Euler problem 23.3

Watermelon 3

.

isAbundant[n_] :=
    Module[{sumDivisors},
        sumDivisors = Total[Divisors[n]] - n;
        sumDivisors > n]

abundantNumbers[a_, b_] :=
    Select[
        Range[a, b],
        isAbundant]

L := 28123

abundantList := abundantNumbers[1, L]

firstHalfAbundantList :=
    abundantNumbers[1, Quotient[L,2]]

abundantSums[aList_, bList_] :=
    DeleteDuplicates[
        Flatten[
            Table[
                aList[[i]] + bList[[j]],
                {i, Length[aList]},
                {j, Length[bList]}]]]

nonAbundantSums[limit_] :=
    Total[
        Complement[
            Range[1, limit],
            abundantSums[
                firstHalfAbundantList,
                abundantList]]]

result = Timing[nonAbundantSums[L]]

{1.73236, 4179871}

— Me@2024-10-20 01:49:38 PM

.

.

2024.10.25 Friday (c) All rights reserved by ACHK

Table border

所有第一次, 2

.

<table border="1" style="border-collapse: collapse;width: 70%;margin: 20px auto">
  <thead>
    <tr>
      <th style="border: 1px solid #ddd;padding: 12px;text-align: left;background-color: #f2f2f2;color: #333">Duration</th>
      <th style="border: 1px solid #ddd;padding: 12px;text-align: left;background-color: #f2f2f2;color: #333">Subject</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td style="border: 1px solid #ddd;padding: 12px">3 hours</td>
      <td style="border: 1px solid #ddd;padding: 12px">理科</td>
    </tr>
    <tr>
      <td style="border: 1px solid #ddd;padding: 12px">1 hour</td>
      <td style="border: 1px solid #ddd;padding: 12px">語文</td>
    </tr>
    <tr>
      <td style="border: 1px solid #ddd;padding: 12px">1 hour</td>
      <td style="border: 1px solid #ddd;padding: 12px"></td>
    </tr>
    <tr>
      <td style="border: 1px solid #ddd;padding: 12px">3 hours</td>
      <td style="border: 1px solid #ddd;padding: 12px">文科</td>
    </tr>
  </tbody>
</table>
Duration Subject
3 hours 理科
1 hour 語文
1 hour
3 hours 文科

這個心結的本質就在於,和維珍在一起的時候,她所有的經歷,都是[只]跟你有關。那從青澀到成熟,都是只有你的參與。這種經歷,是可以支撐後面,漫漫的人生路。

維珍是把她自己,所有的一切,[都]給了你。而面對英航的時候,腦海當中,想起那段感情和經歷,不是跟自己的,就會覺得憋屈難受。

— 軒語說媒

.

.

2024.10.23 Wednesday ACHK

Euler problem 23.2

import Data.Array as Array ( Array )
import Data.Array.Base ( (!), listArray )
import Data.List (group)

primes :: [Int]
primes = 2 : filter (null . tail . (pfa !)) [3, 5 ..]

ub :: Int
ub = 28123

pfa :: Array Int [Int]
pfa = listArray (2, ub) $ map pf [2..ub]

pf :: Int -> [Int]
pf n = factor n primes
  where
    factor n (p : ps)
      | p * p > n = [n]
      | n `mod` p == 0 = p : pfa ! (n `div` p)
      | otherwise = factor n ps

groupFactors :: [Int] -> [[Int]]
groupFactors xs = [[head x, length x] | x <- group xs]

genDiv :: [[Int]] -> [Int]
genDiv = foldr combine [1]
  where
    combine [p, n] acc
      = [x * p^e | x <- acc, e <- [0..n]]

sumProperDivisors :: Int -> Int
sumProperDivisors n
  = -n + sum (genDiv $ groupFactors $ pfa ! n)

isAbundant :: Int -> Bool
isAbundant n = sumProperDivisors n > n

abdSieve :: Array Int Bool
abdSieve = listArray (2, ub) $ map isAbundant [2..ub]

abundantNumbers :: Int -> [Int]
abundantNumbers m
  | m > ub = filter isAbundant [2..m]
  | otherwise = filter (abdSieve !) [2..m]

abundantList :: [Int]
abundantList = abundantNumbers ub

isAbundantSum :: Int -> Bool
isAbundantSum n 
  = any (\a -> abdSieve ! (n-a))
    $ takeWhile (<= n `div` 2) abundantList

notAbundantSumsList :: Int -> [Int]
notAbundantSumsList m
  = filter (not . isAbundantSum) [1..m]

λ> :set +s
λ> sum (notAbundantSumsList ub)
4179871
(1.14 secs, 1,772,425,584 bytes)
λ> 

— Me@2024-10-19 09:53:30 PM

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.

