DO

Posted on January 30, 2019 by rpflee

While DOLIST and DOTIMES are convenient and easy to use, they aren’t flexible enough to use for all loops. For instance, what if you want to step multiple variables in parallel?

(do (variable-definition*)
    (end-test-form result-form*)
  statement*)

— Practical Common Lisp

— Peter Seibel

(defun split-if (fn lst)
  (let ((acc nil))
    (do ((src lst (cdr src)))
        ((or (null src) (funcall fn (car src)))
         (values (nreverse acc) src))
      (push (car src) acc))))

>(split-if #'(lambda (x) (> x 4)) '(1 2 3 4 5 6 7 8 9 10))

(1 2 3 4) (5 6 7 8 9 10)

— p.50

— On Lisp

— Paul Graham

Exercise 4.5

Implement the function split-if without using the macro do.

— Me@2019-01-30 09:58:30 PM

2019.01.30 Wednesday ACHK

Problem 14.5d3

Posted on January 27, 2019 by rpflee

Counting states in heterotic SO(32) string theory | A First Course in String Theory

d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$ .

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$ .

~~~

— This answer is my guess. —

~~~

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{ \begin{aligned} &f_{L, NS'+}(x) \\ &= a_{NS'+} (r) x^r \\ &= \frac{1}{2x} \left[ \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \right] \\ & \\ &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\ \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ & \\ &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5} + ... \end{aligned}}$

$\displaystyle{ \begin{aligned} \alpha' M_R^2: \end{aligned}}$

$\displaystyle{ \begin{aligned} f_{NS+}(x) &= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ... \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} f_{R-}(x) &= 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ... \\ \end{aligned}}$

So the total number of states in heterotic string theory at $\displaystyle{ \begin{aligned} \alpha' M^2 = 8 \end{aligned}}$ is

$\displaystyle{ \begin{aligned} &\left(1384320 + 1310720 \right) \times \left(1152 + 1152\right) \\ \end{aligned}}$ .

$\displaystyle{ \begin{aligned} &= 6209372160 \\ \end{aligned}}$ .

~~~

— This answer is my guess. —

— Me@2019-01-26 04:49:37 PM

Quantum logic, 3

Posted on January 26, 2019 by rpflee

The more common view regarding quantum logic, however, is that it provides a formalism for relating observables, system preparation filters and states. $^\text{[citation needed]}$ In this view, the quantum logic approach resembles more closely the C*-algebraic approach to quantum mechanics. The similarities of the quantum logic formalism to a system of deductive logic may then be regarded more as a curiosity than as a fact of fundamental philosophical importance. A more modern approach to the structure of quantum logic is to assume that it is a diagram – in the sense of category theory – of classical logics (see David Edwards).

— Wikipedia on Quantum logic

2019.01.26 Saturday ACHK

You

Posted on January 25, 2019 by rpflee

Now + you = your future

— Me@2011.07.05

PhD, 3.2

Posted on January 25, 2019 by rpflee

故事連線 1.1.5.2 | 碩士 4.2 | On Keeping Your Soul, 2.2.2 | Release early. Release often, 3.2

這段改編自 2010 年 4 月 18 日的對話。

同理，合理一點的博士論文編寫模式，同樣是「release early, release often」（極速（而）頻繁（地）出版）。如果我攻讀博士課程，從課程之開頭，我就應該（例如）每一出版一篇網誌短文。日日如是，累績三年。然後，選當中最好之十篇文章合體，再打磨成學術文章出版。

（問：那樣，會不會在你成功，出版學術文章之前，就已經被同行之中的害群之馬，盜取了你的原創概念，從而捷足先登，出版了文章？）

絕對有可能。所以，你未必要跟足這個例子。重點是，你根據其精神，來設計你的策略。

（問：你的「其精神」，是指「release early, release often」？）

無錯。以下把它簡稱為「頻繁出版」。有了這個精神，你自然可以適當地修改這個例子，避免被壞人抄襲。例如，你每天學術短文，不必發表於公開網誌上。你可以改為，只發表在半公開的網誌上，導致只有知己朋友可以看到。

而最重點是，有了這個「『極速頻繁出版』遠比『延緩完美出版』優勝」的心態後，你會知道，在博士課程中途，甚至是之初，在專業學術期刊發表，每篇幾頁的文章，遠比在博士課程後期，在大學研究院之內發表，每本三百頁的論文，來得重要。

試想想，如果已經出版過，幾篇學術文章，把它們合體成博士論文，又有何難度呢？

其實，你博士論文中，最重要的想法，就已經在那幾篇文章之中。你需要做的，就只是串聯、擴展和打磨罷了。

（問：你的意思是，要透過每天出版一篇短文，累積成每數個月一篇，可出版的學術期刊文章；然後，再透過那數篇學術期刊文章，累積成你那本博士論文？）

— Me@2019-01-22 06:46:23 PM

duplicate

Posted on January 21, 2019 by rpflee

If (member o l) finds o in the list l, it also returns the cdr of l beginning with o. This return value can be used, for example, to test for duplication. If o is duplicated in l, then it will also be found in the cdr of the list returned by member. This idiom is embodied in the next utility, duplicate:
>(duplicate ’a ’(a b c a d)) (A D)

(defun duplicate (obj lst &key (test #’eql))
    (member obj (cdr (member obj lst :test test))
            :test test))

— p.51

— On Lisp

— Paul Graham

Exercise 4.4

Without using the existing function member, define duplicate as in
>(duplicate ’a ’(a b c a d)) (A D)

— This answer is my guess. —


(defun my-member (obj lst)
    (cond ((not lst) NIL)
          ((eq obj (car lst)) lst)
          (t (my-member obj (cdr lst)))))

— This answer is my guess. —

— Me@2019-01-21 06:34:46 AM

Problem 14.5d2

Posted on January 20, 2019 by rpflee

Counting states in heterotic SO(32) string theory | A First Course in String Theory

d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$ .

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$ .

