# Ken Chan 時光機 1.3

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— Me@2018-10-31 09:39:05 AM

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# funcall

In Common Lisp, apply can take any number of arguments, and the function given first will be applied to the list made by consing the rest of the arguments onto the list given last. So the expression

(apply #’+ 1 ’(2))

is equivalent to the preceding four. If it is inconvenient to give the arguments as
a list, we can use funcall, which differs from apply only in this respect. This expression

(funcall #’+ 1 2)

has the same effect as those above.

— p.13

— On Lisp

— Paul Graham

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Exercise 7.1

Define funcall.

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(defmacro our-funcall (f &rest p)
(apply ,f (list ,@p)))



— Me@2018-10-30 03:24:05 PM

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# Problem 14.5b2

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

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

We have 32 zero modes $\displaystyle{\lambda_0^A}$ and 16 linear combinations behave as creation operators.

As usual half of the ground states have $\displaystyle{(-1)^{F_L} = +1}$ and the other half have $\displaystyle{(-1)^{F_L} = -1}$.

Let $\displaystyle{|R_\alpha \rangle_L}$ denote ground states with $\displaystyle{(-1)^{F_L} = +1}$.

How many ground states $\displaystyle{|R_\alpha \rangle_L}$ are there?

~~~

— This answer is my guess. — \displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states. Postulating a unique vacuum $\displaystyle{|0 \rangle}$, the creation operators allow us to construct $\displaystyle{16 = 2^4}$ degenerate Ramond ground states.”

Following the same logic:

Postulating a unique vacuum $\displaystyle{|0 \rangle}$, the creation operators allow us to construct $\displaystyle{2^{16}}$ degenerate Ramond ground states.

Therefore, there are $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha \rangle_L}$.

— This answer is my guess. —

— Me@2018-10-29 03:11:07 PM

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# Visualizing higher dimensions

The trick of visualizing higher dimension is: not to visualize it.

— Wikipedia

— Me@2011.08.19

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Besides trying to visualize, there are other methods to understand higher dimensions.

— Me@2018-10-28 04:28:01 PM

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What is the meaning of visualization?

— Me@2018-09-02 4:35 pm

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feel ~ receive all the data at once

(This definition is not totally correct, but is useful in the meantime.)

visual ~ feel at once through eyes

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you can visualize a 3D object ~ you can see all of a 3D object at once

you cannot visualize a 4D object ~ you cannot see all of a 4D object at once

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Actually, you can only visualize a 2D object, such as a square.

You cannot visualize a 3D object, such as a cube.

That’s why the screen of any computer monitor is 2 dimensional, not 3.

— Me@2018-10-28 04:32:41 PM

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# A pretty girl, 2

Women are directly fitted for acting as the nurses and teachers of our early childhood by the fact that they are themselves childish, frivolous and short-sighted; in a word, they are big children all their life long–a kind of intermediate stage between the child and the full-grown man, who is man in the strict sense of the word. See how a girl will fondle a child for days together, dance with it and sing to it; and then think what a man, with the best will in the world, could do if he were put in her place.

— Of Women

— Arthur Schopenhauer

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When the elderly Schopenhauer sat for a sculpture portrait by the Prussian sculptor Elisabet Ney in 1859, he was much impressed by the young woman’s wit and independence, as well as by her skill as a visual artist. After his time with Ney, he told Richard Wagner’s friend Malwida von Meysenbug, “I have not yet spoken my last word about women. I believe that if a woman succeeds in withdrawing from the mass, or rather raising herself above the mass, she grows ceaselessly and more than a man.

— Wikipedia on Arthur Schopenhauer

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Anybody can look at a pretty girl and see a pretty girl. An artist can look at a pretty girl and see the old woman she will become. A better artist can look at an old woman and see the pretty girl that she used to be. But a great artist — a master — and that is what Auguste Rodin was — can look at an old woman, portray her exactly as she is… and force the viewer to see the pretty girl she used to be…. and more than that, he can make anyone with the sensitivity of an armadillo, or even you, see that this lovely young girl is still alive, not old and ugly at all, but simply prisoned inside her ruined body. He can make you feel the quiet, endless tragedy that there was never a girl born who ever grew older than eighteen in her heart…. no matter what the merciless hours have done to her. Look at her, Ben. Growing old doesn’t matter to you and me; we were never meant to be admired — but it does to them. Look at her! (UC)

— Robert A. Heinlein

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2018.10.27 Saturday ACHK

