC 6-8 年级的中学生。想要报名、或了解更多，请查看下一篇博文：中文戏剧社正式成立！Season 3 开始报名

戏剧社的申请没通过时，一切都感觉还是幻想。每次给戏剧社做计划时，都在想：通过不通过都还没知道呢，这么早做计划干什么？现在一切都在表格上清清楚楚地写着，压力瞬间就来了。

目前，我们已经找到了几个备选的剧本，audition也筹划得差不多了，还做出了一张海报，接下来要做的，就是找一个可以教中文drama的老师。

所以，欢迎6-8年级的同学前来参加！也请大家多多宣传！

## “好奇心是最好的老师”

—— but in a bad sort of way

Macs are known for their superb security system. I personally think that Windows are easier to get viruses than macs based on my experience of killing my first Windows laptop by allowing some strange dialogue box to update my windows system. However, my mac still got viruses just a day ago. And I realized that the biggest danger to my mac isn't a bad security system -- instead it was my curiosity that kills it.

So it was a cool Sunday afternoon. I had just finished all of my homework (or maybe that was my excuse for browsing on the web) and started looking for a crack for Sibelius. There was a really good melody in my head that I was in an urge to record it but I could not find any tool. The basic version for Sibelius was actually quite good already but the song that I was thinking of includes so many different musical instruments that Sibelius Basic does not have. And so therefore I was on my journey to a crack for Sibelius Pro.

As usually I went in and out of website, but Sibelius seemed to protect its copyright really well and I could not find even one pirated version of it. The only ones that I could find were the websites that automatically generate webpages that says "XXX's crack" or "XXX free down" but are actually another app when you actually download and open the package. I got really frustrated seeing all these unzipped files that says FlashPlayer (remember they are supposed to be cracks for Sibelius?) and the RAR files scattered around my desktop.

I was just going to hurl these into the trash bin in the corner of my screen when I suddenly thought: Why not see what's inside? It was a stupid thought, an extremely stupid idea indeed, but there was the devil in my heart continuously prompting me to do the "double-click". The icon was there. Open it up and install it and I will see what it is and what it does, I was thinking. I knew it's a scam but what’s there to be afraid of? At the worst, it would just not run and I would just move it to trash. A package file ending up in the trash was just a matter of time, why not try something fun if I couldn't install Sibelius Pro anyways?

So then the process started. I knew exactly how to start installing, to authenticate the app, to key in my password and to drag things around. It was awesome. The fun of seeing the process bar slowly reaching for the end at the right was more enjoyable than everything else.

Installation ends. While I was closing the window I suddenly realized that some other softwares were also installed: there was one called "Phase Search" -- an app with a green icon and the sign with a magnifier in the middle, another app called something doctor, and some so-called photo editor.
 (I clearly remembered phase search was something like this)
This was a really common case if it was in windows. Some apps just want to promote some other apps so that they make you accidentally download something else when you are downloading the apps you want. Just delete the ones you don't want. It was that easy. So I proceeded to the "Applications" folder and searched for the five extra apps that I downloaded by chance.

And then it started. Mac's default browser Safari opened on its own. I was quite surprised. Instinctively I did ⌘ ⌥ esc to quit Safari, but something was faster than me. A series of apps suddenly jumps out and blocked my sight to the Force Quit Window. I tried to reselect the window, but I couldn't move my mouse. It seemed to get stuck.

Oops.

I used my trackpad instead and tried to move the cursor on the screen. It was no use. My cursor was controlled by something.

Ooooops.

Instantly my sight went to the opened apps. There were five of them, exactly the ones I unexpectedly installed. The "Phase Search", "Doctor", "Photo Editor", "Image editor".

No!! ⌘ ⌥ esc! My cursor was as still as a statue in the middle of the screen.

⌘ ⌥ esc

⌘ ⌥ esc!

Windows started jumping out. At a rate of several per second. 10 of them, 20 of them, 30 of them, gradually covering my whole 15-inch screen. NOOOOOO!!!!

My mac started fanning, beeping irregularly like a siren. It was a mix of red and orange and yellow on the screen. "100 windows opened!" One of the windows says, I didn't know which. It was a huge mess everywhere. Everything appeared as a flash in a fraction of a second and then being covered by another new window again.

So force-quit did not work, cursor did not work, staring at the screen and be at a loss at what to do was the least helpful thing of the all. I stared at the button at the top right part of my keyboard, thinking of my unsaved documents, my Econs and Compsci essays unsaved in Microsoft word. It was the only choice.

I pressed the button.

My sight turned black abruptly in a brief second, and then I drew my self back to the reality. The fanning without any warning stopped "phew" it said.

It was just 5 minutes ago since I double-clicked and put my mac's fate in the mouth of a shark. 6:15, the clock said as I sat back with my head blank. Perhaps the virus drained all of my energy too. It was always summer in Singapore, yet it feels so cold sitting in an aircon room hearing wind whistling violently outside.

## 人们为什么要把氯加入饮用水？

今天是化学考试。其中有这么一道题让我十分困惑：Why do people add Chlorine in drinking water? 这道题的基本意思就是：人们为什么要把氯加入饮用水？

## 批评与表扬式教育

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/main/2019/11/25/批评与表扬式教育

## 每周歌曲推荐

我已经喜欢匈牙利圆舞曲很久了，相信很多人也是。其中我最喜欢的版本就是上面的这个，其他有一些是没有情感的，有一些是节奏杂乱的即兴创作，也有一些变得现代化了——不但加入了架子鼓，还有吉他，这好好地经典名曲被糟蹋得惨不忍睹...

## [转载] 2019年8月巴厘岛之旅

还记得三个月前，我写得两篇关于巴厘岛旅行的文章吗？

然而，由于太忙了，最后只写了两天的...于是，我就转载了妈妈博客中巴厘岛旅行的游记！Enjoy！

## 远亲不如近邻

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/main/2019/11/17/远亲不如近邻/

## [转载] 孤独

在匆忙的城市里，我们都是孤独的个体。有时候我站在路边看着人来人往，觉得城市像沙漠一样荒凉，大家都靠得那么近，但完全不知道彼此的心事；朋友们说的是同一种语言，但还常常产生误会。

## 每周歌曲推荐

第一次听这首歌是在爸爸的车上，但当时不知道它叫什么名字，也没听见歌词。真正认真听是在今年九月初时，九年级的中文课上。九年级中文课和想象中的不太一样。在学期初，老师并没有带领我们用IGCSE的教材上课，而是尝试组织一些活动。

## 2019年10月澳洲骑马之旅

介绍：10月18日，我们前往机场，即将开始我们的澳洲骑马之旅。以下是我在澳洲帕斯写的日记，我将它按原文电子化了。在澳洲的每天都非常充实，也非常累，日记里难免会出现一些语病和尴尬的描述。请大家不要介意：）

## 每周歌曲推荐

这首歌本来听起来不是很好听，没什么情感，平铺直叙，但因为是肖邦的，打算再听听，结果在虾米音乐上听了这首歌的另一个版本（就是上面这个）。这是我第一次体会到不同的演奏方式有多么大的差异。

## 满招损，谦受益

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/main/2019/11/01/满招损，谦受益/

## 一个残疾人的经历

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/main/2019/11/02/一个残疾人的经历/

## 奶茶工厂之旅

从澳洲回来，在新加坡享受假期的最后一个星期。这天，忙着写数学功课的我突然接到了朋友的一条消息：
顿时整个人都兴奋了起来！周末 / 假期最兴奋地事情，莫过于和朋友一起出去Hangout！于是我微微一皱眉，一挥笔将数学作业规划到了明天，然后“爽快”地答应了

## 关灯了，谁在那儿？

Celia成功从澳洲骑马户外旅行活着回来啦！这是我写的一篇英文恐怖小说，在这里分享给大家。Celia的博客的博文以中文为主，更多英文博文，请访问Nightingale（nightingale.becomingcelia.com）。
Inspired by Lights out, who's there?
Watch here

11 pm, her normal bedtime. She filled herself a glass of water on the dining table. Swallowing, she felt the sound of water sliding down her throat. After finishing she looked around. Things were strangely quiet in an eerie way. She shivered.

She had graduated from university three months ago. Mum had wanted her to work in her hometown so that she would be able to return home everyday, but thinking of getting a job with more potential, she had decided to go to a foreign first-tier city, alone. Considering her situation, she had rented a cheap house near the central of the city, paying an unacceptably high rent.

A corridor, long and narrow, was at the corner of the dinning room. The end of the corridor was the door to her bedroom. It had been left ajar, light shone out from it, an ice-blue color. She preferred warmer colors, but this light would do for her first night in the cheap rental house. Light switches for the whole house were on the left of the doorframe, yellowed with age, which she hesitated before touching it.

With a few clicks the lights were off. She got into the room walking directly towards her bed. Sitting down, she took her phone and opened WhatsApp. 0 messages. Refresh. 0 messages. It was not until then she realized that there was no signal in the rental house, and as she never turned on her data, she was completely separated from the world in the past few hours when hurrying to get things organized in the rental house. Mum should have already texted her a thousand messages now. Worriedly, she turned on her data and called Mum.

“Hello?” The warm and steady voice put her a bit more at ease.

“Hi Mum, it’s me.” She had a sudden impulse to cry.

“Fibie! Oh my god I… Why did you not call me! It seemed like you have disappeared. How was everything going?”

She hesitated, looking around the old, yellowish rental house. “Oh I am living with my friend! Remember Maggie? My primary school’s friend.” She started making it up. “Never mind, everything is going w—” Crack! Something is wrong with the bed. She struggled to stand up, but it was too late. Boom! The whole bed crashed onto the filthy floor. She felt extreme pain in her waist. She tried to stand up. Two legs at the end of the bed had snapped. She stepped back in terror. “What happened?!” Mum yelled on the other end of the call. “I’m fine. I’m fine.” She tried to speak in her normal voice. “Good night mum.” She said quickly and abruptly ended the call.

Dread was spreading in her body. The broken bed seemed to be unexpected and normal. It was getting cold. Wind was whistling by the window violently, vehemently, fiercely. I should rest, she thought. I should rest and tomorrow will be a sunny day again.

She walked towards the door and stretch out a hand lazily and pulled on it. The door did not move. She dragged the door with both hands, hard this time. The wooden door started to move and squeaked as it turns. Crack! The poor-quality doorframe had a tiny spine sticking out from the top. The door hit the spine and stopped moving, as if setting itself against her deliberately.

She pulled even harder. The door did not move. She put her whole weight on the door, staring at the end of the corridor hard, as if that would make her stronger or ease the pain on her hands. A gust of cold wind blew onto her cheeks. She opened her eyes wide, but suddenly, it seemed, there was something in the darkness. She searched in the darkness for the same color and contour that she just saw, and there, at the end of the corridor, was a lump of shadow, shadow with a dark color, much darker than its surrounding area. She widened her eyes in horror, with her sudden realization that it was a silhouette, a still silhouette of a woman. Restraining herself from screaming, she immediately fumbled for the light switches on her right. Click click. Lights were on, nothing was there. “Ah.” She felt a sense of relief. Out of pure curiosity, she turned the lights back off. Click. The shadow was still there. Her heart tensed again. Click. Lights on, nothing. Click. Lights off, the shadow appeared. She fiddled the switch for a few times, started to feel bored and turned the lights off the last time, deciding to go to bed. Click. The shadow was not there. She searched for the shadow again and again at the end of the corridor, and in a sudden, she saw it.

The silhouette was midway through the corridor.

Lights on. Lights off again, just as her hand got away from the switches, the air coagulated.

The shadow was in front of her, next to the doorframe.

She screamed.

Lights on. She took some tape to fix the light switches, making sure that they are not going to turn off. I need to sleep, she thought. Corridor lights on, desk lamp on, the whole house bright and safe. She crawled onto the broken bed, hiding her head underneath the blankets. She watched the bright lights through the blanket, starting to feel dizzy and sleepy.

A soft click. Losing her consciousness, she did not hear it. A loud click. She was quickly alerted, shrinking into the blanket fearfully.

Click. Her heart started racing. She held her breath.