~~~

— This answer is my guess. —

~~~

The left R’+ sector:

$\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_{\alpha} \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&|{\bar \alpha}_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}$

$\displaystyle{N^\perp:}$

$\displaystyle{\begin{aligned} \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 \left( 1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ... \right)^8 ... \left( 1 + \lambda_{-1} x^{1} \right)^{32} \left( 1 + \lambda_{-2} x^{2} \right)^{32} ... \\ \end{aligned}}$

However, there are $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha\rangle_L}$ and $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha \rangle_R}$ :

$\displaystyle{\begin{aligned} (2^{15} + 2^{15}) \left[ \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 ... \left( 1 + \lambda_{-1} x^{1} \right)^{32} ... \right] \\ \end{aligned}}$

$\displaystyle{\begin{aligned} 2^{16} \prod_{r=1}^\infty \frac{1}{(1 - x^r)^8} (1 + x^{r})^{32} \\ \end{aligned}}$

“Keep only states with $\displaystyle{(-1)^{F_L} = +1}$ ; this defines the left R’+ sector.”

$\displaystyle{\begin{aligned} \frac{2^{16}}{2} \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

~~~

— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

Logical arrow of time, 6.4

Posted on January 19, 2019 by rpflee

The source of the macroscopic time asymmetry, aka the second law of thermodynamics, is the difference of prediction and retrodiction.

In a prediction, the deduction direction is the same as the physical/observer time direction.

In a retrodiction, the deduction direction is opposite to the physical/observer time direction.

.

— guess —

If a retrodiction is done by a time-opposite observer, he will see the entropy increasing. For him, he is really doing a prediction.

— guess —

.

— Me@2013-10-25 3:33 AM

The existence of the so-called “the paradox of the arrow of time” is fundamentally due to the fact that some people insist that physics is about an observer-independent objective truth of reality.

However, it is not the case. Physics is not about “objective” reality. Instead, physics is always about what an observer would observe.

— Lubos Motl

— paraphrased

— Me@2019-01-19 10:25:15 PM

Good intentions

Posted on January 19, 2019 by rpflee

地獄之路 2

The road to hell is paved with good intentions.

Hell is full of good meanings, but heaven is full of good works.