2024.10.21 Monday (c) All rights reserved by ACHK

Ex 1.32 Path functions and state functions, 1

Structure and Interpretation of Classical Mechanics

.

In classical mechanics, there are scenarios where the path function cannot be reconstructed from a finite number of local tuples, such as time t, position q, velocity v, and acceleration a. One notable example involves non-conservative forces and chaotic systems.

Example: Chaotic Motion

Consider a system exhibiting chaotic behavior, such as the double pendulum. In this system, the motion is highly sensitive to initial conditions, meaning that even small changes in the initial state can lead to vastly different trajectories over time.

Local Information Limitations

  1. Finite Local Tuples: If you only have a finite number of measurements (e.g., position and velocity at discrete time intervals), you may not capture the full complexity of the system’s dynamics. For instance, knowing the position and velocity at a few points does not provide enough information to reconstruct the entire path due to the chaotic nature of the motion.

  2. Path Reconstruction Failure: In chaotic systems, the trajectory can diverge significantly from nearby trajectories. Thus, even if you have the position q(t) and velocity v(t) at several time points, the underlying dynamics can lead to a completely different path that cannot be predicted or reconstructed from those finite local tuples.

Conclusion

In summary, chaotic systems like the double pendulum illustrate how the path function cannot be reconstructed from a finite number of local tuples due to their sensitive dependence on initial conditions and the complexity of their dynamics. This highlights the limitations of local information in capturing the full behavior of certain mechanical systems.

— AI

.

.

2024.10.18 Friday (c) All rights reserved by ACHK

DSE 數學補邊個好?(神器五)

這篇文章繼續討論,選擇數學補習老師的一些準則。而這裡的數學,是指香港中學文憑試的數學科,包括核心,M1 及 M2。

如果你問我物理補習推介,特別是物理補習天王的話,我會答:「補 physics,補 Ken Chan!」但是,如果你問我數學補習天王的話,恕沒有推介,因為本人沒有遇過。

那樣,數學科補習界如果有 Ken Chan,他又必須具備,什麼條件呢?

上文提到:

1. 背誦篇

理科是文科,但不是一般的文科,而是極超巨大級的文科。

1.1 他會提醒你,數學其實是,極超大量的背誦。

……

1.2 極超大量的背誦,來自極端深刻的理解。

……

2. 年月時間篇:歷屆試題

2.1 他會提議你,考試前要操練 10 年、20 年 或 30 年的歷屆試題。

……

2.1.2 他會告訴你「八二法則」——頭八十分,只需兩成的付出;而尾二十分,則需佔八成的努力。亦即是話,付出和收穫,不成比例。

……

2.2 最近五年的題目,則從來不要看,因為你要保留它們,在公開試前的休假月時,來作模擬考試用。

2.3 最近五年以前的題目,則可以在課程的那三年間,隨時隨地做。

留意,做試題有兩大主要方法:「分課題」或「分年份」。

……

2.4 而歷屆試題的來源是:……

3. 星期時間篇

他會告訴你,如何從平衡到駕馭,各科溫習時間之爭。

……

4.0 分秒時間篇

除了日常生活中的,溫習時間管理外,他亦會教你,考試答題時的時間策略。那都是要在事先,學習和訓練的。

有關詳情,請點擊本綱誌之文章分類

exam time | 戰鬥意志

然後於網址結尾,加以下字串:

?order=asc

例如,剛才提到的「八二原則」指出了:同樣的分數,在不同的部分,難易不同。所以,考試時應先做最易的題目。

留意,題目的預設次序,未必是由易至深。其中一個原因是,各人對即使是同一道題目,深淺的感覺,未必相同。所以,必須在考試前鐵定,「唔識就飛」的好習慣:遇到不懂的題目時,不要思考,改為立刻跳過,去先做懂做的題目;最後還有時間剩餘時,才回到不懂的題目。

4.1 他會教你如何高速作答。

留意,高速不是急促。高速的原因在於,你掌握的數學知識,深於常人所及;繼而,你駕馭的運算技巧,高過他人所能。

每個課題的深知識,和對應的速算技巧,也可以不同,也可以是該課獨家。亦即是話,每種題目的速算技巧,都要遠在考試前就,事先學習和訓練好。

4.2 同理,每課題目的驗算技巧,亦都是那樣。

4.3 他亦會教你,考試作答時的表達技巧。自以為「數學和理科,毋須語文能力」的,是狂妄。

4.\infty 為了方便記憶,背了這公式:

數學考試

= 兩成數學 + 八成人格(速度+準確度+表達)

公式中的「準確度」,是指「數字準確度」或「運算準確度」。而「表達」,則指「文字準確度」或「語文準確度」。數學考試成績反映的,大概而言,只有兩成是數學中的,所謂「純知識」;其餘八成,都是人格的考驗。

有些數學知識,只在考試有益;而人格才能,則一生受用。

5.