Click. It was the desk lamp. She screamed, becoming weak and limp, feeling unable to control her body. The blanket slid off down the slope of the broken bed.

Now she was exposed to the infinite darkness.

## 为何印尼烧芭，坡县天空会变灰？

最近几天，有到了一年一度的“印尼烧芭日”了。印度尼西亚烧起了芭蕉，导致新加坡严重的雾霾，这是年年都有的。然而，今年的雾霾貌似非常严重，UWC的同学们甚至需要带着口罩上学。于是，某天在车上，宝宝问了个问题。

## Designs: 科学课还是打怪？

Why would I think of making this? Look carefully in the flames, you will find something.

Yes, we are doing a science experiment in science class. There was a balloon filled with Lithium gas and one of our classmates volunteered to burn the balloon with a long stick with fire on its end. And then...

The balloon popped and exploded.

The reason why our classmate looked so calm is that the explosion just happened and no one had reacted to it yet.

This is the full video of what happened in Chemistry class. Enjoy.

## 黄昏中的校园

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/main/2019/10/10/黄昏中的校园/

## Designs: 背影

Designs: 背影:     这本是语文的作业，因为我们正在学习朱自清的《背影》，因此要给自己的亲人拍一张他们的背影，然后分享自己对于这张照片的感受。     宝宝这张是我第一次将图P到这么融入景色的。自我感觉水波纹做的不错，但是就是边缘还是有些粗糙，而且光线貌似还是有点不太对。感受嘛，...

*点击上方蓝字阅读

## [续写] 永远的蝴蝶

又是一个雨天，雨点“咚咚”地砸在窗棂上，盖过了楼下车辆的喇叭声和刹车声。我摊在沙发上，望着一条条雨滴汇成小溪，不断地流下来。

## 论旅游业的发展对国家和人民的影响

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/2019/09/30/论旅游业的发展对国家和人民的影响/

## 一处我最熟悉的地方

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/2019/09/30/一处我最熟悉的地方/

## 苦涩的眼泪

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/2019/09/30/苦涩的眼泪/

## 境遇与成才

请点击此链接查看文章：http://lockedbecomingcelia.pcriot.com/forbidden/2019/09/30/境遇与成才-2/

## [转载] 永不录用

已是凌晨三点，当他从网吧里出来时，满脑子仍然充满游戏里的刀光剑影。他如网游中的大侠般麻利地翻过校园墙头，突然一个黑影横亘眼前，他意识到：这次真要挂了。

班主任只冷冷地丢下一句话：回去收拾收拾，准备卷铺盖回家吧。

## Summer

Exams are coming, summer is ending.

## Class Reunion？

时间飞逝，转眼间，八年级的小朋友们又开始互相想念了。八年级班级群"Chewdren"安静了两个月后，突然有人发话了。

（为了大家的隐私，已把手机号码抠掉）
接着有人就有了一个好主意。

## 英文课主题——六种经典的情节设置

经过了Orientation，开学的第一天，IGCSE英语课开始了。作为学校IGCSE必选课程之一——英语课自然格外重要。我们学习的第一个单元，是阅读叙述性文章。第一堂课，英语老师Mr. Minford给我们讲了一个非常有趣的概念：情节设置。

## 妇好

今天的语文课上，我们阅读了一篇著名作家峻青写的描写文，名为《海滨仲夏夜》。在这个文章中，作者将所有的劳动人民都称为“英雄”。我们也因此谈起了这个话题：你最喜欢的古代女英雄是谁呢？

## 每周歌曲推荐

这首歌，和其他一样，是我从新加坡电台中发现滴。这首歌总体的感觉是比较轻松愉快的，中间有一部分在抒情，最后又回归愉快。

## 质数与合数

最近在数学课上正在学习指数运算和开根号 (indices and surds)。这时，要知道一个数字是不是质数就非常重要了。这个小工具能快速告诉你一个数字是否是质数，以及它所有的因子。只要点击绿色的按钮开始运行程序，再输入数字就可以啦！

## New College Information Page Design

So, what is this about? High school is starting, and I have to frequently visit the website where our school information is stored. However, look at the first image, the actual website's sidebar, I couldn't bear its terrible (for me) look after seeing it so many times.
The designer of the webpage probably want to make the website look like material design, but the result was not quite of a material design project. As a result, I decided to redesign it in material design (well the background behind the sidebar is a screenshot).
Looks better right? Hope that the real Information page could be redesigned someday. For myself, I could only use my photoshop skills and wish so. I will not have so much time to spend on changing (or learn to change) this design into the real code.

lol

## 新生培训日！

从巴厘岛回来后，一看日历，发现马上就要开学了。今天再一看，离开学前的高中培训日竟只剩下了四天的时间。高中培训日这天，除了多熟悉高中的things（例如教学楼，上课方式，etc），还有一个重要的任务，那就是：

2019年8月4日

2019年8月3日

## [Survey] Making Choices

Hi everyone, here is a survey about choice making! If you are enrolled in an extracurricular activity or any other kind of activity, you are welcome to click the button below to answer our survey (only 2 questions)!


text:
# Math Help - Partial fraction method (in s-space)

1. ## Partial fraction method (in s-space)

The following is a part of a current I(s) (in $s$-space):

What am I doing wrong in using this formula for finding $A$ and $B$:
; where $a$ is a pole.

PROBLEM
I easily find A, but not B:

end of PROBLEM

The solution (which is correct) is

What am I doing wrong inside the PROBLEM section? Or is the formula for $A$ (or $B$) wrong?

2. The mistake is at the beginning:

$B/L=\displaystyle{\lim_{s\rightarrow b}(s-b)F(s)}}$.

3. You wrote this, right?

So, is the formula I used for $A$ wrong? Why do you divide $B$ by $L$ ( B/L )?

Seeing the correct solution, shouldn't you multiply by $L$?

4. Yes, I wrote that. For some reason Math Help Forum wasn't compiling Latex at the time I wrote it. I'm still having trouble with their Latex and so I won't use it.

No, the formula for A is correct. You divide B by L because sL is in the denominator of F(s).

You don't multiply by L; B/L gives the correct answer.

5. Originally Posted by ojones
No, the formula for A is correct. You divide B by L because sL is in the denominator of F(s).
Aren't s and ( $R_2$ + sL) the two denominators of $F(s)$? Or is $F(s)$ something entirely different (than the one I used inside PROBLEM tags)?

In short, what am I doing wrong inside the PROBLEM section?

Thank you for the help.

6. Why are you bothering with the formalism of limits? This is just a (very trivial) partial fraction decomposition you should have learned in precalculus --

You have the following after clearing fractions:

(1). A * (R + sL) + B * (s) = U

Let s = 0 => A = U/R
Let s = -R/L => B = -UL/R

Therefore, after substitution for A and B into your original decomposition, we obtain:

(A/s) + (B/(R + sL)) = (U/(Rs)) - (UL/(R(R+sL))).

The answer is easily verifiable by recombining fractions.

7. I'm not questioning whether the solution you (and ojones) gave is correct (and that the problem is (very) trivial).

But what the heck is wrong with "the limit formula" (or my way of (blindly) using it)?

8. Originally Posted by courteous
I'm not questioning whether the solution you (and ojones) gave is correct (and that the problem is (very) trivial).

But what the heck is wrong with "the limit formula" (or my way of (blindly) using it)?
Because you're not setting up the correct expression for the limit. It should be the limit as s => -R/L of the expression [U/s] because the factors (R + sL) will cancel. Where you obtain the factor (s + R/L) I have no idea; this is why you don't get a perfect cancellation, have to combine denominators and introduce an extra L.

By the way, this is known as the Heaviside "cover up" method. Here is a helpful resource you should consider: http://www.math-cs.gordon.edu/course.../heavyside.pdf. SO now, you can forget about this limit formalism and just "cover up" the factors (only applies when all factors are linear).

9. I am looking at what's in the PROBLEM section. The statement B=lim (s-b)F(s) is not correct. Why do you think it should be? (Latex still not working. Fix it MHF!)

format: plain


text:
# Toronto Math Forum

## MAT334-2018F => MAT334--Tests => Quiz-7 => Topic started by: Victor Ivrii on November 30, 2018, 03:58:52 PM

Title: Q7 TUT 5101
Post by: Victor Ivrii on November 30, 2018, 03:58:52 PM
Using argument principle along line on the picture, calculate the number of zeroes of the following function in the upper half-plane:
$$z^4 + 3iz^2 + z - 2 + i.$$
Title: Re: Q7 TUT 5101
Post by: Yuechen Huang on November 30, 2018, 04:12:50 PM
$1. f(z)=2z^{4}-2iz^{5}+z^{2}+2iz-1$
$f(x)=2x^{4}-3ix^{3}+x^{2}+2ix-1$
$x: [-\infty, \infty]$
$f(-\infty)\rightarrow \infty$
$f(\infty)\rightarrow \infty$ arg$(f(z))=0$

2. $Re^{it}$   $0\leq t \leq \pi$
$f(t)=2R^{4}e^{i4t}- 2iR^{3}e^{i3t}+R^{2}e^{i2t}+2iRe^{it}-1$
$0\leq 4t \leq 4\pi$   arg$(f(z))=4\pi$

The net change of argument is $4\pi$ , so that there are four solutions
Title: Re: Q7 TUT 5101
Post by: Jeffery Mcbride on November 30, 2018, 04:24:14 PM
$\displaystyle \begin{array}{{>{\displaystyle}l}} f( z) \ =\ z^{4} \ +\ 3iz^{2} \ +\ z\ -\ 2\ +\ i\\ \\ f( iy) \ is\ always\ positive\\ \\ f( x) \ =\ x^{4} \ +\ 3ix^{2} \ +\ x\ -\ 2\ +\ i\\ \\ Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2 \end{array}$

This is positive at x = R, then negative when it hits x = 1 to x = -1 then it becomes positive again as x approaches -R
.
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$ This is always positive.

So, we are always in the upper half plane. Change in Arg = 0.

$\displaystyle f\left( Re^{it}\right) \ =\ e^{4it} \$from [0, $\displaystyle \pi$] because the 4 term domindates. So the change in argument is 4$\displaystyle \pi$.

$\displaystyle \frac{0+4\pi }{2\pi } \ =\ 2$ zeros total

Title: Re: Q7 TUT 5101
Post by: Siying Li on November 30, 2018, 05:28:32 PM
Solve with argument principle

$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}+3{\mathrm{i}}{\mathrm{z}}^{\mathrm{2}}-2+{\mathrm{i}}$

When $\mathrm{z\ }$is on Real axis, let $\mathrm{z=x+iy}$, then $\mathrm{y=0}$, $\mathrm{z=x}$
$\mathrm{f}\left(\mathrm{z}\right)\mathrm{=}{\mathrm{4z}}^{\mathrm{4}}+3{\mathrm{i}}{\mathrm{z}}^{\mathrm{2}}-2+{\mathrm{i}}\mathrm{=4}{\mathrm{x}}^{\mathrm{4}}+3{\mathrm{i}}{\mathrm{x}}^{\mathrm{2}}-2+{\mathrm{i}}$
${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=}{\mathrm{arctan} \left(\frac{3x^2+1}{4x^4+x-2}\right)\ }$
As R goes to infinity, $\frac{3x^2+1}{4x^4+x-2}\mathrm{=}0$
Then ${\mathrm{arctan} \left(\frac{3x^2+1}{4x^4+x-2}\right)\ }\mathrm{=}0$

Let $\mathrm{z=}{\mathrm{Re}}^{\mathrm{it}}\mathrm{,\ \ 0}\mathrm{\le }\mathrm{t}\mathrm{\le }\mathrm{\pi }$
Then $\mathrm{f}\left(\mathrm{z}\right)\mathrm{=4}{({\mathrm{Re}}^{\mathrm{it}})}^{\mathrm{4}}+3i{\left({\mathrm{Re}}^{\mathrm{it}}\right)}^2-2+i\mathrm{=4}{\mathrm{R}}^{\mathrm{4}}e^{i4t}+3iR^2e^{i2t}-2+i$
$\mathrm{arg}\mathrm{}\mathrm{(f}\left(\mathrm{z}\right)\mathrm{)}\mathrm{\cong }\mathrm{4t}$
When $\mathrm{t=0}$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4*0=0}$

When $\mathrm{t=}\mathrm{\pi }$, ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }\mathrm{=4}\ \mathrm{\pi }\mathrm{=4}\mathrm{\pi }$

Then the overall net change in ${\mathrm{arg} \left(\mathrm{f}\left(\mathrm{z}\right)\right)\ }$ is $\left(\mathrm{4}\mathrm{\pi }\mathrm{-0}\right)\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{+}\mathrm{(}0\mathrm{-}\mathrm{0)}\mathrm{=4}\mathrm{\pi }$

Then the number of zeros in $\mathrm{f}\left(\mathrm{z}\right)$ is $\frac{1}{2\pi }*\left(4\pi \right)=2$
Title: Re: Q7 TUT 5101
Post by: Victor Ivrii on December 01, 2018, 01:09:35 PM
Siying, Yuechen,

Why are you bringing here other problems? It is called offtop and some can it spam .