—

2019.01.19 Saturday ACHK

（反對）開夜車 2.1

Posted on January 19, 2019 by rpflee

Ken Chan 時光機 1.4.2.1

之前提到，Ken Chan 所讀的中學不正常：

「

在中四和中五的校內考試測驗中，不斷地考核高考課程。所以，他在中四時代開始，已經要鑽研，高考程度和大學程度的物理。

」

那樣，他何來有，那麼多的時間呢？

那時，他往往要讀書，至通宵達旦。

那樣，他的公開試成績，又如何呢？

根據他所講，他事後發現，會考物理科中的 MC（多項選擇題）部分，錯了兩題；他達不到他原本,「全部科目全部滿分」的理想。

那樣，他又做不做到「狀元」，全部科目「奪 A」呢？

他在另一天的課堂中，暗示自己當年，差一點才做到「狀元」：

（我暫時不記得，他以下說話的上半句是什麼。）

「

… 如何不是那樣，我就毋須於，放榜當天的晚上，在家裡哭。唉！還是不說了。

」

當時，我想知道詳情。可惜，他真的沒有說下去，我也沒有辦法。

雖然，主觀而言，從他自己的角度，成績不是理想，因為，那不是「全部科目全部滿分」；但是，如果不是對自己，那麼苛刻的話，客觀來說，他的成績是上乘的。

可能，因為他「溫習到凌晨」式的方法，對他來說有效，他亦鼓勵學生那樣做。

我現在的記憶，暫時未能確定，他在課堂中，有沒有明示推介過，用這個方法。但是，我記得在他派發的筆記中，其中一頁，有一個正在深夜讀書的漫畫。而在漫畫下面，有一句「study to 3 a.m.」（讀書至深夜三點鐘）。

長話短說，我是反對這個方法的，…

— Me@2019-01-18 03:47:50 PM

Equality Predicates

Posted on January 16, 2019 by rpflee

6.3. Equality Predicates

Common Lisp provides a spectrum of predicates for testing for equality of two objects: eq (the most specific), eql, equal, and equalp (the most general).

eq and equal have the meanings traditional in Lisp.

eql was added because it is frequently needed, and equalp was added primarily in order to have a version of equal that would ignore type differences when comparing numbers and case differences when comparing characters.

If two objects satisfy any one of these equality predicates, then they also satisfy all those that are more general.

[Function]
eq x y

(eq x y) is true if and only if x and y are the same identical object. (Implementationally, x and y are usually eq if and only if they address the same identical memory location.)

The predicate eql is the same as eq, except that if the arguments are characters or numbers of the same type then their values are compared. Thus eql tells whether two objects are conceptually the same, whereas eq tells whether two objects are implementationally identical. It is for this reason that eql, not eq, is the default comparison predicate for the sequence functions defined in chapter 14.

[Function]
eql x y

The eql predicate is true if its arguments are eq, or if they are numbers of the same type with the same value, or if they are character objects that represent the same character.

[Function]
equal x y

The equal predicate is true if its arguments are structurally similar (isomorphic) objects. A rough rule of thumb is that two objects are equal if and only if their printed representations are the same.

Numbers and characters are compared as for eql. Symbols are compared as for eq. This method of comparing symbols can violate the rule of thumb for equal and printed representations, but only in the infrequently occurring case of two distinct symbols with the same print name.

[Function]
equalp x y

Two objects are equalp if they are equal; if they are characters and satisfy char-equal, which ignores alphabetic case and certain other attributes of characters; if they are numbers and have the same numerical value, even if they are of different types; or if they have components that are all equalp.

— Common Lisp the Language, 2nd Edition

— Guy L. Steele Jr.

Conrad’s Rule of Thumb for Comparing Stuff:

1. Use eq to compare symbols

2. Use equal for everything else

— Land of Lisp, p.63

— Conrad Barski, M. D.

2019.01.16 Wednesday ACHK

Problem 14.5d1.1.2

Posted on January 15, 2019 by rpflee

Counting states in heterotic SO(32) string theory | A First Course in String Theory

d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$ .

