數學時間論 4

Posted on May 4, 2022 by rpflee

生物時間論 4

2022.05.04 Wednesday ACHK

數學教育 7.5.1

Posted on May 3, 2022 by rpflee

Genius 4.2.1 | A Fraction of Algebra, 2.1

這段改編自 2010 年 4 月 24 日的對話。

另外，他提的另一個，有關學習數學的要點是，即使假設你在大學中，學到的數學，在日常生活中沒有用，單單是為獲取，那些嶄新的元素概念本身，就已經能夠令你有超能力；令你有一些，常人沒有的思考工具、比喻語言。

（安：但是，這個講法可能有一個問題。

雖然，你剛才列舉了數個例子，來示範如何將高深數學，間接應用到人生處世，但是，一般人未必有那種能力。所以我想問，你又是如何去跨過這個難關呢？）

什麼難關？

（安：去翻譯那些抽象數學概念，到其他範疇，或者日常生活。）

那不是「難關」。你的意思是，一般人也沒有那個能力，而我有。所以，那是超能力；我當年一定是，用了一些秘技，才獲取之。

天才之道，點滴累積。其實並沒有所謂的「秘技」。只要一步一步地，學習數學，就自然建構出，一個相對接近完整的數學思考體系，生成「翻譯抽象數學概念到其他範疇」等能力。

所以，我猜想你的疑問是，其實我所講的「點滴累積」，或者「一步一步地」，雖然理想上是，基本的要求，但是現實中是，大部人也做不到。那就代表著，大部人可能也會遇到，一個共通的「難關」。那個「難關」究竟是什麼？我又是如何克服它，而做到「一步一步地」「點滴累積」的呢？

你問題的最簡化版本是：「如何學習數學，開創人生？」

有起碼以下三個先決條件：

1. 對數學（及其他學問人生），有極大興趣；

2. 遇到合理的老師和書籍：

重點是，數學概念或運算上的主要步驟，亳無違漏。支節可免，但主旨必須。細節可以無師自通，大節必靠前人指點。平地自己行，斜地靠梯級。平地可跳步，梯不可跳級。

3. 極超大量的背誦和練習：

數學是理科，所以其背誦方法，不是「死背」零碎隨機的資料，而是「生背」息息相關的訊息。融匯貫通地背誦的唯一方法是，極超大量的操練。

你所講的「難關」，就是以上的第二點。老師有分好老師和差老師。大部分也是差老師。而差老師再分兩類：不懂數學和不懂教學。

— Me@2022.05.02 11:48 PM

AutoKey

Posted on April 28, 2022 by rpflee

First photo, 3 | 1990, 3


output = system.exec_command("date +%Y.%m.%d")
keyboard.send_keys(output + " ")
output = system.exec_command("date +%I:%M")
keyboard.send_keys(output + " ")
output = system.exec_command("date +%p")
keyboard.send_keys(output)


output = system.exec_command("date +%Y.%m.%d")
keyboard.send_keys(output + " ")
output = system.exec_command("date +%A")
keyboard.send_keys(output)
output = " (c) All rights reserved by "
keyboard.send_keys(output)


output = system.exec_command("date +_%Y_%m_%d_")
keyboard.send_keys(output)
output = system.exec_command("date +_%H_%M_%S_%p")
keyboard.send_keys(output)

— Me@2022.04.28 09:04 AM

Ex 1.25 Properties of Dt

Posted on April 21, 2022 by rpflee

Structure and Interpretation of Classical Mechanics

The total time derivative $\displaystyle{D_t F}$ is not the derivative of the function $\displaystyle{F}$ . Nevertheless, the total time derivative shares many properties with the derivative. Demonstrate that $\displaystyle{D_t}$ has the following properties …