Jeffery,
If you plug in $z=yi$ you will get not real valued , so "$f(iy)$ 𝑖is always positive" is both incorrect and irrelevant. But something (tell us what) is along $(-\infty, \infty)$ and this would save the day

Title: Re: Q7 TUT 5101
Post by: Jeffery Mcbride on December 01, 2018, 04:54:19 PM
You're right, $\displaystyle f( iy)$ is totally irrelevant here. My mistake. Do you mean the fact that f([-R, R]) lies completely in the upper half plane. Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$

When $\displaystyle z\ \rightarrow \ \pm \infty$, Re(f(z)) $\displaystyle \rightarrow \ \infty$ and Im(f(z)) $\displaystyle \rightarrow \ \infty \$and the endpoints of $\displaystyle f([ -R,\ R]) \$lie in the first quadrant for R really big. So the change in argument is going to be 0.

This can be coupled with what I said about the curve.

$\displaystyle f\left( Re^{it}\right)$ =  $\displaystyle e^{4it} \$from [0, $\displaystyle \pi$] because the 4 term dominates. So the change in argument is 4$\displaystyle \pi$.

So in total we get 2 zeros as the total change in argument is 4$\displaystyle \pi$.

Title: Re: Q7 TUT 5101
Post by: Victor Ivrii on December 01, 2018, 05:51:45 PM
Yes, $f(x)$ belongs to upper half-plane for $x\in (-\infty,\infty)$. But why?
Title: Re: Q7 TUT 5101
Post by: Jeffery Mcbride on December 01, 2018, 08:49:16 PM
Because
$\displaystyle Re( f( x)) \ =\ x^{4} \ +\ x\ -\ 2$
$\displaystyle Im( f( x)) \ =\ 3x^{2} \ +\ 1$
$\displaystyle Im( f( x)) \ >\ 0$


text:
# Maximal subcoalgebras of an $F+1$-coalgebra corresponding to an $F$-coalgebra

This context of this question is Rutten's Universal Coalgebra, used for modelling systems. I'm interested in finding a description of a functor between different types of coalgebras corresponding to finding a certain subcoalgebra.

An $F+1$-coalgebra $\langle S,\alpha:S\to F(S)+1\rangle$ can be thought of as a system whose transition shape is given by functor F plus the possibility of an error/termination, given by the +1. Assume that $F$ is a so-called Kripke polynomial functor: $F::=Id ~|~ B ~|~ F+F ~|~ F\times F ~|~ F^ A ~|~ \mathcal{P}_\omega F$, thus it preserves pullbacks.

A subcoalgebra of $\langle S,\alpha:S\to F(S)+1\rangle$ is coalgebra $\langle S',\alpha':S'\to F(S')+1\rangle$, where $S'\subseteq S$ and $\alpha'$ is $\alpha'$ restricted to $S'$, such that its range falls within $F(S')+1$.

I want to find the maximal subcoalgebra of this coalgebra which corresponds to an $F$-coalgebra. In terms of my application, this means I'm looking for the subset of states $S'\subseteq S$ that do not lead to the error state.

Clearly, I can take the pullback of the functions $\alpha:S\to F(S)+1$ and $inl:F(S)\to F(S)+1$ to get a set $S_0\subseteq S$ which do not lead to an error in the first step. Iterating this process for functors $F^i+1$ seems to lead to progressively smaller subsets $S_i$ of $S$ each avoiding the error state for $i$ steps. What I'm lacking is a coherent description of the process.

Is whether there is a more universal description of this construction in terms of limits or colimits, or at least, some known approaches to the problem?

I asked this question on the computer science theory overflow (here), but am re-asking it here as I received no feedback.

-
Incidentally, do you mean that Kripke polynomial functors are finitary, i.e. preserve filtered colimits, rather than preserving pullbacks? $\newcommand{\Pw}{\mathcal{P}_\omega} \Pw$ doesn't preserve pullbacks, if I'm not mistaken: the map $\Pw(2 \times_1 2) \rightarrow \Pw(2) \times_{\Pw(1)} \Pw(2)$ isn't injective, since $\{(x,y),(x',y')\}$ and $\{(x,y'),(x',y)\}$ both get sent to $(\{x,x'\},\{y,y'\})$. –  Peter LeFanu Lumsdaine Sep 8 '10 at 14:42

## 1 Answer

I think the construction you're looking for can be seen as a right adjoint, and hence the details of the construction can be seen as coming from general transfinite constructions of adjoints.

Now, if I'm understanding right, the construction you're looking at, the "error-free core" of an $F+1$ coalgebra, is the right adjoint to this.

Moreover, I think there should be theorems that show automagically why this can be computed by the construction you give, as an $\omega$-long limit of pullbacks — but I'm not sure exactly where, I'm afraid. It's almost certainly deducible from the Kelly "Unified treatment of transfinite constructions" paper, well-described by Tom Leinster here; the constructions of that have a very similar flavour.

Possibly relevant well-known constructions to compare (in Kelly and elsewhere): the construction of an algebraically-(co)free (co)monad on an endofunctor; the construction of the free $T$-algebra on a $T$-graph; the free $S$-algebra on a $T$-algebra, given a monad map $S \to T$.

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That seems to be just what the doctor ordered. –  supercooldave Sep 14 '10 at 4:49
Great! One other thing I should probably have mentioned, actually: the (proofs of the) adjoint functor theorems (well-expounded in eg Mac Lane) also give explicit constructions for adjoints, under reasonable conditions; but they give an "all-at-once" (co)limit rather than a transfinite sequence, so aren't as closely related to the construction you give as the others I mentioned might be. –  Peter LeFanu Lumsdaine Sep 14 '10 at 14:44


text:
# How do you use the quadratic formula to solve 5s^2+6x+3=0?

Jan 30, 2018

x = 0.2*(-3 + sqrt6color(red) i), (0.2 * (-3 - sqrt6color(red) i)

$x = 0.2 \cdot \left(- 3 \pm 2.4495 \textcolor{red}{i}\right)$

#### Explanation:

Given Equation : $5 {x}^{2} + 6 x + 3 = 0$

Formula for finding the routes : $\frac{- b \pm \sqrt{{b}^{2} - \left(4 a c\right)}}{2 a}$

where a, b c are the coefficients of ${x}^{2} , x \mathmr{and} c o n s t$ terms respy.

$x = \frac{- 6 \pm \sqrt{{6}^{2} - \left(4 \cdot 5 \cdot 3\right)}}{2 \cdot 5}$

$x = \frac{- 6 \pm \sqrt{- 24}}{10}$

$x = - 0.6 \pm 0.2 \sqrt{6} i$

$x = - 0.6 + 0.2 \sqrt{6} \textcolor{red}{i} , - 0.6 - 0.2 \sqrt{6} \textcolor{red}{i}$


text:
The use of Bernoulli's equation with a siphon and similar problems?

There are a variety of problems in fluid mechanics that involve the flow out off, for example, a beaker. A typical example is that of a siphon shown below:

(From wikipedia, by Tomia, CC BY 2.5)

The typical way to approach such a problem is to use Bernoulli's equation at the top of one of the beakers and e.g. within the tube. I am yet to find a satisfactory justification for its use in such a situation. I guess we are not assuming the flow is irrotational. In which case a streamline must connect the points of consideration which seems unlikely. So what is the resolution to this i.e. why can we use Bernoulli's equation in problems such as this one?

• The Bernouill differential equation? en.wikipedia.org/wiki/Bernoulli_differential_equation This is a mathematical form of equation for which it is easy to find solutions. You have to use something else to define the physical laws of the system to get the equation you want to solve, in order to then apply Bernoulli to actually solve it. If you want to know flow rate you can resolve newtonian forces due to gravity in the siphon acting acros it's cross-sectional area to get the equation. – JMLCarter Dec 26 '16 at 22:15
• @JMLCarter No I am talking about the equation given in en.wikipedia.org/wiki/… – Quantum spaghettification Dec 27 '16 at 10:32
• back to square 1, then, (find time to re-interpret the question :-( – JMLCarter Dec 27 '16 at 21:58


text:
Let G be an algebraic group acting on a scheme X. Then f: X --> Y is a categorical quotient if it is constant on G-orbits and any other G-invariant morphism factors through it in a unique fashion. We say f is a 'good' categorical quotient if:

1) f is a surjective open submersion (i.e. the topology on Y is induced from X).

2) for any open U ⊂ Y, the induced map from functions on U to G-invariant functions on f^-1(U) is an isomorphism.

Does anyone know an example of a 'bad' categorical quotient (by which I mean...well...a not good one).

-
I don't think you mean open immersion. –  David Zureick-Brown Oct 16 '09 at 21:06
Correct; it's sub- not im- –  Harold Williams Oct 16 '09 at 22:52

Note that if f: X→Y is a categorical quotient in the category of schemes which is stable under base change by open immersions, then the second condition (ie. OY→(f* OX)G is an isomorphism) is automatically satisfied.

In the paper "Examples and counterexamples for existence of categorical quotients" by A'Campo-Neuen and Hausen, there is an example of a categorical quotient f: X→A1 such that f-1(A1 - 0)→A1 - 0 is not a categorical quotient. I haven't checked but I believe this should also give an example where condition (2) fails.

I don't know of example of a categorical quotient where condition (1) fails.

-
It seems to me that you are trying to get a useful notion of quotient for group operations. Points in such a quotient $Y$ correspond to invariant closed subsets of $X$. So, if you have only a closed orbit in $X$ or if you have only one open orbit the good quotient you described is a point and there is not much useful information on $X$ that you can see there.
Once you acknowledged this, you can try to fix this. The direction choosed my Mumford is to look for invariant principal open subsets: by discarding the zero locus of some $G$-invariant homogeneous form, it is possible to construct a quotient.


text:
# Counting by fives

The time I spend doing the dishes is frequently productive. Tabitha (6 years old) or Griffin (9 years old) will often linger nearby playing and talking. Sometimes they talk to me. Sometimes they talk about math, as on a recent evening.

Tabitha (6 years old): 5, 10, 15, 20, 25, 30…

She continues until 170, at which point she becomes bored.

T: Dad! I can count by hundreds. Want to hear?

Me: Yes. Yes I do.

T: One hundred, two hundred, …

She continues to nine hundred, when she pauses. I wait a few beats to see what will happen. When nothing does, I ask.

Me: What comes next?

T: Ten hundred.

Then…

T: Hey Dad, when do you get to one thousand?

Me: Awesome. Ten hundred is one thousand. That’s the word for ten hundreds.

T: What comes next?

Me: One thousand one hundred.

T: One thousand two hundred, one thousand three hundred,…

She continues. She reaches one thousand nine hundred.

Long pause.

T: Two thousand. [She smiles broadly.]

Later that evening, she is doing her weekly homework, which involves counting by fives. It is not clear from the directions whether she is to start over for each kind of five, or whether she is to continue counting all the way down the page.

Tabitha decides to continue the counting, and this seems the more appropriate challenge for her.

She gets to 95.

T: How do you write one hundred?

Me: A one and two zeroes.

T: How do you write one hundred five?

Me: You tell me. You try, and I’ll tell you whether it’s right.