# 凌晨舊戲 2

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（問：你閱讀過很多有關「瀕死經驗」的文章？）

（問：那你怎樣分辨，「瀕死經驗」的文章之中，哪些是真，哪些為假？）

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——

（問：那亦可能只是你的誤會。按常理，程度較低的人不會知道，所謂「程度高的人或說話」的真假。

——

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（留意，這裡的「判別其真假」只是「評價其可信度」的簡寫。剛才你也提到，要百分百地證明，某些言論是真的，是沒有可能的。）

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（問：但是，道理又怎樣為之「境界極高」呢？）

— Me@2018-10-26 08:48:21 PM

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# A Road to Common Lisp

tumba 57 days ago [-]

My advice is this: as you learn Common Lisp and look for libraries, try to suppress the voice in the back of your head that says “This project was last updated six years ago? That’s probably abandoned and broken.” The stability of Common Lisp means that sometimes libraries can just be done, not abandoned, so don’t dismiss them out of hand.

I have found this to be true in my own experience. The perception of stagnation is, however, a common initial objection to folks working in CL for the first time.

mike_ivanov 57 days ago [-]

My personal problem with CL libraries is not that, but rather the lack of documentation. More often than not, there is no documentation at all, not even a readme file.. It feels like some library authors simply don’t care. I’d say this attitude has a negative impact on how people perceive viability of those libraries — and by extension, of the language.

armitron 56 days ago [-]

A lot of libraries that don’t have separate documentation in the form of HTML/PDF/README.. are actually well-documented at source level in the form of docstrings.
Since Common Lisp is an interactive programming language and is meant to be used interactively (think Smalltalk, not Python) it is common practice to (interactively) load a library in a Common Lisp image and explore it (interactively). One can see what symbols are exported from packages, what these symbols are used for and (interactively) retrieve their documentation. All of this takes place within the editing environment (ideally Emacs) in a rapid feedback loop (again think Smalltalk, not Python) that feels seamless and tremendously empowering.

stevelosh 56 days ago [-]

I agree with this — it’s a real problem. My only solution has been to try to be the change I want to see in the world and document all of my own libraries before I officially release them.

— A Road to Common Lisp

— Hacker News

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2018.10.24 Wednesday ACHK

# Problem 14.5b

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

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

— This answer is my guess. —

The naive mass formula in the left R’ sector: \displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\ &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\ \end{aligned}}

. \displaystyle{ \begin{aligned} \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = -1, -2, ...} n \lambda_{-n}^A \lambda_n^A + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\ \end{aligned}}

p.316 Equation (14.51): \displaystyle{ \begin{aligned} \frac{1}{2} \sum_{n = -1. -2, ...} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} (D - 2) \\ \end{aligned}} \displaystyle{ \begin{aligned} \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A &= \left[ \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} \left[(D - 2)\right]_A \right] + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\ &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{32}{24} \\ &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \\ \\ \end{aligned}}

. \displaystyle{ \begin{aligned} \alpha' M_L^2 &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\ &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \left[ \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \right] \\ &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

. \displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

— This answer is my guess. —

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# Relational quantum mechanics

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Relational quantum mechanics (RQM) is an interpretation of quantum mechanics which treats the state of a quantum system as being observer-dependent, that is, the state is the relation between the observer and the system. This interpretation was first delineated by Carlo Rovelli in a 1994 preprint, and has since been expanded upon by a number of theorists. It is inspired by the key idea behind special relativity, that the details of an observation depend on the reference frame of the observer, and uses some ideas from Wheeler on quantum information.

,,,

Relational solution

In RQM, an interaction between a system and an observer is necessary for the system to have clearly defined properties relative to that observer. Since the two measurement events take place at spacelike separation, they do not lie in the intersection of Alice’s and Bob’s light cones. Indeed, there is no observer who can instantaneously measure both electrons’ spin.

The key to the RQM analysis is to remember that the results obtained on each “wing” of the experiment only become determinate for a given observer once that observer has interacted with the other observer involved. As far as Alice is concerned, the specific results obtained on Bob’s wing of the experiment are indeterminate for her, although she will know that Bob has a definite result. In order to find out what result Bob has, she has to interact with him at some time $t_{3}$ in their future light cones, through ordinary classical information channels.

