Author Archives: rpflee

Path-distinguishing function

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So we will try to arrange that the path-distinguishing function, constructed as an integral of a local property along the path, assumes a stationary value for any realizable path. Such a path-distinguishing function is traditionally called an action for the system. We use the word “action” to be consistent with common usage. Perhaps it would be clearer to continue to call it “path-distinguishing function,” but then it would be more difficult for others to know what we were talking about.

— 1.3 The Principle of Stationary Action

— Structure and Interpretation of Classical Mechanics

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2019.02.17 Sunday ACHK

Logical arrow of time, 7

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When we imagine that we know and keep track of all the exact information about the physical system – which, in practice, we can only do for small microscopic physical systems – the microscopic laws are time-reversal-symmetric (or at least CPT-symmetric) and we don’t see any arrow. There is a one-to-one unitary map between the states at times “t1” and “t2” and it doesn’t matter which of them is the past and which of them is the future.

A problem is that with this microscopic description where everything is exact, no thermodynamic concepts such as the entropy “emerge” at all. You might say that the entropy is zero if the pure state is exactly known all the time – at any rate, a definition of the entropy that would make it identically zero would be completely useless, too. By “entropy”, I never mean a quantity that is allowed to be zero for macroscopic systems at room temperature.

But whenever we deal with incomplete information, this one-to-one map inevitably disappears and the simple rules break down. Macroscopic laws of physics are irreversible. If friction brings your car to a halt and you wait for days, you won’t be able to say when the car stopped. The information disappears: it dissipates.

— The arrow of time: understood for 100 years

— Lubos Motl

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If there is a god-view, there is no time arrow.

Time arrow only exists from a macroscopic point of view. Microscopically, there is no time arrow.

If there is a god-view that can observe all the pieces of the exact information, including the microscopic ones, there is no time arrow.

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Also, if there is a god-view, there will be paradoxes, such as the black hole information paradox.

Black hole complementarity is a conjectured solution to the black hole information paradox, proposed by Leonard Susskind, Larus Thorlacius, and Gerard ‘t Hooft.

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Leonard Susskind proposed a radical resolution to this problem by claiming that the information is both reflected at the event horizon and passes through the event horizon and cannot escape, with the catch being no observer can confirm both stories simultaneously.

— Wikipedia on Black hole complementarity

The spirit of black hole complementarity is that there is no god-view. Instead, physics is always about what an observer can observe.

— Me@2018-06-21 01:09:05 PM

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2019.02.11 Monday (c) All rights reserved by ACHK

Pandemonium

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E: Can you just, you know, like, just tell me the answer?

J: Sorry?

E: You know, the answer. To everything.

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What’s the point of love if it’s just disappear?

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There has to be meaning to existence, otherwise the universe is made of pain and I don’t like the thought of that.

So, tell me the answer!

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J: The more human I become, the less things make sense.

But that’s part of the fun. Right?

E: What do you mean?

J: If there were an answer I can give you to, how the universe works, it wouldn’t be special. It would be just a machinery fulfilling its cosmic design. It would be just a big, dumb food processor.

But, since nothing seems to make sense, when you find something or someone that does, it’s euphoria.

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In all of this randomness, in this pandemonium, you and Chidi found each other and you had a life together.

Isn’t that remarkable?

E: Pandemonium is from Paradise Lost. Milton called the center of hell “pandemonium”, meaning “place of all demons”.

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I guess all I can do is to embrace the Pandemonium.

Find happiness in the unique insanity of being here, now.