$\displaystyle{D_t (F + G) = D_t F + D_t G}$

~~~

Eq. (1.108):

$\displaystyle{ D(F \circ \Gamma[q]) = (DF\circ\Gamma[q])D\Gamma[q] }$

Eq. (1.109):

$\displaystyle{ DF \circ \Gamma[q] = \left[ \partial_0 F \circ \Gamma[q], \partial_1 F \circ \Gamma[q], \partial_2 F \circ \Gamma[q], ... \right] }$

Eq. (1.110):

$\displaystyle{ \left(D \Gamma[q] \right)(t) = \left( 1, Dq(t), D^2 q(t), ... \right) = \begin{bmatrix} 1 \\ Dq(t) \\ D^2 q(t) \\ ... \\ \end{bmatrix} \\ }$

$\displaystyle{ \begin{aligned} D(F \circ \Gamma[q])(t) &= (DF\circ\Gamma[q])D\Gamma[q](t) \\ &= \left[ \partial_0 F \circ \Gamma[q], \partial_1 F \circ \Gamma[q], \partial_2 F \circ \Gamma[q], ... \right] \begin{bmatrix} 1 \\ Dq(t) \\ D^2 q(t) \\ ... \\ \end{bmatrix} \\ &= \partial_0 F \circ \Gamma[q] + \partial_1 F \circ \Gamma[q] D q(t) + \partial_2 F \circ \Gamma[q] D^2 q(t) + ... \\ &= \partial_0 F \circ \Gamma[q] u(t) + \partial_1 F \circ \Gamma[q] D q(t) + \partial_2 F \circ \Gamma[q] D^2 q(t) + ... \\ \end{aligned} }$ ,

where $\displaystyle{u(t) \equiv 1}$ .

$\displaystyle{ \begin{aligned} D(F \circ \Gamma[q]) &= \partial_0 F \circ \Gamma[q] u + \partial_1 F \circ \Gamma[q] D q + \partial_2 F \circ \Gamma[q] D^2 q + ... \\ &= \partial_0 F \circ \Gamma[q] J_0 \circ \Gamma[q] + \partial_1 F \circ \Gamma[q] J_1 \circ \Gamma[q] + \partial_2 F \circ \Gamma[q] J_2 \circ \Gamma[q] + ... \\ \end{aligned} }$

where

$\displaystyle{ \begin{aligned} I_0 \circ \Gamma[q] &= t \\ \\ I_{n>0} \circ \Gamma[q] &= I_{n>0} (t, q, v, a, ...) \\ &= I_{n>0} (t, q, Dq, D^2 q, ...) \\ &= D^{(n-1)} q \\ \\ J_{n} \circ \Gamma[q] &= D(I_n (t, q, v, a, ...)) \\ \end{aligned} }$

The meaning of $\displaystyle{\delta_\eta (fg)[q]}$ is

$\displaystyle{\delta_\eta (f[q]g[q])}$

— Me@2019-04-27 07:02:38 PM

$\displaystyle{ \begin{aligned} D(F \circ \Gamma[q]) &= \partial_0 F \circ \Gamma[q] J_0 \circ \Gamma[q] + \partial_1 F \circ \Gamma[q] J_1 \circ \Gamma[q] + \partial_2 F \circ \Gamma[q] J_2 \circ \Gamma[q] + ... \\ &= \left[(\partial_0 F) J_0 + (\partial_1 F) J_1 + (\partial_2 F) J_2 + ... \right] \circ \Gamma[q] \\ \end{aligned} }$

.
Eq. (1.113):

$\displaystyle{ D_t F \circ \Gamma[q] = D(F \circ \Gamma[q]) }$

$\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] &= \left[(\partial_0 F) J_0 + (\partial_1 F) J_1 + (\partial_2 F) J_2 + ... \right] \circ \Gamma[q] \\ \\ D_t F &= (\partial_0 F) J_0 + (\partial_1 F) J_1 + (\partial_2 F) J_2 + ... \\ \end{aligned} }$

Eq. (1.114):