……

— Me@2024-10-10 08:01:07 PM

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2024.10.12 Saturday (c) All rights reserved by ACHK

Euler problem 23.1.2

(defmacro abundant-numbers (limit)
  `(filter (macro-to-function is-abundant)
           (gen-list 1 ,limit)))

(defmacro none (predicate lst)
  `(not (some ,predicate ,lst)))

(defmacro take-while (predicate lst)
  `(labels ((take-while-iter (a-lst acc)
              (if (null a-lst)
                  (nreverse acc)
                  (let ((head (car a-lst))
                        (tail (cdr a-lst)))
                    (if (funcall ,predicate head)
                        (take-while-iter tail (cons head acc))
                        (nreverse acc))))))
     (take-while-iter ,lst '())))

(defmacro drop-while (predicate lst)
  `(labels ((drop-while-iter (a-lst)
              (if (null a-lst)
                  a-lst
                  (let ((head (car a-lst))
                        (tail (cdr a-lst)))
                    (if (funcall ,predicate head)
                        (drop-while-iter tail)
                        a-lst)))))
     (drop-while-iter ,lst)))

(defun create-hash-from-list (lst)
  (let ((hash (make-hash-table :test 'equal)))
    (dolist (item lst)
      (setf (gethash item hash) t))
    hash))

(defun is-not-abundant-sum (n abundant-list)
  (let* ((half-n-floor (floor (/ n 2)))
         (first-half (take-while #'(lambda (x) (<= x half-n-floor)) abundant-list))
         (second-half (drop-while #'(lambda (x) (< x half-n-floor)) abundant-list))
         (second-half-to-n (take-while #'(lambda (x) (< x n)) second-half))
         (second-half-hash (create-hash-from-list second-half-to-n)))
    (none (lambda (a)
            (let ((b (- n a)))
              (gethash b second-half-hash)))
          first-half)))

(defun not-abundant-sums-list (limit)
  (let ((abundant-list (abundant-numbers limit)))
    (filter #'(lambda (n)
                (is-not-abundant-sum n abundant-list))
            (gen-list 1 limit))))

CL-USER> (time (sum (not-abundant-sums-list 28123)))
 Evaluation took:
  2.481 seconds of real time
  2.477531 seconds of total run time (2.389778 user, 0.087753 system)
  [ Run times consist of 0.152 seconds GC time, and 2.326 seconds non-GC time. ]
  99.88% CPU
  6,194,100,940 processor cycles
  7,411,226,352 bytes consed
  
4179871

— Me@2024-10-04 03:38:12 PM

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2024.10.05 Saturday (c) All rights reserved by ACHK

xrandr

Quantum Mechanics: End Game

.

physical definition

~ define microscopic events in terms of observable physical phenomena, such as the changes in readings of the measuring device

~ define unobservable events in terms of observable events

— Me@2022-01-31 08:33:01 AM
 

.

I give you the final answer to this topic. You should continue to study this topic until you understand that.

A mixed state exists due to the lack of information.

A superposition state exists not due to that.

A superposition state exists due to the lack of a physical definition.

An eigenstate and a superposition state are both pure states.

A superposition state is neither an AND nor an OR because it is a pure state, meaning that it is one single quantum state, not a mixture of two.

A mixed state is an OR.

xrandr --output DP-0 --auto --output HDMI-0 --left-of DP-0

In the realm of knowledge, I share with you,
The final answer, clear and true.
Study deeply, let curiosity flow,
Until this truth, you truly know.

A mixed state emerges, shrouded in doubt,
Born from the shadows where facts are left out.
Yet a superposition, distinct in its grace,
Doesn’t arise from that uncertain place.

It finds its definition in the absence of form,
A concept elusive, where ideas transform.
Both eigenstates and superpositions, pure,
In the quantum dance, their essence is sure.

Neither an AND nor an OR, it stands alone,
One single state, in the vastness, it’s grown.
Not a blend of two, but a singular hue,
While a mixed state proclaims, “I am an OR,” too.

— Me@2024-08-19 03:23:59 PM

— Me@2024-09-30 03:40:25 PM

.

.

2024.10.01 Tuesday (c) All rights reserved by ACHK

DSE 數學補邊個好?(神器四)

這篇文章繼續討論,選擇數學補習老師的一些準則。而這裡的數學,是指香港中學文憑試的數學科,包括核心,M1 及 M2。

如果你問我物理補習推介,特別是物理補習天王的話,我會答:「補 physics,補 Ken Chan!」但是,如果你問我數學補習天王的話,恕沒有推介,因為本人沒有遇過。

那樣,數學科補習界如果有 Ken Chan,他又必須具備,什麼條件呢?