— The Good Place

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2019.01.27 Sunday ACHK

（反對）開夜車 2.2

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Ken Chan 時光機 1.4.2.2

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The light that burns twice as bright burns half as long, …

— Blade Runner (1982)

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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）。

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長話短說，我是反對這個方法的，理由有很多，例如：

第一，長期睡眠不足，會導致腦袋呆了，大大減低下一天的工作效率。

第二，長期睡眠不足，會導致身體時常病。

在 Ken Chan 的部分課堂，特別是假期的補課中，我就見到他時常咳嗽。可能，他在工作時代，仍然過著，極度忙碌的生活。

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還記得在九月，在課程的第一課，他就已經遲到，遲了大概 15 至 30 分鐘之間。那課原定於早上八時半開始。上課中途，他有點咳嗽。他說，因為當天早上之前，吸入了太多的氮氣。他懷疑一位同事，作弄了他。

話說，他工作的地方，有一部儀器壞了。他的同事跟他說：「部機好似壞咗，漏緊啲 nitrogen 喎，你聞下係唔係？」（那儀器疑似壞了，正在洩漏氮氣。你聞一聞，看看是不是。）

Ken Chan 事後回心一想，才醒起：「Nitrogen 又點會有味咖？」（氮氣又何來會有味道呢？）

結果，他那天只睡了，一個半小時。

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我那時以為，那沒有什麼大不了。但是現在，我也已一把年紀了，才驚覺，只要有一點不幸，那就可以造成，嚴重的後果。

— Me@2019-02-03 07:02:42 AM

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2019.02.03 Sunday (c) All rights reserved by ACHK

SICM

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d_2019_02_02__12_27_49_PM_

Structure and Interpretation of Classical Mechanics (SICM) is a classical mechanics textbook written by Gerald Jay Sussman and Jack Wisdom with Meinhard E. Mayer. The first edition was published by MIT Press in 2001, and [the] second edition was released in 2015. The book is used at the Massachusetts Institute of Technology to teach a class in advanced classical mechanics, starting with Lagrange’s equations and proceeding through canonical perturbation theory.

— Wikipedia on Structure and Interpretation of Classical Mechanics

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2019.02.02 Saturday ACHK

DO

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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

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(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

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Exercise 4.5

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

— Me@2019-01-30 09:58:30 PM

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2019.01.30 Wednesday ACHK

Problem 14.5d3

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

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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}.

~~~

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— This answer is my guess. —

~~~

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

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\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}}

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\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}}

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\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}}

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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

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2019.01.27 Sunday (c) All rights reserved by ACHK

Quantum logic, 3

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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

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2019.01.26 Saturday ACHK

PhD, 3.2

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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。

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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

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2019.01.22 Tuesday (c) All rights reserved by ACHK

duplicate

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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

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Exercise 4.4

Without using the existing function member, define duplicate as in

>(duplicate ’a ’(a b c a d))
(A D)

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— 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

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2019.01.21 Monday (c) All rights reserved by ACHK

Problem 14.5d2

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

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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}.

~~~

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— This answer is my guess. —

~~~

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The left R’+ sector:

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\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}}

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\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}}

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\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}}

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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}}

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“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}}

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\displaystyle{ \begin{aligned} \alpha' M_L^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}+39739981824 \, x^{6} + ... \end{aligned}}

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~~~

— This answer is my guess. —

— Me@2019-01-20 09:09:37 PM

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2019.01.20 Sunday (c) All rights reserved by ACHK

Logical arrow of time, 6.4

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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.

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— 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 —

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— Me@2013-10-25 3:33 AM

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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

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2019.01.19 Saturday (c) All rights reserved by ACHK

（反對）開夜車 2.1

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Ken Chan 時光機 1.4.2.1

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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。

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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。

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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。

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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）。

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長話短說，我是反對這個方法的，…

— Me@2019-01-18 03:47:50 PM

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2019.01.19 Saturday (c) All rights reserved by ACHK

Equality Predicates

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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.

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[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.)

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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.

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[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.

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[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.

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[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.

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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.

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2019.01.16 Wednesday ACHK

Problem 14.5d1.1.2

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

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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}.

~~~

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— 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}

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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}}

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\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}}

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“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}.

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\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}} + ...}

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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}}

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\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}}

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The left R’+ sector:

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

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— This answer is my guess. —

— Me@2019-01-14 04:28:10 PM

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2019.01.14 Monday (c) All rights reserved by ACHK

EPR paradox, 5.3

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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

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2019.01.14 Monday (c) All rights reserved by ACHK