T: One-zero-five.

Me: How did you know that?

T: Replace the zero with a five. How do you write one hundred ten?

Me: What do you think?

T: One-one-zero.

# So What Do We Learn?

The first activity was about number language. Tabitha was following the patterns in number language and puzzling over the places where the patterns break down. One hundred, two hundred, three hundred…how does one thousand pop out of this sequence?

The second activity was about writing numbers (or numeration). Tabitha had no trouble counting through one hundred five, but she wasn’t sure how to write it.

A common error is to write this: 1005. The thinking goes like this: One hundred is 100, so I’ll write that, then put a 5 at the end.

# Starting the conversation

Any time a child asks if you want to hear her count, the correct answer is yes. Then listen. Listen for the challenging bits (the teens are difficult, as is anything following a ‘9’—such as 19 to 20, or 49 to 50). Be ready to help if they get stuck but also to let them think it through and try for themselves. Then talk about how they knew to do what they did. Talk about their thinking.

# Post-Halloween Math Talk

Waiting for the school bus this morning, the two adults and three children discussed last night’s Halloween events.

The neighbor girl, W (9 years old), announced that her brother, E (six years old), had gotten 90 pieces of candy for his trick-or-treating efforts. Griffin (9 years old) announced his haul of 51 pieces.

Me: Did E count each Nerd as one?

Image from Wikipedia

W: Oooo…maybe he did!

P (who is W and E’s father): We were at a house last night that had a bowl with a Take one sign. E went up, then came back and announced that he had taken three.

We told him he had to put two back.

He smiled and said, It’s a package of three!

I love this boy!

I thought for a moment about how various Halloween candies are packaged.

Me:Whoppers?

P: Yeah.

Image courtesy of Free Photo of the Day

I am not proud that I know this sort of thing. But on the rare occasion that my extensive candy knowledge is useful, I am not going to hide it either.

# So what do we learn?

We learn that there is always a follow-up question, and that the follow-up question can bring out fun stories and ideas.

The conversation could have died after E’s 90 and Griffin’s 51 pieces were announced. But I got fun stuff by asking exactly what was being counted.

We have had fun with the question of what counts as one before, when Tabitha and I talked about Eggo mini-wafflesfor example.

# Starting the conversation

North American residents probably don’t need my help here. Your children probably know yesterday’s candy count cold. Ask whether the Nerds (or Whoppers or Smarties or…) count as one piece.

If Halloween isn’t a thing where you are, keep an eye and an ear open for when your children are counting things that are packaged in groups.

# Counting fingers

## Counting fingers

A while back I met a mathematician. He is the husband of a colleague. He found my Talking Math with Your Kids project fascinating and asked repeatedly for additional examples of the conversations I have had with Griffin and Tabitha.

He referred to my work as brainwashing, using the term with great delight.

He shared a story of a young child who, when asked Do you have more fingers on your right hand, or on your left hand? responded without counting, but by matching the fingers thumb-to-thumb, index-to-index, et cetera.

The child invented one-to-one correspondence! my mathematician friend exclaimed with pleasure.

In a sense this is true.

There are things that we tell children. And there are ideas they have on their own, without knowing that anyone has had these ideas before. These really are inventions.

Children can invent more than we sometimes suppose they can.

In any case, this mathematician friend of mine was very curious to know what Tabitha would make of this story. I promised him I would ask. Here is what happened.

We were lying on the bed one evening, having just finished a book and with a few minutes left before beginning the remaining bedtime rituals.

Me: Tabitha, I want to ask you a question.

I told her that I had met a mathematician who was curious to know what she thought about something, and that this something had to do with an interesting answer that another child had once supplied to a question.

Me: The question asked of this child was, “Do you have more fingers on your left hand or on your right hand?”

Tabitha (six years old): That question doesn’t make any sense!

Me: But it’s the question that was asked.

T: But it doesn’t make any sense. Look.

[She counted the fingers first on her left hand, then on her right]

T: 1, 2, 3, 4, 5…1, 2, 3, 4, 5.

Me: So it’s the same on both hands.

T: Right, so the question doesn’t make any sense.

Me: OK. But that’s not how the child answered it. The child did this.

Above, you see what the child did originally.
Tabitha re-enacted it later for the purposes of this post. We regret any confusion.

Me: The question I want to ask you is, what do you think the child was thinking?

T: Oh, I know what she was thinking!

Me: Really?

T: Yeah. It’s the same. If they all touch it’s the same number.

Me: I wonder if that would work with toes.

Tabitha proceeded to demonstrate that it does in fact work with toes.

Me: Ha! I was thinking about comparing the fingers on one hand to the toes on one foot.

T: Well, it would be hard because the toes are all squished together.

We spend a few moments playing with our fingers and toes, trying to match them up, noting their relative cleanliness, and then we get on with the rest of our evening.

## So what do we learn?

The technique of asking what a child thinks of an idea is a powerful one. I use it in class all the time: What do you think the person was thinking who got a different answer from you? How do you think Brianna knew to do that?

Asking children to evaluate and comment on the ideas of others helps them also to think about their own thinking.

The specific idea we discussed here is that of one-to-one correspondence. We discussed this in the recent conversation about holding hands at the farmers’ market.

## Starting the conversation

This is an easy one. It doesn’t depend on your child providing an idea or knowing any particular fact of mathematics. Sometime soon, you will have a quiet moment together. Maybe it will be at the end of an all-out living room danceathon, or after reading a big pile of books. Tell your child about the mathematician’s question. Show your child the answer that so impressed the mathematician and ask, What do you think the child was thinking?

I had this same conversation with a highly precocious three-year old recently. She insisted that you needed to count the fingers in order to be sure. We had a fine time doing that. Tabitha was within earshot of the conversation with a wry smile.

# An evening ride on the A train

I was in New York City for a conference recently. I arrived at JFK airport on a Monday evening and took public transportation into the city. I observed the following scene.

At about 8:30 p.m., a mother and her (roughly) 6-year old twin girls board the train and sit quite near me.

Mom opens the first girl’s backpack to check whether her homework has been completed.

Child 1: I did it. Oh…on that one, I forgot to write the 6.

[Child 1 proceeds to talk aloud and hold up her fingers]

These are my daughter’s fingers. I did not photograph the children on the subway.

Child 1: Two and four is…1, 2, 3, 4, 5, 6!

Mom checks a couple of other things and puts the materials back in Child 1’s backpack. She opens Child 2’s backpack.

Mom: Why don’t you have a math book?

Child 2: I do! It’s Go Math!

Mom: Where is it?

Child 2: At school. Teacher says it never leaves school.

Mom: So you get worksheets?

Child 2: Uh huh.

Mom: What’s the point of having a math book, then? Oh well.

She flips through the worksheets, but does not discuss their content with the children. The worksheets appear to her (and to me) to be a disconnected jumble, which the child has completed.

About ten minutes later, the girls are acting up a little bit. Nothing major, but they are getting loud and silly. They are clearly getting on Mom’s last nerve. Mom has expressed to them how tired she is.

Another ten minutes pass, with the girls just barely keeping it together. They get off the train at their stop.

So what do we learn?

A major goal of my writing here and in my book is to help parents notice opportunities to support their children’s mathematical development, and to take advantage of these opportunities.

Traditionally as parents, we look over our children’s homework and keep an eye our for errors. We help our children when they are stuck. If everything is correct and complete, we move on. All of that is good. It is important to help our children when they are stuck. It is important to check their homework. All good.

I want to point out the opportunity this mother had when her daughter was talking about the neglected six.

Child 1 said, “I forgot to put the 6,” and then demonstrated the relevant addition problem on her fingers.

One way to seize this opportunity is to say, “Oh, good! I like that you showed how you know 2 and 4 is 6. What are some other ways you can make 6?”

That conversational technique will bear fruit with kids 95% of the time. When you notice an idea, praise the child for expressing it and ask a follow-up question, the child will nearly always answer that question. If the question suggests that there are multiple answers, the child will usually keep thinking beyond their first idea.

Continuing the conversation would provide the mother in my story with two things:

(1) Increased mathematics thinking on the part of her children, and

(2) less misbehavior later on.

Yes it takes energy to carry on a conversation of any kind with your children when everyone is tired. But it takes less energy to do that than to keep them quiet when they don’t want to be.

I understand, of course, that not everyone knows how to do this. That is what we’re working on here.

## Starting the conversation

This is an example of starting a conversation by listening and asking follow up questions.

Some conversations parents initiate. Others children initiate. When children initiate them, the conversation only moves forward if we listen and ask questions.

I will restate an important conversational move in this scenario:

Oh, good! I like that you showed how you know 2 and 4 is 6. What are some other ways you can make 6?

# Holding hands at the market

I take both kids grocery shopping pretty much every weekend, and I have since each was an infant. It’s a routine for us in which Mommy gets some quiet time around the house and I get some extended time around town with my little ones.

This time of year, the excursion includes the farmer’s market. (Which, by the way, if you are ever in St Paul on a Saturday morning, you must attend; it’s one of the best in the country for sure.)

There is a tremendous amount of construction in the area right now, so the walk from where we park is circuitous and requires sharing a short stretch of street with an occasional slow-moving automobile.

Me: Can you guys grab my hands please? A car is coming.

Tabitha (five years old): We each get a hand!

Me: Yeah. Good thing I only have two kids, huh?

T: Yeah, if there were more kids, there wouldn’t be enough hands. Like Yusef [our next-door neighbor who has three children].

Me: Oh, right. Good point. What if Natalie came too, though?

T: Then there would be an extra hand to carry a bag. You don’t have that.

Me: Right. Sorry. That means you’re going to get hit by the bag a few times. At least it’s not full yet, though.

[pause]

Me: Do we know anyone with fewer children than hands?

T: Dawn!

Me: Good. I hadn’t gotten to her yet. I was thinking about Jenn, but she has Wynne and Emmett; and I was thinking about Addie, but she has August and Leo. Then I thought about Jimmy, but he has Leila and Otis. I hadn’t thought about Dawn yet. She just has Mateo, doesn’t she?

T: Yeah. So she would have an extra hand for a bag.

Number of children = number of hands

## So What Do We Learn?

An important thing to notice here is that there is only one number word in this whole conversation. I say the word two. That’s it.

The rest is about whether Set A (children) has more or fewer members than Set B (parental hands available for holding). This is a remarkably sophisticated idea. The fancy math term for what we are talking about here is one-to-one correspondence. It refers to the fact that when two collections (A and B) have the same number of things, we can match them up; one thing from set A and one from B, with no leftovers in either collection.

The mind-blowing part of one-to-one correspondence is that it’s true the other way around. If we can match up with no leftovers, then the sets are the same size. Even if we don’t know how big either set is. That is what this conversation works with—comparing sizes of sets without stating the size of either one.

I am quite certain that Tabitha pictured Yusef holding the hands of two children, leaving one who held Natalie’s hand. That left one hand unheld, available for a bag. I do not think (although it’s possible) that she thought 4 hands minus 3 children leaves one.

She matched kids to hands in her mind. One-to-one correspondence.

## Starting the Conversation

Listen, and notice when your child is comparing two quantities. maybe they are equal (as in this case), maybe they are not.

As in this conversation, you don’t need to discuss actual numbers to compare two quantities. More, fewer, same…there are many times your children use these words. Follow up with some what if questions and see where they lead.

## Beyond the Conversation

One-to-one correspondence, and its implication that we can compare two sets without counting either one, was the idea that Georg Cantor exploited in the late 1800’s to prove that some infinities are “bigger” than others. Cantor demonstrated that there are just as many whole numbers as fractions (because we can cleverly match them up in one-to-one correspondence), but more real numbers than fractions (because it is impossible to match them up in one-to-one correspondence; any attempt you might make will leave out some real numbers).

Five year olds do not need to know this. Nor do you, probably. Middle school kids will find it fascinating. Cantor’s argument is accessible to most high schoolers (but would never occur to them, nor to me—it’s a brilliant insight).

Marilyn vos Savant is wrong when she writes, “Math doesn’t enlighten us the way literature, social studies, or art appreciation do.”