$\displaystyle{ \begin{aligned} D_t F (t, q, v, a, ...) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v + \partial_2 F(t, q, v, a, ...) a + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} D_t F \circ \Gamma[q] (t) &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &D_t (F + G) \circ \Gamma[q] (t) \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= \partial_0 \left[F(t, q, v, a, ...) + G(t, q, v, a, ...)\right] \\ &+ \partial_1 \left[F(t, q, v, a, ...) + G(t, q, v, a, ...)\right] v(t) \\ &+ \partial_2 \left[F(t, q, v, a, ...) + G(t, q, v, a, ...)\right] a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= \left[ \partial_0 F(t, q, v, a, ...) + \partial_0 G(t, q, v, a, ...)\right] \\ &+ \left[ \partial_1 F(t, q, v, a, ...) + \partial_1 G(t, q, v, a, ...)\right] v(t) \\ &+ \left[ \partial_2 F(t, q, v, a, ...) + \partial_2 G(t, q, v, a, ...)\right] a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= \partial_0 F(t, q, v, a, ...) + \partial_1 F(t, q, v, a, ...) v(t) + \partial_2 F(t, q, v, a, ...) a(t) + ... \\ &+ \partial_0 G(t, q, v, a, ...) + \partial_1 G(t, q, v, a, ...) v(t) + \partial_2 G(t, q, v, a, ...) a(t) + ... \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} &= D_t F \circ \Gamma[q] (t) + D_t G \circ \Gamma[q] (t) \\ \end{aligned} }$

In short,

$\displaystyle{ \begin{aligned} D_t (F + G) \circ \Gamma[q] (t) &= D_t F \circ \Gamma[q] (t) + D_t G \circ \Gamma[q] (t) \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} D_t (F + G) \circ \Gamma[q] (t) &= (D_t F + D_t G) \circ \Gamma[q] (t) \\ \\ D_t (F + G) \circ \Gamma[q] &= (D_t F + D_t G) \circ \Gamma[q] \\ \\ D_t (F + G) &= D_t F + D_t G \\ \end{aligned} }$

— Me@2022-04-20 11:42:52 AM

Y Combinator, 2

Posted on April 19, 2022 by rpflee

arethuza 7 hours ago [-]

It’s over 30 years since I was taught about the Y combinator and I still find it amazing – it allows you to define recursion in a language which has no concept of recursion or named functions – such as λ-calculus.

The fact that you then define this in terms of amazingly simple functions like the S and K combinators just, in my opinion, adds to how wonderful it is.

— Why Y? Deriving the Y Combinator in JavaScript

— Hacker News

2022.04.19 Tuesday ACHK

Jonny English, 2

Posted on April 19, 2022 by rpflee

A man without hope is a man without fear.

— Daredevil: Born Again

— Frank Miller

— Me@2016-04-09 09:49:28 PM

2022.04.19 Tuesday ACHK

相聚零刻 1.2.2

Posted on April 18, 2022 by rpflee

這段改編自 2021 年 12 月 12 日的對話。

你需要的，不是一個口頭的承諾，而是一個關係［中］，實質的進展。在朋友的外殼下，關係慢慢的，向情侶靠近。明修棧道，暗度陳倉。

當你們的關係，越來越像情侶的時候，你們自然就是情侶。表白呢，只是給這段關係，一個正式的名份。

不單是表白，在建立起吸引之前，暴露需求感，都會降低你的吸引力。 …　而表白呢，其實就是「暴露需求感」的最高形式。

如果愛情是一個朴實的話，那你就一定要，在成熟的時候，［才］去採摘。如果在此之前採摘，那這個果子採下來之後呢，它就不會再成長了；直接爛掉。

— 為什麽遇到喜歡的女生，千萬別去表白！

— 楚兒戀愛說

朋友之道，在乎平等，無所顧忌地聊天。

市面上，有一個害人不淺的思想，就是：「我向一個女仔表白，被拒絶後，我倆仍然可像之前，正常朋友般相處。」

那有可能，但機會極微，因為表白後，我倆再也不是平等的交往了。除非，雙方也很大方，性格異於常人地高級。但是，那樣的話，原本她就大概，不會拒絶我。

— Me@2022-04-18 12:29:00 PM

All men are born free.

Posted on April 17, 2022 by rpflee

Then some get married.

—

2022.04.17 Sunday ACHK

Quick Calculation 3.3

Posted on April 16, 2022 by rpflee

A First Course in String Theory

Verify the equations in (3.26).