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$ .

~~~

— This answer is my guess. —

~~~

p.322

$\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}$

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

$\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}$

The left NS’+ sector:

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}$

$\displaystyle{N^\perp:}$

$\displaystyle{\begin{aligned} \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 \left( 1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ... \right)^8 ... \left( 1 + \lambda_{-\frac{1}{2}} x^{\frac{1}{2}} \right)^{32} \left( 1 + \lambda_{-\frac{3}{2}} x^{\frac{3}{2}} \right)^{32} ... \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \prod_{r=1}^\infty \frac{1}{(1 - x^r)^8} (1 + x^{r-\frac{1}{2}})^{32} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, NS'}(x) \\ &= a_{NS'} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

“The left NS’ sector is built with oscillators $\displaystyle{\bar \alpha_{-n}^I}$ and $\displaystyle{\lambda_{-r}^A}$ acting on the vacuum $\displaystyle{|NS' \rangle_L}$ , declared to have $\displaystyle{(-1)^{F_L} = + 1}$ :”

$\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}$

So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = +1}$ .

$\displaystyle{ \begin{aligned} &f_{L, NS'}(x) \\ \end{aligned}}$

$\displaystyle{ = \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}$

Let

$\displaystyle{ \begin{aligned} &g (\sqrt{x}) \\ &= \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ &= 1+32 \, \sqrt{x}+504 \, x+5248 \, x^{\frac{3}{2}}+40996 \, x^{2}+258624 \, x^{\frac{5}{2}}+1384320 \, x^{3}+6512384 \, x^{\frac{7}{2}} + ... \\ \end{aligned}}$

Then

$\displaystyle{ \begin{aligned} &g (-\sqrt{x}) \\ &= \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ &= 1 - 32 \sqrt{x} + 504 x - 5248 \, x^{\frac{3}{2}} + 40996 \, x^{2} - 258624 \, x^{\frac{5}{2}}+1384320 \, x^{3} - 6512384 x^{\frac{7}{2}} + ... \\ \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} &f_{L, NS'+}(x) \\ &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\ &= \frac{1}{2x} \left[ g(\sqrt{x}) + g(-\sqrt{x}) \right] \\ &= \frac{1}{2x} \left[ \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \right] \\ \end{aligned}}$

The left R’+ sector:

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}$

~~~

— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

EPR paradox, 5.3

Posted on January 14, 2019 by rpflee

According to special relativity, in EPR, which of Alice and Bob collapses the wavefunction is not absolute. In other words, they do not have any causal relations.

— Me@2012-04-12 10:42:22 PM

接觸永恆 2

Posted on January 13, 2019 by rpflee

Publish! 11

每個人，都需要有作品。

你的作品，就是你的人生意義。

每個人，都依靠自己的作品來生存。

作品，就是一些，不會隨自己的消失而消失的東西。

你，有沒有自己的作品？

— Me@2011.07.03

PhD, 3.1

Posted on January 13, 2019 by rpflee

故事連線 1.1.5.1 | 碩士 4.1 | On Keeping Your Soul, 2.2.1 | Release early. Release often, 3

這段改編自 2010 年 4 月 18 日的對話。

一般寫博士論文的方法過程，有如 closed source software（閉源軟件）的開發一樣，是不合理的。

那個方法是，花數年時間，編寫一個軟件程式，然後才正式出版。這個方法的危險之處是，如果軟件上市以後，才發現不受市場認同的話，那數年的投資，基本上是血本無歸的。

合理一點的開發模式是，open source software（開源軟件）開發時，一個經常採用的方法，叫做「release early, release often」（極速（而）頻繁（地）出版）。意思是，與其花數年時間，去編寫一個軟件程式的「完美」版，然後才正式出版，倒不如，先把那個軟件的「最粗疏但已經可用」版本，極速完成，然後立刻出版。

那樣，那公司就可以馬上得到，市場的初步反應；從而決定：是否繼續開發那個軟件：如果繼續的話，又應該改進哪些部分。

如果決定繼續開發，就立刻重覆這個「極速出版」流程。亦即是話，再把那個軟件改進部分的，「最粗疏但已經可用」版本，極速完成，然後立刻出版。

— Me@2019-01-13 06:22:43 PM

Problem 14.5d1.2 | SageMath

Posted on January 11, 2019 by rpflee

The generating function is an infinite product:

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

To evaluate the infinite product, you can use SageMath with the following commands:

typeset_mode(True)
(1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..oo)) a = (1/x)*prod(((1+x^(n-1/2))^(32)/(1-x^n)^8) for n in (1..200)) F = a.taylor(x,0,6) g = "+".join(map(latex, sorted([f for f in F.operands()], key=lambda exp:exp.degree(x))))
g

$\displaystyle{ \begin{aligned} &f_{L, NS+}(x) \\ \end{aligned}}$

$\displaystyle{ \approx \frac{1}{x}+\frac{32}{\sqrt{x}}+504+5248 \, \sqrt{x}+40996 \, x+258624 \, x^{\frac{3}{2}}+1384320 \, x^{2}+6512384 \, x^{\frac{5}{2}} + ...}$

— Me@2019-01-11 11:52:33 AM

Problem 14.5d1

Posted on January 10, 2019 by rpflee

Counting states in heterotic SO(32) string theory | A First Course in String Theory

d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$ .

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$ .