# Learning to count

I am fascinated by watching children learn to count. There are many surprising twists and turns kids take along the way.

Even more surprising, perhaps, is that what seem like crazy mistakes to us adults are completely sensible attempts at getting it right for kids.

For example…

• In English, the pattern that occurs in the teens is complicated. “Thirteen” doesn’t sound very much like what it is: three plus ten, while “Fourteen” does.
• Likewise, the names for the “decades”: “Twenty” means two tens and “Thirty” means three tens.
• But once you get to twenty-one, the pattern is regular until twenty-nine.
• We start counting at 1 (not 0), but we don’t start the decades at 21 or 31, so kids following the 1, 2, 3 pattern will often skip 20 and 30.

If you put all of this together, you might expect a typical young child who is counting “as high as I can” to:

1. Have trouble in the teens
2. Skip 20 in favor of 21
3. Have more success in the twenties than in the teens, and
4. End the count at or about twenty-nine (since the word thirty is not very predictable from the previous language patterns).

Here goes…

# More fun with board games [Reports from the field]

Of particular interest is his observation of the difference between how we structure hundreds grids in classrooms and how the Chutes and Ladders board is structured (each, notice with a ten-by-ten grid). I had not thought about that before and now have something new to play with.

Standard hundreds grid. Left-to-right, top-to-bottom.

Chutes and ladders board. Snaking back-and-forth from bottom to top.

Oh. Here is an important thing about Talking Math with Your Kids: When you notice differences like these, see them as opportunities to talk, ask questions, wonder and even to argue good-naturedly. It is not a problem that these grids are set up differently; it is an opportunity.

Here is rjbrow‘s report. Enjoy.

I try to play games whenever I can with my kids (ages 7, 5, and 2). Great practice with turn taking, understanding rules, making decisions etc. But like you mentioned here, games involving numbers provide a great opportunity to talk math with your kids.

I’m glad you mentioned Chutes and Ladders here too. I had written this game off because in playing it with my two older boys, frustration would ensue by the abundance of chutes that would prolong the game and sabotage their progress. After reading through your Uncle Wiggly post, I played a game of Chutes and Ladders with my 5 year old yesterday and found a couple of really nice opportunities to talk numbers.

First, each time he traveled a chute or a ladder, we had the conversation of which way to go on the new row. For example, he’d land on space 51 and travel the ladder up to space 67. So we’d talk about whether his guy should be traveling to the left or to the right. Jake figured out that if he was on space 67 he should be pointing his guy toward the 68 since it was bigger than 66. We’d also practice reading the numbers out loud. Like you mentioned, Jake can count to 100 but recognizing the written numbers is not necessarily the case. This was good practice.

You also mention that learning to count can be messy. We have a write-on number grid at home that counts 1 to 100. It’s really nice at looking at number patterns when you count by 2′s, 5′s, 10′s, etc. The numbers on this grid wrap to the next row so that all the like digits in the ones place line up vertically. Very nice. We’ve done some counting and pattern recognition using this grid. The difference with Chutes and Ladders is that the numbers wind back and forth on the way up the board. This was a shift to the counting we had done on the other grid. While this provided some nice conversation about how to work our way through this board, it did point out some messiness in how we present elementary number concepts to kids.

I hope my description makes sense. I have some pictures of our activity here:

Now who has a Chutes and Ladders board I can borrow to play with Tabitha and Griffin?

# Uncle Wiggily

Tabitha was $3\frac{1}{2}$  years old when we were playing a game of Uncle Wiggily.

In case you are not familiar with the game, I’ll briefly describe it. Uncle Wiggily is a board game with 100 spaces along a twisty path. Players draw cards; each card has a number and a brief poem. Perils and bonuses are judiciously spaced along the path. Uncle Wiggily is approximately 10% more complicated than Candy Land (which is to say, not very complicated at all!)

Tabitha: (Drawing a card for her first turn-it’s an 8) Got one Daddy!

Me: Mmm-hmmm.

T: What is it?

Me: Can you guess? Look closely.

T: (Quickly and with a big, eager smile on her face) Six!

Me: Good guess. It’s eight.

T: Oh!

Me: Can you count to eight?

T: (Bouncing her piece along the path, ending near the henhouse on the farm-themed board) One, two, three, four, five, six, seven, eight. By the cluck-cluck house!

Me: My turn. (Drawing a card-it’s a 10) What card did I choose?

T: Ten!

Me: Good. (Testing a hypothesis, I skip eight as I count) One, two, three, four, five, six, seven, nine, ten.

T: (Oblivious) My turn.

# So what do we learn?

Learning to count is messy. Many things we might expect to be true about how children learn to count are not true at all.

We might expect children to learn the numerals (8) at the same time that they learn the words (eight). They do not. Notice that Tabitha counted flawlessly to eight, but did not recognize the symbol “8”.

We might expect children to learn the numerals in order, with all multi-digit numbers coming only after mastering the single-digit numbers. They do not. Tabitha recognized “10” but not “8”.

When I counted to ten, I intentionally left out eight to see whether she would notice. She did not. She could count to eight, but didn’t notice when it got left out on the way to ten. Mathematics is logical and orderly. The ways children learn mathematics are not.

This conversation came from a short video I made one day. I watched this video a year and a half later, when Tabitha was 5. After watching it, I immediately went into the kitchen where she was having a snack and counted: 1, 2, 3, 4, 5, 6, 7, 9, 10. She smiled and asked Why did you do that? (referring to counting in her ear). Then, a moment later, she said, Hey! You skipped eight!

# Starting the conversation

Play games that involve numbers. Uncle Wiggily is great. So is Chutes and Ladders. Or Hopscotch. Any game that involves counting and reading numerals will give you the chance to practice these early number ideas.

While you’re playing, ask your child what number he drew, and what number you drew. If he doesn’t know, have him guess. Don’t worry about precision or correctness. Model good counting for your child. Help him count out some of his turns and let him count incorrectly on others. Have fun and don’t worry too much if he gets bored before the game is finished.

Have fun with it. Whatever you need to do to stay engaged in a couple of rounds of Uncle Wiggily is worth the effort. You can see the effort I invested in keeping myself entertained; I formulated a hypothesis about whether she would notice my own incorrect counting and tested that hypothesis.

Don’t get carried away with the hypothesis testing, though. Children do need models of correct counting. They won’t be damaged by a few experiments, of course. But you don’t want to become an unreliable source of knowledge.

# The unit is the thing that you count

Griffin (eight years old) and Tabitha (five years old) were discussing the day’s activities. The feature activity had been making brownies with Mommy. This occurred while Griffin was out of the house.

Griffin: How many brownies did you make?

Tabitha: One big one! Mommy cut it up.

# So What Do We Learn?

What makes this more than just a funny story is that Griffin and Tabitha are clearly counting different things. They are talking about different units.

When we make cookies, everyone agrees on the unit; we know what one cookie is.

But brownies are different. Tabitha seems to think that a brownie is the thing that comes out of the oven. Griffin seems to think that a brownie is what you eat in one serving.

One brownie according to Tabitha.

One brownie, according to Griffin

I have emphasized elsewhere the importance of the unit; that one is a more flexible concept than we might think.

Fun follow-up question: Does the thing in this video count as one brownie?

# Starting the Conversation

Anytime there are things in groups—or things being cut—is a good time to talk about units.

Grocery stores usually have express lanes where you have to have Ten items or fewerAsk your child whether someone with a dozen eggs could use that lane. What about someone with 12 apples in a bag? What if the apples are loose?

When your child asks for two slices of pizza, take one slice, cut it down the middle, smile wryly and ask whether that’s OK.

In all of these cases, the central question is What counts as one? Play with that question.

Also, watch that video together. It’s a ton of fun.


text:
Markov_Based_Methods

# Spatially Explicit Markov Methods

Author: Serge Rey sjsrey@gmail.com, Wei Kang weikang9009@gmail.com

## Introduction

This notebook introduces Discrete Markov Chains (DMC) model and its two variants which explicitly incorporate spatial effects. We will demonstrate the usage of these methods by an empirical study for understanding regional income dynamics in the US. The dataset is the per capita incomes observed annually from 1929 to 2009 for the lower 48 US states.

Note that a full execution of this notebook requires pandas, matplotlib and light-weight geovisualization package pysal-splot.

### Classic Markov

giddy.markov.Markov(self, class_ids, classes=None)

We start with a look at a simple example of classic DMC methods implemented in PySAL-giddy. A Markov chain may be in one of $k$ different states/classes at any point in time. These states are exhaustive and mutually exclusive. If one had a time series of remote sensing images used to develop land use classifications, then the states could be defined as the specific land use classes and interest would center on the transitions in and out of different classes for each pixel.

For example, suppose there are 5 pixels, each of which takes on one of 3 states (a,b,c) at 3 consecutive periods:

import numpy as np
c = np.array([['b','a','c'],['c','c','a'],['c','b','c'],['a','a','b'],['a','b','c']])

So the first pixel was in state ‘b’ in period 1, state ‘a’ in period 2, and state ‘c’ in period 3. Each pixel's trajectory (row) owns Markov property, meaning that which state a pixel takes on today is only dependent on its immediate past.

Let's suppose that all the 5 pixels are governed by the same transition dynamics rule. That is, each trajectory is a realization of a Discrete Markov Chain process. We could pool all the 5 trajectories from which to estimate a transition probability matrix. To do that, we utlize the Markov class in giddy:

import giddy
m = giddy.markov.Markov(c)

In this way, we create a Markov instance - $m$. Its attribute $classes$ gives 3 unique classes these pixels can take on, which are 'a','b' and 'c'.

print(m.classes)

['a' 'b' 'c']

print(len(m.classes))

3

In addition to extracting the unique states as an attribute, our Markov instance will also have the attribute trnasitions which is a transition matrix counting the number of transitions from one state to another. Since there are 3 unique states, we will have a $(3,3)$ transtion matrix:

print(m.transitions)

[[1. 2. 1.]
[1. 0. 2.]
[1. 1. 1.]]

The above transition matrix indicates that of the four pixels that began a transition interval in state ‘a’, 1 remained in that state, 2 transitioned to state ‘b’ and 1 transitioned to state ‘c’. Another attribute $p$ gives the transtion probability matrix which is the transition dynamics rule ubiquitous to all the 5 pixels across the 3 periods. The maximum likehood estimator for each element $p_{i,j}$ is shown below where $n_{i,j}$ is the number of transitions from state $i$ to state $j$ and $k$ is the number of states (here $k=3$):

$$\hat{p}_{i,j} = \frac{n_{i,j}}{\sum_{q=1}^k n_{i,q} }$$
print(m.p)

[[0.25       0.5        0.25      ]
[0.33333333 0.         0.66666667]
[0.33333333 0.33333333 0.33333333]]

This means that if any of the 5 pixels was in state 'c', the probability of staying at 'c' or transitioning to any other states ('a', 'b') in the next period is the same (0.333). If a pixel was in state 'b', there is a high possibility that it would take on state 'c' in the next period because $\hat{p}_{2,3}=0.667$.

m.steady_state  # steady state distribution

array([0.30769231, 0.28846154, 0.40384615])

This simple example illustrates the basic creation of a Markov instance, but the small sample size makes it unrealistic for the more advanced features of this approach. For a larger example, we will look at an application of Markov methods to understanding regional income dynamics in the US. Here we will load in data on per capita incomes observed annually from 1929 to 2010 for the lower 48 US states:

#### Regional income dynamics in the US

Firstly, we load in data on per capita incomes observed annually from 1929 to 2009 for the lower 48 US states. We use the example dataset in libpysal which was downloaded from US Bureau of Economic Analysis.

import libpysal
f = libpysal.io.open(libpysal.examples.get_path("usjoin.csv"))
pci = np.array([f.by_col[str(y)] for y in range(1929,2010)])
print(pci.shape)

(81, 48)

The first row of the array is the per capita incomes for the 48 US states for the year 1929:

print(pci[0, :])

[ 323  600  310  991  634 1024 1032  518  347  507  948  607  581  532
393  414  601  768  906  790  599  286  621  592  596  868  686  918
410 1152  332  382  771  455  668  772  874  271  426  378  479  551
634  434  741  460  673  675]

In order to apply the classic Markov approach to this series, we first have to discretize the distribution by defining our classes. There are many ways to do this including quantiles classification scheme, equal interval classification scheme, Fisher Jenks classification scheme, etc. For a list of classification methods, please refer to the pysal package mapclassify.