~~~

Eq. (3.26):

$\displaystyle{ \begin{aligned} T_{\lambda \mu \nu} &= - T_{\mu \lambda \nu} \\ T_{\lambda \mu \nu} &= - T_{\lambda \nu \mu} \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} T_{\lambda \mu \nu} &= \partial_\lambda F_{\mu \nu} + \partial_\mu F_{\nu \lambda} + \partial_\nu F_{\lambda \mu} \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} T_{\mu \lambda \nu} &= \partial_\mu F_{\lambda \nu} + \partial_\lambda F_{\nu \mu} + \partial_\nu F_{\mu \lambda} \\ &= - \partial_\mu F_{\nu \lambda} - \partial_\lambda F_{\mu \nu} - \partial_\nu F_{\lambda \mu} \\ &= - T_{\lambda \mu \nu} \\ \end{aligned} }$

— Me@2022-04-15 05:10:09 PM

Logical arrow of time, 11

Posted on April 15, 2022 by rpflee

The initial microstates should be averaged, because it forms an ensemble for the initial macrostate.

Note that a macrostate is actually one particular microstate, not a collection of microstates; it is just that we don’t know which particular microstate.

But how come the final possible states should be summed over, not be averaged?

— Me@2013-08-13 05:16 PM

a macrostate = (a microstate in) a set of macroscopically-indistinguishable microstates

— Me@2022-01-09 07:43 AM

Note that, by definition, two macroscopically-indistinguishable microstates will never separate into two distinct macrostates.

— Me@2022-04-14 05:55 PM

The initial macrostate is with probability one, because it is already known. So the summation of the probabilities of all possible mutually exclusive initial microstates that are corresponding to that initial macrostate is one, such as

$\displaystyle{P(I_1) + P(I_2) = 1}$

By definition, the final macrostate is not known yet. Each possible final macrostate is not with probability one.

The probability of getting a particular final macrostate from that initial macrostate is the summation of the probabilities of all possible mutually exclusive final microstates that are corresponding to that final macrostate.

$\displaystyle{P(F_1~\text{or}~F_2) = P(F_1) + P(F_2)}$

$\displaystyle{P(I\to F) = \frac{1}{N_{\text{initial}}} \sum_{ij} P(I_i \to F_j)}$

— Me@2022-04-13 01:09 PM

The only assumptions I made are those about the addition of probabilities of assumptions and their effects – and these logical rules are fundamentally asymmetric when it comes to the role of the assumptions and their consequences. This logical arrow of time can’t be removed from any reasoning about a world that depends on time – time only copies the logical relationship of implication. And this logical arrow of time is the source of the thermodynamic arrow of time as well.