~~~

— This answer is my guess. —

~~~

p.322

$\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}$

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

$\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}$

The left NS’+ sector:

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}$

$\displaystyle{N^\perp:}$

$\displaystyle{\begin{aligned} \left( 1 + \bar \alpha_1 x + (\bar \alpha_1)^2 x^2 + ... \right)^8 \left( 1 + \bar \alpha_2 x + (\bar \alpha_2)^2 x^4 + ... \right)^8 ... \left( 1 + \lambda_{-\frac{1}{2}} x^{-\frac{1}{2}} \right)^{32} \left( 1 + \lambda_{-\frac{3}{2}} x^{-\frac{3}{2}} \right)^{32} ... \\ \end{aligned}}$

$\displaystyle{\begin{aligned} \prod_{r=1}^\infty \frac{1}{(1 - x^r)^8} (1 + x^{r-\frac{1}{2}})^{32} \\ \end{aligned}}$

$\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}$

$\displaystyle{\begin{aligned} &f_{L, NS+}(x) \\ &= a_{NS+} (r) x^r \\ &= \frac{1}{x} \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \\ \end{aligned}}$

The left R’+ sector:

$\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}$

~~~

— This answer is my guess. —

— Me@2019-01-10 01:49:43 PM

Consistent histories, 6

Posted on January 7, 2019 by rpflee

an observer ~ a consistent history

— Me@2019-01-05 04:02:43 PM

Guiding Star

Posted on January 7, 2019 by rpflee

Set your course by the stars, not by the lights of every passing ship.

– Omar Bradley

2019.01.07 Monday ACHK

Ken Chan 時光機 2.2

Posted on January 6, 2019 by rpflee

聖誕假後的常規課程中，Ken Chan 預計應該不會有時間，教光學和熱學。但是，他認為那兩個課題很淺易，所以，期望我們即使自修，也沒什麼難度。

但是，他後來改變主意，還是決定於，農歷新年的年初一、二、三，為我們補課，教回光學和熱學。

今次，那三天的課，他不單不收我們的學費，他還要自己，花時間找課室、花金錢付租金。

後來，他改為每天收 20 元，即是總共 60 元。他解釋道，補課社的職員因為他的補課，要犧牲那三天的農曆新年假期。那每人 60 元的學費是，全數慰勞他們的。

那三天的補課，好處是光學和熱學；壞處是，由於每天也補多個小時，我少了大量，各科的溫習時間。但是，那個農曆新年長假，因為我自己時間管理不善，而損失的時間，遠多於那三天的補課時間。

感謝 Ken Chan 的額外付出。我亦慶幸，生於那個時空，還是有真人授課的年代。出生遲十年，就已經只能透過錄影帶，來獲得 Ken Chan 的教導。

— Me@2019-01-06 02:18:47 PM

Physics Town

continues

Monthly Archives: January 2019

DO

Problem 14.5d3

Quantum logic, 3

You

PhD, 3.2

duplicate

Problem 14.5d2

Logical arrow of time, 6.4

Good intentions

（反對）開夜車 2.1

Equality Predicates

Problem 14.5d1.1.2

EPR paradox, 5.3

接觸永恆 2

PhD, 3.1

Problem 14.5d1.2 | SageMath

Problem 14.5d1

Consistent histories, 6

Guiding Star

Ken Chan 時光機 2.2