Here we will use the quintiles for each annual income distribution to define the classes. It should be noted that using quintiles for the pooled income distribution to define the classes will result in a different interpretation of the income dynamics. Quintiles for each annual income distribution (the former) will reveal more of relative income dynamics while those for the pooled income distribution (the latter) will provide insights in absolute dynamics.

import matplotlib.pyplot as plt
%matplotlib inline
years = range(1929,2010)
names = np.array(f.by_col("Name"))
order1929 = np.argsort(pci[0,:])
order2009 = np.argsort(pci[-1,:])
names1929 = names[order1929[::-1]]
names2009 = names[order2009[::-1]]
first_last = np.vstack((names1929,names2009))
from pylab import rcParams
rcParams['figure.figsize'] = 15,10
plt.plot(years,pci)
for i in range(48):
plt.text(1915,54530-(i*1159), first_last[0][i],fontsize=12)
plt.text(2010.5,54530-(i*1159), first_last[1][i],fontsize=12)
plt.xlim((years[0], years[-1]))
plt.ylim((0, 54530))
plt.ylabel(r"$y_{i,t}$",fontsize=14)
plt.xlabel('Years',fontsize=12)
plt.title('Absolute Dynamics',fontsize=18)

Text(0.5,1,'Absolute Dynamics')
years = range(1929,2010)
rpci= (pci.T / pci.mean(axis=1)).T
names = np.array(f.by_col("Name"))
order1929 = np.argsort(rpci[0,:])
order2009 = np.argsort(rpci[-1,:])
names1929 = names[order1929[::-1]]
names2009 = names[order2009[::-1]]
first_last = np.vstack((names1929,names2009))
from pylab import rcParams
rcParams['figure.figsize'] = 15,10
plt.plot(years,rpci)
for i in range(48):
plt.text(1915,1.91-(i*0.041), first_last[0][i],fontsize=12)
plt.text(2010.5,1.91-(i*0.041), first_last[1][i],fontsize=12)
plt.xlim((years[0], years[-1]))
plt.ylim((0, 1.94))
plt.ylabel(r"$y_{i,t}/\bar{y}_t$",fontsize=14)
plt.xlabel('Years',fontsize=12)
plt.title('Relative Dynamics',fontsize=18)

Text(0.5,1,'Relative Dynamics')
import mapclassify as mc
q5 = np.array([mc.Quantiles(y,k=5).yb for y in pci]).transpose()
print(q5[:, 0])

[0 2 0 4 2 4 4 1 0 1 4 2 2 1 0 1 2 3 4 4 2 0 2 2 2 4 3 4 0 4 0 0 3 1 3 3 4
0 1 0 1 2 2 1 3 1 3 3]

/Users/weikang/anaconda3/lib/python3.6/site-packages/scipy/stats/stats.py:1713: FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated; use arr[tuple(seq)] instead of arr[seq]. In the future this will be interpreted as an array index, arr[np.array(seq)], which will result either in an error or a different result.
return np.add.reduce(sorted[indexer] * weights, axis=axis) / sumval

print(f.by_col("Name"))

['Alabama', 'Arizona', 'Arkansas', 'California', 'Colorado', 'Connecticut', 'Delaware', 'Florida', 'Georgia', 'Idaho', 'Illinois', 'Indiana', 'Iowa', 'Kansas', 'Kentucky', 'Louisiana', 'Maine', 'Maryland', 'Massachusetts', 'Michigan', 'Minnesota', 'Mississippi', 'Missouri', 'Montana', 'Nebraska', 'Nevada', 'New Hampshire', 'New Jersey', 'New Mexico', 'New York', 'North Carolina', 'North Dakota', 'Ohio', 'Oklahoma', 'Oregon', 'Pennsylvania', 'Rhode Island', 'South Carolina', 'South Dakota', 'Tennessee', 'Texas', 'Utah', 'Vermont', 'Virginia', 'Washington', 'West Virginia', 'Wisconsin', 'Wyoming']

A number of things need to be noted here. First, we are relying on the classification methods in mapclassify for defining our quintiles. The class Quantiles uses quintiles ($k=5$) as the default and will create an instance of this class that has multiple attributes, the one we are extracting in the first line is $yb$ - the class id for each observation. The second thing to note is the transpose operator which gets our resulting array $q5$ in the proper structure required for use of Markov. Thus we see that the first spatial unit (Alabama with an income of 323) fell in the first quintile in 1929, while the last unit (Wyoming with an income of 675) fell in the fourth quintile.

So now we have a time series for each state of its quintile membership. For example, Colorado’s quintile time series is:

print(q5[4, :])

[2 3 2 2 3 2 2 3 2 2 2 2 2 2 2 2 3 2 3 2 3 2 3 3 3 2 2 3 3 3 3 3 3 3 3 3 3
2 2 2 3 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4
3 3 3 4 3 3 3]

indicating that it has occupied the 3rd, 4th and 5th quintiles in the distribution at the first 3 periods. To summarize the transition dynamics for all units, we instantiate a Markov object:

m5 = giddy.markov.Markov(q5)

The number of transitions between any two quintile classes could be counted:

print(m5.transitions)

[[729.  71.   1.   0.   0.]
[ 72. 567.  80.   3.   0.]
[  0.  81. 631.  86.   2.]
[  0.   3.  86. 573.  56.]
[  0.   0.   1.  57. 741.]]

By assuming the first-order Markov property, time homogeneity, spatial homogeneity and spatial independence, a transition probability matrix could be estimated which holds for all the 48 US states across 1929-2010:

print(m5.p)

[[0.91011236 0.0886392  0.00124844 0.         0.        ]
[0.09972299 0.78531856 0.11080332 0.00415512 0.        ]
[0.         0.10125    0.78875    0.1075     0.0025    ]
[0.         0.00417827 0.11977716 0.79805014 0.07799443]
[0.         0.         0.00125156 0.07133917 0.92740926]]

The fact that each of the 5 diagonal elements is larger than $0.78$ indicates a high stability of US regional income dynamics system.

Another very important feature of DMC model is the steady state distribution $\pi$ (also called limiting distribution) defined as $\pi p = \pi$. The attribute $steady\_state$ gives $\pi$ as follows:

print(m5.steady_state)

[0.20774716 0.18725774 0.20740537 0.18821787 0.20937187]

If the distribution at $t$ is a steady state distribution as shown above, then any distribution afterwards is the same distribution.

With the transition probability matrix in hand, we can estimate the first mean passage time which is the average number of steps to go from a state/class to another state for the first time:

print(giddy.ergodic.fmpt(m5.p))

[[  4.81354357  11.50292712  29.60921231  53.38594954 103.59816743]
[ 42.04774505   5.34023324  18.74455332  42.50023268  92.71316899]
[ 69.25849753  27.21075248   4.82147603  25.27184624  75.43305672]
[ 84.90689329  42.85914824  17.18082642   5.31299186  51.60953369]
[ 98.41295543  56.36521038  30.66046735  14.21158356   4.77619083]]

Thus, for a state with income in the first quintile, it takes on average 11.5 years for it to first enter the second quintile, 29.6 to get to the third quintile, 53.4 years to enter the fourth, and 103.6 years to reach the richest quintile.

#### Regional context and Moran's Is

Thus far we have treated all the spatial units as independent to estimate the transition probabilities. This hides an implicit assumption: the movement of a spatial unit in the income distribution is independent of the movement of its neighbors or the position of the neighbors in the distribution. But what if spatial context matters??

We could plot the choropleth maps of per capita incomes of US states to get a first impression of the spatial distribution.

import geopandas as gpd
import pandas as pd

geo_table = gpd.read_file(libpysal.examples.get_path('us48.shp'))
complete_table = geo_table.merge(income_table,left_on='STATE_NAME',right_on='Name')

AREA PERIMETER STATE_ STATE_ID STATE_NAME STATE_FIPS_x SUB_REGION STATE_ABBR geometry Name ... 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009
0 20.750 34.956 1 1 Washington 53 Pacific WA (POLYGON ((-122.400749206543 48.22539520263672... Washington ... 31528 32053 32206 32934 34984 35738 38477 40782 41588 40619
1 45.132 34.527 2 2 Montana 30 Mtn MT POLYGON ((-111.4746322631836 44.70223999023438... Montana ... 22569 24342 24699 25963 27517 28987 30942 32625 33293 32699
2 9.571 18.899 3 3 Maine 23 N Eng ME (POLYGON ((-69.77778625488281 44.0740737915039... Maine ... 25623 27068 27731 28727 30201 30721 32340 33620 34906 35268
3 21.874 21.353 4 4 North Dakota 38 W N Cen ND POLYGON ((-98.73005676269531 45.93829727172852... North Dakota ... 25068 26118 26770 29109 29676 31644 32856 35882 39009 38672
4 22.598 22.746 5 5 South Dakota 46 W N Cen SD POLYGON ((-102.7879333496094 42.99532318115234... South Dakota ... 26115 27531 27727 30072 31765 32726 33320 35998 38188 36499

5 rows × 92 columns

index_year = range(1929,2010,15)
fig, axes = plt.subplots(nrows=2, ncols=3,figsize = (15,7))
for i in range(2):
for j in range(3):
ax = axes[i,j]
complete_table.plot(ax=ax, column=str(index_year[i*3+j]), cmap='OrRd', scheme='quantiles', legend=True)
ax.set_title('Per Capita Income %s Quintiles'%str(index_year[i*3+j]))
ax.axis('off')
leg = ax.get_legend()
leg.set_bbox_to_anchor((0.8, 0.15, 0.16, 0.2))
plt.tight_layout()

/Users/weikang/anaconda3/lib/python3.6/site-packages/pysal/__init__.py:65: VisibleDeprecationWarning: PySAL's API will be changed on 2018-12-31. The last release made with this API is version 1.14.4. A preview of the next API version is provided in the pysal 2.0 prelease candidate. The API changes and a guide on how to change imports is provided at https://pysal.org/about
), VisibleDeprecationWarning)
/Users/weikang/anaconda3/lib/python3.6/site-packages/scipy/stats/stats.py:1713: FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated; use arr[tuple(seq)] instead of arr[seq]. In the future this will be interpreted as an array index, arr[np.array(seq)], which will result either in an error or a different result.
return np.add.reduce(sorted[indexer] * weights, axis=axis) / sumval

It is quite obvious that the per capita incomes are not randomly distributed: we could spot clusters in the mid-south, south-east and north-east. Let's proceed to calculate Moran's I, a widely used measure of global spatial autocorrelation, to aid the visual interpretation.

from esda.moran import Moran
import matplotlib.pyplot as plt
%matplotlib inline
w.transform = 'R'
mits = [Moran(cs, w) for cs in pci]
res = np.array([(mi.I, mi.EI, mi.seI_norm, mi.sim[974]) for mi in mits])
years = np.arange(1929,2010)
fig, ax = plt.subplots(nrows=1, ncols=1,figsize = (10,5) )
ax.plot(years, res[:,0], label='Moran\'s I')
#plot(years, res[:,1], label='E[I]')
ax.plot(years, res[:,1]+1.96*res[:,2], label='Upper bound',linestyle='dashed')
ax.plot(years, res[:,1]-1.96*res[:,2], label='Lower bound',linestyle='dashed')
ax.set_title("Global spatial autocorrelation for annual US per capita incomes",fontdict={'fontsize':15})
ax.set_xlim([1929,2009])
ax.legend()

<matplotlib.legend.Legend at 0x1a27ef4390>

From the above figure we could observe that Moran's I value was always positive and significant for each year across 1929-2009. In other words, US regional income series are not independent of each other and regional context could be important in shaping the regional income dynamics. However, the classic Markov approach is silent on this issue. We turn to the spatially explict Markov methods - Spatial Markov and LISA Markov - for an explicit incorporation of space in understanding US regional income distribution dynamics.