— edited Feb 2, 2011 at 15:23

— answered Jan 14, 2011 at 11:42

— Luboš Motl

— Calculation of the cross section

— Physics StackExchange

地獄篇 10

Posted on April 13, 2022 by rpflee

外在世界是地獄，心中宇宙是天堂；把知識和想像化成現實，就是把地獄天堂化。

— Me@2011.10.05

方法一：

創造或發掘有用的資訊，然後發佈。

例如，公佈騙子的身份和行騙方法，就可否定其害處。

又例如，天花的預防方法和醫治方法，否定了其害處之餘，甚至還否定了其存在。

— Me@2011.10.05

— Me@2022-04-13

伏線驅動程式 1.4

Posted on April 12, 2022 by rpflee

這段改編自 2010 年 10 月 14 日的對話。

答覆我一句，聘請還是不聘請，並不是問題。最滋擾人的是，不置可否。

「究竟那間補習社會聘請我，我應該等待那份工作，還是立刻找其他呢？」

記住，大概而然：

需要對方行動的事情，對方「不置可否」這行動本身，就已代表了「否」，即婉拒。婉拒者，婉轉地拒絶也。例如，你向心上人表白，但她很長時間後，例如一星期，仍然沒有回覆。

需要你方主動的項目，對方「不置可否」這行動本身，就已代表了「可」，即默許。默許者，沉默地允許也。例如，你看到一枚戒指很美，問女朋友：「買給你好嗎？」她不出聲的話，就即是「要」。

這大原則未必準確，但亦所差無幾。

當然，最終我也只能，找其他工作。我在報紙中的招聘廣告中，找到這裡的一份數學助教工作。我其實不是找「中學教師」的工作。我曾經立志，不做中學教師。

我在中六預科，一堂純數學課中，覺得老師講得很好，心裡有感而發：「大個千萬不要做中學教師。」否則，在事業上，我將要花數十年的時間，循環同一年。

我當時找助教工作，只視作臨時的唯生措施。只不過到簽約時，校長問如果給三班給我教好不好。薪金高一節，我當然說好。

先教兩年中學，之後我才回去讀物理，都算是可取的人生軌跡。

— Me@2022-04-12 11:34:16 AM

Radio-controlled car

Posted on April 11, 2022 by rpflee

1990s, 19

— Me@2022-04-11 11:04:37 AM

Ex 1.24 Constraint forces, 1.3

Posted on April 10, 2022 by rpflee

Structure and Interpretation of Classical Mechanics

Find the tension in an undriven planar pendulum.

~~~

Tangential component:

$\displaystyle{\begin{aligned} \left(F_{\text{net}}\right)_t &= m a_t \\ - mg \sin \theta &= m l \ddot \theta \\ \end{aligned}}$

Radial component:

$\displaystyle{\begin{aligned} \left(F_{\text{net}}\right)_r &= m a_r \\ F(t) - mg \cos \theta &= m l \dot \theta^2 \\ \end{aligned}}$

— Me@2022-04-10 04:22:27 PM

Remove time from physics

Posted on April 10, 2022 by rpflee

motohagiography on May 14, 2018 | next [–]

I once saw a fridge magnet that said “time is natures way of making sure everything doesn’t happen all at once,” and it’s stuck with me.

The concept of time not being “real,” can be useful as an exercise for modelling problems where to fully explore the problem space, you need to decouple your solutions from needing them to occur in an order or sequence.

From an engineering perspective, “removing” time means you can model problems abstractly by stepping back from a problem and asking, what are all possible states of the mechanism, then which ones are we implementing, and finally, in what order. This is different from the relatively stochastic approach most people take of “given X, what is the necessary next step to get to desired endstate.”

More simply, as a tool, time helps us apprehend the states of a system by reducing the scope of our perception of them to sets of serial, ordered phenomena.

Whether it is “real,” or an artifact of our perception is sort of immaterial when you can choose to reason about things with it, or without it. A friend once joked that math is what you get when you remove time from physics.

I look forward to the author’s new book.

— Hacker News

2022.04.10 Sunday ACHK

Fearing the wolf

Posted on April 10, 2022 by rpflee

The sheep will spend its entire life fearing the wolf only to be eaten by the shepherd.

—

2022.04.09 Saturday ACHK

十萬七千里 3.2

Posted on April 9, 2022 by rpflee

數學教育 7.4 | The least of all evils 8.2 | 眾害取其輕 8.2

這段改編自 2010 年 4 月 24 日的對話。

… began doing just one thing ever which he had control.

— Stephen Covey

這個原理用作，人生哲學的比喻的話，就是指，即使「目標不可能」，也不代表「要完全放棄」。亦即是話，即使你的一些人生目標，不可能實現，那也不代表你，完全沒有東西可以做。那「不可能的目標」或許可以，轉化為一個「可能的近似版本」。

即使有些情況下，現實和理想，還相差十萬八千里，你總可以嘗試，把理實拉近理想一點點。你由 108,000 里，行到 107,999 里之處，才再想下一步，如何行近再一點點。見步行步，行步見步。

達到或接近理想，不是每人都做到。但是，減少理實和理想的距離，那怕只是一點點，則是大部人都，可以執行的任務。

追尋快樂，要機緣巧合。但減輕痛苦，則能力可及。

— Me@2022-04-09 04:20:42 PM

Inspiration

Posted on April 8, 2022 by rpflee

If you need inspiring words, don’t do it.

— Elon Musk

2022.04.08 Friday ACHK

Quick Calculation 3.2

Posted on April 7, 2022 by rpflee

A First Course in String Theory

Verify that the gauge transformation (3.10) are correctly summarized by (3.21).

