# 機遇創生論

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（安：除了創作那個大理論的名字外，還有什麼話題？）

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（安：不行。我都想過這個名字。但它非常誤導。「合體機械人」只是比喻。整個理論和機械人無關。）

「副作用機械人」？

（安：不如直接用你的名字，命名那個理論。）

「種子論」是起點，「果實論」是終點。然後，每個果實內，其實又有很多新種子。

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「緣份管理學」？

（安：我覺得「管理學」好像令個理論降級了一點，因為一般而言，「管理學」並不是，太高深的學問。）

「緣份機械人」？

（安：叫做「人生攻略理論」？）

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「緣份攻略」都不行，因為感覺有點怪。

（安：那就不如叫做「緣份理論」。）

「理論」很空泛。不應把「理論」，視為名字的一部分。

— Me@2020-01-29 12:23:38 AM

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# 1986

I keep this picture here as a backup.

— Me@2020-01-27 06:33:01 PM

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# scmutils, 2.2

Either use an older version of scmutils in order to follow the previous instructions for setting up Emacs for scmutils, or give up using Emacs for scmutils for the time being.

Using command line is the best way to go, so far.

The “using command line” method does not really work.

Within the MIT Scheme environment, it is not the original command line (bash) anymore. I can neither repeat the last command by just pressing the up key once, nor select the last command by mouse in order to copy it.

So I think I have to use an older version of scmutils.

— Me@2020-01-26 07:38:31 PM

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# Frequency probability and Bayesian probability, 2.2

The probability frequentist vs Bayesian debate can be transcended by realizing that an observer cannot be separated from the observed.

— Me@2011.07.23

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# Bus Stop 5

I am the first in the queue because I have missed the last bus.

— Me@2009.09.13

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If you are the smartest person in the room, you are in the wrong room,

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for you should have gone to a better room.

— Me@2019-12-21 07:11:21 PM

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for you deserve a better room.

— Me@2020-01-25 04:47:13 PM

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# Ken Chan 時光機 3.3

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（問：你研究院碩士畢業時，才二十四歲。為什麼要在二十九歲時，才有所領悟？）

「二十九歲時才有所領悟」的意思是，我在那時才有一個，（我覺得）完整的讀書策略體系，適合大部分人使用。

（問：你又怎樣知道，你的讀書策略體系完整，適合大部分人使用？）

What is most personal is most general.

— Carl Rogers

（問：那些是什麼問題？）

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— Me@2020-01-20 10:26:20 PM

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# 1990

First photo

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— Me@2020-01-08 12:11:16 AM

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# Quick Calculation 13.2

A First Course in String Theory

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Verify that

$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I + {X^I}' \right)}$,

$\displaystyle{\left[ L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I - {X^I}' \right)}$,

~~~

Equation (13.24):

$\displaystyle{X^{\mu} (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}$

Equation (13.39):

$\displaystyle{{\dot{X}}^- + {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \bar \alpha_n^- e^{-in (\tau + \sigma)}}$

$\displaystyle{{\dot{X}}^- - {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \alpha_n^- e^{-in (\tau - \sigma)}}$

Equation (13.51):

$\displaystyle{\left[\bar L_m^\perp, \bar \alpha_n^J \right] = - n \bar{\alpha}_{m+n}^J}$

$\displaystyle{\left[L_m^\perp, \alpha_n^J \right] = - n \alpha_{m+n}^J}$

Equation (13.52):

$\displaystyle{\left[L_m^\perp, \bar \alpha_n^J \right] = 0}$

$\displaystyle{\left[\bar L_m^\perp, \alpha_n^J \right] = 0}$

Equation (13.53):

$\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m}$

$\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}$

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$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right]}$,

$\displaystyle{= \left[ \bar L_0^\perp, x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^I e^{i n \sigma} + \bar \alpha_n^I e^{-in \sigma}) \right]}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} e^{-in \sigma} (-n \bar \alpha_n^I)}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 - i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} e^{-in(\tau + \sigma)} ( \bar \alpha_n^I)}$

$\displaystyle{= - \frac{i}{2} \sqrt{2\alpha'} \sum_{n \in \mathbb Z} \bar \alpha_n^I e^{-in(\tau + \sigma)}}$

— Me@2020-01-06 11:30:38 PM

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# Compare results

Within a universe, any two observers can, at least in principle, compare results.

When they compare, they will have consistent results for any observables/measurables.

— Me@2018-02-06 08:48:24 PM

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being entangled ~ being consistent, with respect to any two observers, when they compare the results

— Me@2018-02-05 10:11:45 PM

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# 時間止血 1.6.2

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— Me@2013.06.25
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— Me@2019-12-24 11:15:13 AM

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