The question then becomes one of whether the expected correlations in results will appear: will the two particles behave in accordance with the laws of quantum mechanics? Let us denote by $M_{A}(\alpha )$ the idea that the observer $A$ (Alice) measures the state of the system $\alpha$ (Alice’s particle).

So, at time $t_{2}$, Alice knows the value of $M_{A}(\alpha )$: the spin of her particle, relative to herself. But, since the particles are in a singlet state, she knows that $M_{A}(\alpha )+M_{A}(\beta )=0,$

and so if she measures her particle’s spin to be $\sigma$, she can predict that Bob’s particle ( $\beta$ ) will have spin $-\sigma$. All this follows from standard quantum mechanics, and there is no “spooky action at a distance” yet. From the “coherence-operator” discussed above, Alice also knows that if at $t_{3}$ she measures Bob’s particle and then measures Bob (that is asks him what result he got) — or vice versa — the results will be consistent: $M_{A}(B)=M_{A}(\beta )$

Finally, if a third observer (Charles, say) comes along and measures Alice, Bob, and their respective particles, he will find that everyone still agrees, because his own “coherence-operator” demands that $M_{C}(A)=M_{C}(\alpha )$ and $M_{C}(B)=M_{C}(\beta )$

while knowledge that the particles were in a singlet state tells him that $M_{C}(\alpha )+M_{C}(\beta )=0.$

Thus the relational interpretation, by shedding the notion of an “absolute state” of the system, allows for an analysis of the EPR paradox which neither violates traditional locality constraints, nor implies superluminal information transfer, since we can assume that all observers are moving at comfortable sub-light velocities. And, most importantly, the results of every observer are in full accordance with those expected by conventional quantum mechanics.

— Wikipedia on Relational quantum mechanics

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2018.10.22 Monday ACHK

# Ken Chan 時光機 1.2

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Ken Chan 所提及，其中一個技巧是，在物理科，如果新學一個課題，就應該先做，大量該個課題的 MC（多項選擇題）。那樣，你就可以極速釐清，該個課題中的新概念。

— Me@2018-10-20 11:46:04 AM

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# SLIME

SLIME, the Superior Lisp Interaction Mode for Emacs, is an Emacs mode for developing Common Lisp applications. SLIME originates in an Emacs mode called SLIM written by Eric Marsden. It is developed as an open-source public domain software project by Luke Gorrie and Helmut Eller. Over 100 Lisp developers have contributed code to SLIME since the project was started in 2003. SLIME uses a backend called Swank that is loaded into Common Lisp.

— Wikipedia on SLIME

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C-x o

Window-Move to other

C-x C-e

Evaluate last expression

C-c C-r`

Evaluate region

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2018.10.19 Friday (c) ACHK

# Problem 14.5a4

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

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

The GSO projection here keeps the states with $\displaystyle{(-1)^{F_L} = + 1}$; this defines the left NS’+ sector.

Write explicitly and count the states we keep for the three lowest mass levels, indicating the corresponding values of $\displaystyle{\alpha' M_L^2}$. [This is a long list.]

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p.314 “Let us declare that number to be minus one, thus making the ground states fermionic:”

Equation (14.39): $\displaystyle{(-1)^F |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle = - |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle}$

Equation (14.40): $\displaystyle{(-1)^F |\lambda \rangle = -(-1)^{\sum_{r,J} \rho_{r,J}} |\lambda \rangle}$

p.315 “So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = -1}$; they are fermionic states.”

However, in this problem:

“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 $\displaystyle{(-1)^F = +1}$.

— This answer is my guess. — \displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

If we define $N^\perp$ in the way similar to equation (14.37), we have \displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}} \displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}

Let $\displaystyle{N(n, k) = {n + k - 1 \choose k - 1}}$, the number of ways to put n indistinguishable balls into k boxes. \displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&1 \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&8 + \frac{32 \times 31}{2} = 504 \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\left( \frac{8 \times 8}{2} + \frac{8}{2} \right) = 36, 8 \times \left( \frac{32 \times 31}{2} \right) = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\ \end{aligned}} \displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&N(2,8) = 36, 8 \times {32 \choose 2} = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~& 36 + 3968 + 1024 + 35960 = 40988 \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-14 03:25:08 PM

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# How far away is tomorrow?

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The cumulative part of spacetime is time.

It is the cumulative nature of time [for an macroscopic scale] that makes the time a minus in the spacetime interval formula? $\displaystyle{\Delta s^{2} = - (c \Delta t)^{2} + (\Delta x)^{2} + (\Delta y)^{2} + (\Delta z)^{2}}$