### Spatial Markov

giddy.markov.Spatial_Markov(y, w, k=4, m=4, permutations=0, fixed=True, discrete=False, cutoffs=None, lag_cutoffs=None, variable_name=None)

Spatial Markov is an extension to class Markov allowing for a more comprehensive analysis of the spatial dimensions of the transitional dynamics (Rey, 2001). Here, whether the transition probabilities are dependent on regional context is investigated and quantified. Rather than estimating one transition probability matrix, spatial Markov requires estimation of $k$ transition probability matrices, each of which is conditional on the regional context at the preceding period. The regional context is usually formalized by spatial lag - the weighted average income level of neighbors:

$$z_{r,t} = \sum_{s=1}^{n} w_{r,s} y_{s,t}$$

where $W$ is the spatial weight matrix and $w_{r,s}$ represents the weight that spatial unit $s$ contributes to the local context of spatial unit $r$ at time period $t$.

Similar to the construction of a Markov instance, we could create a Spatial Markov instance by utilizing the $Spatial\_Markov$ class in giddy. The only difference between the adoption of $Markov$ and $Spatial\_Markov$ class is that the latter accepts the original continuous income data while the former requires a pre-classification/discretization. In other words, here we do not need to apply the classification methods in mapclassify as we did earlier. In fact, the Spatial Markov class nested the quantile classification methods and all we need to do is set the desired number of classes $k$ when creating the $Spatial\_Markov$ instance. Here, we set $k=5$ (quintile classes) as before.

Different from before, quintiles are defined for the pooled relative incomes (by standardizing by each period by the mean). This is achieved by setting the parameter $fixed$ as True.

giddy.markov.Spatial_Markov?

sm = giddy.markov.Spatial_Markov(rpci.T, w, fixed = True, k = 5,m=5) # spatial_markov instance o

/Users/weikang/anaconda3/lib/python3.6/site-packages/scipy/stats/stats.py:1713: FutureWarning: Using a non-tuple sequence for multidimensional indexing is deprecated; use arr[tuple(seq)] instead of arr[seq]. In the future this will be interpreted as an array index, arr[np.array(seq)], which will result either in an error or a different result.
return np.add.reduce(sorted[indexer] * weights, axis=axis) / sumval

We can next examine the global transition probability matrix for relative incomes.

print(sm.p)

[[0.91461837 0.07503234 0.00905563 0.00129366 0.        ]
[0.06570302 0.82654402 0.10512484 0.00131406 0.00131406]
[0.00520833 0.10286458 0.79427083 0.09505208 0.00260417]
[0.         0.00913838 0.09399478 0.84856397 0.04830287]
[0.         0.         0.         0.06217617 0.93782383]]

The Spatial Markov allows us to compare the global transition dynamics to those conditioned on regional context. More specifically, the transition dynamics are split across economies who have spatial lags in different quintiles at the preceding year. In our example we have 5 classes, so 5 different conditioned transition probability matrices are estimated - P(LAG0), P(LAG1), P(LAG2), P(LAG3), and P(LAG4).

sm.summary()

--------------------------------------------------------------
Spatial Markov Test
--------------------------------------------------------------
Number of classes: 5
Number of transitions: 3840
Number of regimes: 5
Regime names: LAG0, LAG1, LAG2, LAG3, LAG4
--------------------------------------------------------------
Test                   LR                Chi-2
Stat.              170.659              200.624
DOF                   60                   60
p-value                0.000                0.000
--------------------------------------------------------------
P(H0)           C0         C1         C2         C3         C4
C0      0.915      0.075      0.009      0.001      0.000
C1      0.066      0.827      0.105      0.001      0.001
C2      0.005      0.103      0.794      0.095      0.003
C3      0.000      0.009      0.094      0.849      0.048
C4      0.000      0.000      0.000      0.062      0.938
--------------------------------------------------------------
P(LAG0)         C0         C1         C2         C3         C4
C0      0.963      0.030      0.006      0.000      0.000
C1      0.060      0.832      0.107      0.000      0.000
C2      0.000      0.140      0.740      0.120      0.000
C3      0.000      0.036      0.321      0.571      0.071
C4      0.000      0.000      0.000      0.167      0.833
--------------------------------------------------------------
P(LAG1)         C0         C1         C2         C3         C4
C0      0.798      0.168      0.034      0.000      0.000
C1      0.075      0.882      0.042      0.000      0.000
C2      0.005      0.070      0.866      0.059      0.000
C3      0.000      0.000      0.064      0.902      0.034
C4      0.000      0.000      0.000      0.194      0.806
--------------------------------------------------------------
P(LAG2)         C0         C1         C2         C3         C4
C0      0.847      0.153      0.000      0.000      0.000
C1      0.081      0.789      0.129      0.000      0.000
C2      0.005      0.098      0.793      0.098      0.005
C3      0.000      0.000      0.094      0.871      0.035
C4      0.000      0.000      0.000      0.102      0.898
--------------------------------------------------------------
P(LAG3)         C0         C1         C2         C3         C4
C0      0.885      0.098      0.000      0.016      0.000
C1      0.039      0.814      0.140      0.000      0.008
C2      0.005      0.094      0.777      0.119      0.005
C3      0.000      0.023      0.129      0.754      0.094
C4      0.000      0.000      0.000      0.097      0.903
--------------------------------------------------------------
P(LAG4)         C0         C1         C2         C3         C4
C0      0.333      0.667      0.000      0.000      0.000
C1      0.048      0.774      0.161      0.016      0.000
C2      0.011      0.161      0.747      0.080      0.000
C3      0.000      0.010      0.062      0.896      0.031
C4      0.000      0.000      0.000      0.024      0.976
--------------------------------------------------------------

#### Visualize the (spatial) Markov transition probability matrix

fig, ax = plt.subplots(figsize = (5,5))
im = ax.imshow(sm.p,cmap = "coolwarm",vmin=0, vmax=1)
# Loop over data dimensions and create text annotations.
for i in range(len(sm.p)):
for j in range(len(sm.p)):
text = ax.text(j, i, round(sm.p[i, j], 2),
ha="center", va="center", color="w")
ax.figure.colorbar(im, ax=ax)

<matplotlib.colorbar.Colorbar at 0x1a2836c588>
fig, axes = plt.subplots(2,3,figsize = (15,10))

for i in range(2):
for j in range(3):
ax = axes[i,j]
if i==1 and j==2:
ax.axis('off')
continue
# Loop over data dimensions and create text annotations.
p_temp = sm.P[i*3+j]
for x in range(len(p_temp)):
for y in range(len(p_temp)):
text = ax.text(y, x, round(p_temp[x, y], 2),
ha="center", va="center", color="w")
im = ax.imshow(p_temp,cmap = "coolwarm",vmin=0, vmax=1)
ax.set_title("Spatial Lag %d"%(i*3+j),fontsize=18)
cbar_ax = fig.add_axes([0.95, 0.228, 0.01, 0.5])
fig.colorbar(im, cax=cbar_ax)

<matplotlib.colorbar.Colorbar at 0x1a286705f8>
fig, axes = plt.subplots(2,3,figsize = (15,10))
for i in range(2):
for j in range(3):
ax = axes[i,j]
if i==0 and j==0:
p_temp = sm.p
im = ax.imshow(p_temp,cmap = "coolwarm",vmin=0, vmax=1)
ax.set_title("Pooled",fontsize=18)
else:
p_temp = sm.P[i*3+j-1]
im = ax.imshow(p_temp,cmap = "coolwarm",vmin=0, vmax=1)
ax.set_title("Spatial Lag %d"%(i*3+j),fontsize=18)
for x in range(len(p_temp)):
for y in range(len(p_temp)):
text = ax.text(y, x, round(p_temp[x, y], 2),
ha="center", va="center", color="w")

cbar_ax = fig.add_axes([0.95, 0.228, 0.01, 0.5])
fig.colorbar(im, cax=cbar_ax)
#fig.savefig('spatial_markov_us.png', dpi = 300)

<matplotlib.colorbar.Colorbar at 0x1a28f9d198>

The probability of a poor state remaining poor is 0.963 if their neighbors are in the 1st quintile and 0.798 if their neighbors are in the 2nd quintile. The probability of a rich economy remaining rich is 0.977 if their neighbors are in the 5th quintile, but if their neighbors are in the 4th quintile this drops to 0.903.

We can also explore the different steady state distributions implied by these different transition probabilities:

print(sm.S)

[[0.43509425 0.2635327  0.20363044 0.06841983 0.02932278]
[0.13391287 0.33993305 0.25153036 0.23343016 0.04119356]
[0.12124869 0.21137444 0.2635101  0.29013417 0.1137326 ]
[0.0776413  0.19748806 0.25352636 0.22480415 0.24654013]
[0.01776781 0.19964349 0.19009833 0.25524697 0.3372434 ]]

The long run distribution for states with poor (rich) neighbors has 0.435 (0.018) of the values in the first quintile, 0.263 (0.200) in the second quintile, 0.204 (0.190) in the third, 0.0684 (0.255) in the fourth and 0.029 (0.337) in the fifth quintile. And, finally the spatially conditional first mean passage times:

print(sm.F)

[[[  2.29835259  28.95614035  46.14285714  80.80952381 279.42857143]
[ 33.86549708   3.79459555  22.57142857  57.23809524 255.85714286]
[ 43.60233918   9.73684211   4.91085714  34.66666667 233.28571429]
[ 46.62865497  12.76315789   6.25714286  14.61564626 198.61904762]
[ 52.62865497  18.76315789  12.25714286   6.          34.1031746 ]]

[[  7.46754205   9.70574606  25.76785714  74.53116883 194.23446197]
[ 27.76691978   2.94175577  24.97142857  73.73474026 193.4380334 ]
[ 53.57477715  28.48447637   3.97566318  48.76331169 168.46660482]
[ 72.03631562  46.94601483  18.46153846   4.28393653 119.70329314]
[ 77.17917276  52.08887197  23.6043956    5.14285714  24.27564033]]

[[  8.24751154   6.53333333  18.38765432  40.70864198 112.76732026]
[ 47.35040872   4.73094099  11.85432099  34.17530864 106.23398693]
[ 69.42288828  24.76666667   3.794921    22.32098765  94.37966594]
[ 83.72288828  39.06666667  14.3          3.44668119  76.36702977]
[ 93.52288828  48.86666667  24.1          9.8          8.79255406]]

[[ 12.87974382  13.34847151  19.83446328  28.47257282  55.82395142]
[ 99.46114206   5.06359731  10.54545198  23.05133495  49.68944423]
[117.76777159  23.03735526   3.94436301  15.0843986   43.57927247]
[127.89752089  32.4393006   14.56853107   4.44831643  31.63099455]
[138.24752089  42.7893006   24.91853107  10.35         4.05613474]]

[[ 56.2815534    1.5         10.57236842  27.02173913 110.54347826]
[ 82.9223301    5.00892857   9.07236842  25.52173913 109.04347826]
[ 97.17718447  19.53125      5.26043557  21.42391304 104.94565217]
[127.1407767   48.74107143  33.29605263   3.91777427  83.52173913]
[169.6407767   91.24107143  75.79605263  42.5          2.96521739]]]

States in the first income quintile with neighbors in the first quintile return to the first quintile after 2.298 years, after leaving the first quintile. They enter the fourth quintile 80.810 years after leaving the first quintile, on average. Poor states within neighbors in the fourth quintile return to the first quintile, on average, after 12.88 years, and would enter the fourth quintile after 28.473 years.