~~~

Eq. (3.21):

$\displaystyle{ \begin{aligned} A_\nu' &= A_\nu + \partial_\nu \epsilon \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} \left( A_0', A_1', ... \right) &= \left( - \Phi + \frac{\partial \epsilon}{\partial x^0}, A^1 + \frac{\partial \epsilon}{\partial x^1}, ... \right) \\ \left( -\Phi', {A^1}', ... \right) &= \left( - \Phi + \frac{1}{c} \frac{\partial \epsilon}{\partial t}, A^1 + \frac{\partial \epsilon}{\partial x^1}, ... \right) \\ \end{aligned} }$

$\displaystyle{ \begin{aligned} \Phi' &= \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} \\ \left( {A^1}', {A^2}', {A^3}' \right) &= \left( {A^1}, {A^2}, {A^3} \right) + \left( \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \frac{\partial}{\partial x^3} \right) \epsilon \\ \end{aligned} }$

Eq. (3.10):

$\displaystyle{ \begin{aligned} \Phi' &= \Phi - \frac{1}{c} \frac{\partial \epsilon}{\partial t} \\ \vec A' &= \vec A + \nabla \epsilon \\ \end{aligned} }$

— Me@2022-04-07 07:05:29 PM

C and Lisp

Posted on April 7, 2022 by rpflee

numeromancer on Jan 25, 2010 [-]

In every art there is a dichotomy between the practical and the theoretical, and each has their fundamentals. In Comp. Sci., those two sets of fundamentals are these: sets of machine instructions, which come in several varieties; and lambda calculus, or one of the equivalent (by Church’s Thesis) formal systems. C and Lisp are similar in that they represent the first steps in each case to reach the other: C is a level above machine code, providing some abstraction and portability to the use of machine code, the fundamental elements of practical computing; lisp is a level above lambda calculus, providing a practical system for using functions, the fundamental elements of theoretical computing.

In short, mastery of C is concomitant with the ability to measure the cost of computation (sometimes, regardless of the value of it); mastery of Lisp is concomitant with the ability to measure the value of computation (sometimes, regardless of the cost).

Since C and Lisp lie on opposite borders of the universe of computation, knowing both will allow you to better measure the scope of that universe.

— Ask HN: Why does learning lisp make you a better C-programmer?

— Hacker News

2022.04.07 Thursday ACHK

Physics Town

continues

數學時間論 4

數學教育 7.5.1

AutoKey

Ex 1.25 Properties of Dt

Y Combinator, 2

Jonny English, 2

相聚零刻 1.2.2

All men are born free.

Quick Calculation 3.3

Logical arrow of time, 11

地獄篇 10

伏線驅動程式 1.4

Radio-controlled car

Ex 1.24 Constraint forces, 1.3

Remove time from physics

Fearing the wolf

十萬七千里 3.2

Inspiration

Quick Calculation 3.2

C and Lisp