— Me@2011.09.21

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Space cannot be cumulative, for two things at two different places at the same time cannot be labelled as “the same thing”.

— Me@2013-06-12 11:41 am

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There is probably no directly relationship between the minus sign and the cumulative nature of time.

Instead, the minus sign is related to fact that the larger the time distance between two events, the causally-closer they are.

— Me@2018-10-13 12:46 am

. — Distance and Special Relativity: How far away is tomorrow?

— minutephysics

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# 凌晨舊戲

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（問：你閱讀過很多有關「瀕死經驗」的文章？）

（問：那你怎樣分辨，「瀕死經驗」的文章之中，哪些是真，哪些為假？）

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— Me@2018-10-12 05:48:23 AM

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# Insomnia

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DrPhish 5 months ago [-]

This is exactly the Zen moment that fixed my lifelong insomnia: lying in bed relaxed with my eyes closed, even if I’m not asleep, leaves me feeling worlds better in the morning than fretting about not being able to sleep. It’s something that happened when I forgot about it completely. Once I realized I could literally just lie in bed for 8 hours and be OK, I was able to let go.

It’s like a trap that closes tighter about you the more you struggle, and releases if you allow yourself to relax[.]

Once I really got that, sleep became regular and uninterrupted[.]

laumars 5 months ago [-]

I discovered this recently too. Even in the instances when I still don’t get to sleep (eg my 1 year old coming into bed and wriggling in her sleep waking me) I’ve still felt more refreshed when I’ve relaxed rather than laid in bed wound up. I’m not into new age / whatever stuff but I like to think of it as night time meditation as while it’s still not as good as sleep it’s still a great deal better than full blown insomnia! And best of all, sometimes it clears your mind enough to actually sleep.

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— Thinking of yourself as an insomniac may be a part of the problem

— Hacker News

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2018.10.11 Thursday ACHK

# Problem 14.5a4

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

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

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. $\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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— This answer is my guess. —

. $\displaystyle{\alpha' M_L^2}$ $\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$ $\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$ $\displaystyle{= \frac{-1}{24} \left[ (D - 2) \right]_{I= 2, 3, ..., 9} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

. $\displaystyle{\alpha' M_L^2 = \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

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Equation (14.34): $\displaystyle{\frac{1}{2} \sum_{r=- \frac{1}{2}, -\frac{3}{2}} r b_{-r}^I b_{r}^I = \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r b_{-r}^I b_{r}^I - \frac{1}{48} (D - 2)}$

. $\displaystyle{\alpha' M_L^2}$ $\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \frac{1}{2} \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$ $\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \left[ \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r \lambda_{-r}^A \lambda_{r}^A - \frac{1}{48} \left[D - 2\right]_A \right]}$ $\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A - \frac{32}{48}}$

. \displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-10 05:38:08 PM

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# Existence and Description

Bertrand Russell, “Existence and Description”

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§1 General Propositions and Existence

“Now when you come to ask what really is asserted in a general proposition, such as ‘All Greeks are men’ for instance, you find that what is asserted is the truth of all values of what I call a propositional function. A propositional function is simply any expression containing an undetermined constituent, or several undetermined constituents, and becoming a proposition as soon as the undetermined constituents are determined.” (24a)

“Much false philosophy has arisen out of confusing propositional functions and propositions.” (24b)

A propositional function can be necessary (when it is always true), possible (when it is sometimes true), and impossible (when it is never true).

“Propositions can only be true or false, but propositional functions have these three possibilities.” (24b)

“When you take any propositional function and assert of it that it is possible, that it is sometimes true, that gives you the fundamental meaning ‘existence’…. Existence is essentially a property of a propositional function. It means that the propositional function is true in at least one instance.” (25a)

— Brandon C. Look

— University Research Professor and Chair

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2018.10.07 Sunday ACHK Simpson’s paradox, or the Yule–Simpson effect, is a phenomenon in probability and statistics, in which a trend appears in several different groups of data but disappears or reverses when these groups are combined. 