Tests for this conditional type of spatial dependence include Likelihood Ratio (LR) test and $\chi^2$ test (Bickenbach and Bode, 2003) as well as a test based on information theory (Kullback et al., 1962). For the first two tests, we could proceed as follows to acquire their statistics, DOF and p-value.

giddy.markov.Homogeneity_Results(sm.T).summary()

--------------------------------------------------
Markov Homogeneity Test
--------------------------------------------------
Number of classes: 5
Number of transitions: 3840
Number of regimes: 5
Regime names: 0, 1, 2, 3, 4
--------------------------------------------------
Test                   LR                Chi-2
Stat.              170.659              200.624
DOF                   60                   60
p-value                0.000                0.000
--------------------------------------------------
P(H0)        0        1        2        3        4
0    0.915    0.075    0.009    0.001    0.000
1    0.066    0.827    0.105    0.001    0.001
2    0.005    0.103    0.794    0.095    0.003
3    0.000    0.009    0.094    0.849    0.048
4    0.000    0.000    0.000    0.062    0.938
--------------------------------------------------
P(0)         0        1        2        3        4
0    0.963    0.030    0.006    0.000    0.000
1    0.060    0.832    0.107    0.000    0.000
2    0.000    0.140    0.740    0.120    0.000
3    0.000    0.036    0.321    0.571    0.071
4    0.000    0.000    0.000    0.167    0.833
--------------------------------------------------
P(1)         0        1        2        3        4
0    0.798    0.168    0.034    0.000    0.000
1    0.075    0.882    0.042    0.000    0.000
2    0.005    0.070    0.866    0.059    0.000
3    0.000    0.000    0.064    0.902    0.034
4    0.000    0.000    0.000    0.194    0.806
--------------------------------------------------
P(2)         0        1        2        3        4
0    0.847    0.153    0.000    0.000    0.000
1    0.081    0.789    0.129    0.000    0.000
2    0.005    0.098    0.793    0.098    0.005
3    0.000    0.000    0.094    0.871    0.035
4    0.000    0.000    0.000    0.102    0.898
--------------------------------------------------
P(3)         0        1        2        3        4
0    0.885    0.098    0.000    0.016    0.000
1    0.039    0.814    0.140    0.000    0.008
2    0.005    0.094    0.777    0.119    0.005
3    0.000    0.023    0.129    0.754    0.094
4    0.000    0.000    0.000    0.097    0.903
--------------------------------------------------
P(4)         0        1        2        3        4
0    0.333    0.667    0.000    0.000    0.000
1    0.048    0.774    0.161    0.016    0.000
2    0.011    0.161    0.747    0.080    0.000
3    0.000    0.010    0.062    0.896    0.031
4    0.000    0.000    0.000    0.024    0.976
--------------------------------------------------

From the above summary table, we can observe that the observed LR test statistic is 170.659 and the observed $\chi^2$ test statistic is 200.624. Their p-values are 0.000, which leads to the rejection of the null hypothesis of conditional spatial independence.

For the last (information theory-based) test, we call the function $kullback$. The result is consistent with LR and $\chi^2$ tests. As shown below, the observed test statistic is 230.03 and its p-value is 2.22e-16, leading to the rejection of the null.

print(giddy.markov.kullback(sm.T))

{'Conditional homogeneity': 230.0266246375395, 'Conditional homogeneity dof': 80, 'Conditional homogeneity pvalue': 2.220446049250313e-16}

### LISA Markov

giddy.markov.LISA_Markov(self, y, w, permutations=0, significance_level=0.05, geoda_quads=False)

The Spatial Markov conditions the transitions on the value of the spatial lag for an observation at the beginning of the transition period. An alternative approach to spatial dynamics is to consider the joint transitions of an observation and its spatial lag in the distribution. By exploiting the form of the static LISA and embedding it in a dynamic context we develop the LISA Markov in which the states of the chain are defined as the four quadrants in the Moran scatter plot, namely, HH(=1), LH(=2), LL(=3), HL(=4). Continuing on with our US example, the LISA transitions are:

lm = giddy.markov.LISA_Markov(pci.T, w)
print(lm.classes)

[1 2 3 4]

The LISA transitions are:

print(lm.transitions)

[[1.087e+03 4.400e+01 4.000e+00 3.400e+01]
[4.100e+01 4.700e+02 3.600e+01 1.000e+00]
[5.000e+00 3.400e+01 1.422e+03 3.900e+01]
[3.000e+01 1.000e+00 4.000e+01 5.520e+02]]

and the estimated transition probability matrix is:

print(lm.p)

[[0.92985458 0.03763901 0.00342173 0.02908469]
[0.07481752 0.85766423 0.06569343 0.00182482]
[0.00333333 0.02266667 0.948      0.026     ]
[0.04815409 0.00160514 0.06420546 0.88603531]]

The diagonal elements indicate the staying probabilities and we see that there is greater mobility for observations in quadrants 2 (LH) and 4 (HL) than 1 (HH) and 3 (LL).

The implied long run steady state distribution of the chain is:

print(lm.steady_state)

[0.28561505 0.14190226 0.40493672 0.16754598]

again reflecting the dominance of quadrants 1 and 3 (positive autocorrelation). The first mean passage time for the LISAs is:

print(giddy.ergodic.fmpt(lm.p))

[[ 3.50121609 37.93025465 40.55772829 43.17412009]
[31.72800152  7.04710419 28.68182751 49.91485137]
[52.44489385 47.42097495  2.46952168 43.75609676]
[38.76794022 51.51755827 26.31568558  5.96851095]]

To test for dependence between the dynamics of the region and its neighbors, we turn to $\chi^2$ test of independence. Here, the $\chi^2$ statistic, its p-value and degrees of freedom can be obtained from the attribute $chi\_2$. As the p-value is 0.0, the null of independence is clearly rejected.

print(lm.chi_2)

(1058.2079036003051, 0.0, 9)

### Next steps

• Simulation/prediction of Markov chain and spatial Markov chain


text:
# Tag Info

Let me begin with your naive estimation of the significance. Sixteen ($o=16$) events are observed over an expected background of $b=4.2\pm1.6$. I presume that you have calculated that $$\frac{o - b}{\delta b} \simeq 7.4,$$ concluding that the excess has a significance of $7.4\sigma$. There are a two main mistakes in this interpretation of the data and ...
I would have thought your graph should be plotted with the axes the other way round, because the length of skid mark is the dependent variable. If you do this you should have some kind of parabola (in theory!), which if you take thinking/reaction time ($T$) into account will have a form something like $$S=vT+kv^2$$ where $k$ is some constant. I don't think ...


text:
# Tag Info

## Hot answers tagged minimax

5

Both algorithms should give the same answer. However, their main difference is that alpha-beta does not explore all paths, like minimax does, but prunes those that are guaranteed not to be an optimal state for the current player, that is max or min. So, alpha-beta is a better implementation of minimax. Here are the time complexities of both algorithms ...

3

Some basic advantages of MCTS over Minimax (and its many extensions, like Alpha-Beta pruning and all the other extensions over that) are: MCTS does not need a heuristic evaluation function for states. It can make meaningful evaluations just from random playouts that reach terminal game states where you can use the loss/draw/win outcome. So if you're faced ...

1

I'm not familiar with your game so I can't tell you what a good heuristic woul be in your specific case, but I can give you some advice on how to look for a good heuristic function. As a rule of thumb, the heuristic function for a MiniMax algorithm is best kept simple and efficient, so you can get deeper into the tree. But it depends on how costly it is to ...

1

When combatting the horizon affect, you want to consider any short term actions that will greatly affect your position evaluation. Thus, in addition to captures, you will also want to include: When the opponent can make a king next move When the current player can make a king next move When the only legal moves left will lead to capture the turn after for ...

1

I will answer the first question question based on information I have gathered so far. The probability of each action for the $\textbf{two player zero sum game}$ need not be the same for both players. It turns out that in the game of Morra, the probability vectors just turn out to be the same value. In general to determine $\textbf{optimal mixed strategies}$ ...

1

There are a few different ways to improve on your simple heuristic approach, but they mostly resolve to these three things: Find a better heuristic. This could be done by calculating probabilities of results, or running loads of training simulations and somehow tuning the heuristic function. Look-ahead search/planning. There are many possible search ...

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


text:
### Instant Insanity

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

### Tree Graphs

A connected graph is a graph in which we can get from any vertex to any other by travelling along the edges. A tree is a connected graph with no closed circuits (or loops. Prove that every tree has exactly one more vertex than it has edges.

### Magic Caterpillars

Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.

# Plum Tree

##### Age 14 to 18 Challenge Level:

Each of the numbers $1$ to $15$ is used to label one of the vertices or one of the edges of this plum tree.

A graph is said to be 'vertex magic' if the sum of the numbers on a vertex and on all the edges joined to that vertex is the same for each vertex. We call this the 'magic sum'.

A graph is said to be 'edge magic' if, for each edge, the sum of the three numbers, on the edge and the two vertices at its ends, is the same. This is called the 'magic constant'.

A graph is said to be 'totally magic' if it is both vertex magic and edge magic. The labellings to make it vertex magic and to make it edge magic do not have to be the same and the magic sum and magic constant do not have to be the same .

Prove that this plum tree can be labelled to make it totally magic. Is there more than one edge magic labelling and more than one vertex magic labelling apart from symmetries and apart from simply exchanging the number on an extreme vertex with the number on its branch?

Can you construct a tree of your own that is totally magic?


text:
# That would take a long time for us to calculate!!Instead,let's try something else.

Level pending

Let $$A=\dfrac{2}{1*3}+\dfrac{2}{3*5}+\dfrac{2}{5*7}+\dfrac{2}{7*9}+\dfrac{2}{9*11}$$ and let $$B=\dfrac{1}{1*2}+\dfrac{1}{2*3}+\dfrac{1}{3*4}+\dfrac{1}{4*5}+\dfrac{1}{5*6}+\dfrac{1}{6*7}+\dfrac{1}{7*8}$$,then evaluate $$33A-32B$$.

×


text:
## Passage Time Distributions for Poisson processes.

Let $(X_t)_{t \geq 0}$ be a standard Poisson process with intensity $\mu$. Let $\tau_b = \inf ( t>0 : X_t= at + b)$, where $a>0$ and $b<0$, and let $\sigma = \inf (t>0 : X_t \geq at)$. Is there any reference for the distributions of $\tau_b$ and $\sigma$, as well as computing $P(\tau_b < \sigma)$?

Thanks!

-
 What is $N_t$ ? – Ori Gurel-Gurevich Sep 29 2011 at 22:50 Sorry, it's fixed. – weakstar Sep 29 2011 at 22:56 You should consider the process $Y_t=X_t-at$, which is still a Lévy process, because then it is just first passage time, and you should find many references. – Raphael L Sep 30 2011 at 10:34


text:
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 1 - 1 / 11 1.Permanent versus determinant over a finite fieldGregor Dolinar, Aleksandr Èmilevič Guterman, Marko Orel, Bojan Kuzma, 2013, published scientific conference contributionAbstract: Let ▫$\mathbb{F}$▫ be a finite field of characteristic different from 2. We study the cardinality of sets of matrices with a given determinant or a given permanent for the set of Hermitian matrices ▫$\mathcal{H}_n(\mathbb{F})$▫ and for the whole matrix space ▫$M_n(\mathbb{F})$▫. It is known that for ▫$n = 2$▫, there are bijective linear maps ▫$\Phi$▫ on ▫$\mathcal{H}_n(\mathbb{F})$▫ and ▫$M_n(\mathbb{F})$▫ satisfying the condition per ▫$A = \det \Phi(A)$▫. As an application of the obtained results, we show that if ▫$n \ge 3$▫, then the situation is completely different and already for ▫$n = 3$▫, there is no pair ofmaps ▫$(\Phi, \phi)$▫, where ▫$\Phi$▫ is an arbitrary bijective map on matrices and ▫$\phi \colon \mathbb{F} \to \mathbb{F}$▫ is an arbitrary map such that per ▫$A = \phi(\det \Phi(A))$▫ for all matrices ▫$A$▫ from the spaces ▫$\mathcal{H}_n(\mathbb{F})$▫ and ▫$M_n(\mathbb{F})$▫, respectively. Moreover, for the space ▫$M_n(\mathbb{F})$▫, we show that such a pair of transformations does not exist also for an arbitrary ▫$n > 3$▫ if the field ▫$\mathbb{F}$▫ contains sufficiently many elements (depending on ▫$n$▫). Our results are illustrated by a number of examples.Found in: ključnih besedahSummary of found: Zadetek v naslovuKeywords: mathematics, linear algebra, matrix theory, permanent, determinantPublished: 03.04.2017; Views: 716; Downloads: 55 Full text (0,00 KB)
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