上文提到:

1. 背誦篇

理科是文科,但不是一般的文科,而是極超巨大級的文科。

1.1 他會提醒你,數學其實是,極超大量的背誦。

……

1.2 極超大量的背誦,來自極端深刻的理解。

……

2. 年月時間篇:歷屆試題

2.1 他會提議你,考試前要操練 10 年、20 年 或 30 年的歷屆試題。

……

2.1.2 他會告訴你「八二法則」——頭八十分,只需兩成的付出;而尾二十分,則需佔八成的努力。亦即是話,付出和收穫,不成比例。

……

2.2 最近五年的題目,則從來不要看,因為你要保留它們,在公開試前的休假月時,來作模擬考試用。

2.3 最近五年以前的題目,則可以在課程的那三年間,隨時隨地做。

留意,做試題有兩大主要方法:「分課題」或「分年份」。

……

2.4 而歷屆試題的來源是:……

3. 星期時間篇

他會告訴你,如何從平衡到駕馭,各科溫習時間之爭。

其方法是,用自動時間表——透過預先設計好的時間程式,每天自動生成的時間表。每天你毋須花額外時間,人手編配時間表,考慮溫什麼。

該機械化時間表,會自動編排和分配,各科的溫習時間和長度。同一天會由,不同類型科目組合而成,導致腦部的不同部分,可以輪流休息,即使整體的工作時數很多。

該系統化時間表,會有適當分割,導致溫習不會辛苦。

有關詳情,請點擊本綱誌之文章分類

in-time-時來運到

然後於網址結尾,加以下字串:

?order=asc

記住,刻苦不能長久,因為你的潛意識,會無時無刻,找機會「逃走」。那就是「疲倦」的真正動機。

你工作系統的目標,並不是「控制自己」——控制不想讀書的自己讀書,勉強自己做不想做的事;而是「釋放自己」——釋放學習動機,爆發學術潛能。

頭腦清醒時,靈感才會自來,天份才能發揮。

4. 分秒時間篇

……

5.

……

6.

……

— Me@2024-09-28 11:11:52 AM

.

.

2024.09.29 Sunday (c) All rights reserved by ACHK

Euler problem 23.1.1

(defmacro sum (lst)
  `(reduce #'+ ,lst))

(defun proper-divisors (n)
  (when (> n 1)  
    (let ((divisors '())
          (limit (floor (sqrt n))))  
      (loop :for i :from 1 :to limit
            :when (zerop (mod n i))  
              :do (progn
                    (push i divisors)  
                    (when (/= n (floor n i))  
                      (push (floor n i)
                            divisors))))  
      (remove-duplicates (sort divisors #'<)
                         :test
                         #'equal))))

(defmacro sum-proper-divisors (n)
  `(sum (proper-divisors ,n)))

(defmacro is-abundant (n)  
  `(> (sum-proper-divisors ,n) ,n))

(defmacro gen-list (min max)
  `(loop :for n :from ,min :to ,max
         :collect n))

(defmacro filter (predicate list)
  `(remove-if-not ,predicate ,list))

(defmacro macro-to-function (macro-name)
  `(lambda (xs) (,macro-name xs)))

(defmacro db (x)
  `(* 2 ,x))

((lambda (xs) (db xs)) 2.1)

(funcall (macro-to-function db) 2.1)

;; Evaluation Context:

;; ((macro-to-function db) 2.1): This line attempts to call the result of the macro macro-to-function directly as if it were a function. However, since macro-to-function returns a lambda expression, this results in an "illegal function call" error because the macro is not expanded in the context of a function call.

;; (funcall (macro-to-function db) 2.1): In this line, funcall is used to invoke the lambda function returned by macro-to-function. This correctly evaluates the lambda and applies it to 2.1, allowing the macro to be expanded properly.

(defmacro abundant-numbers (limit)
  `(filter (macro-to-function is-abundant) (gen-list 1 ,limit)))

— Me@2024-09-20 11:53:30 PM

.

.

2024.09.21 Saturday (c) All rights reserved by ACHK

Sub ExportCalendarToOrgMode()

Sub ExportCalendarToOrgMode()
    Dim objFolder As Outlook.Folder
    Dim objAppointment As Outlook.AppointmentItem
    Dim strOutput As String
    Dim objFile As Object
    Dim strFilePath As String
    Dim totalAppointments As Integer

    strFilePath = "Z:\media\d\Outl.org"

    Set objFolder = Application.Session.GetDefaultFolder(olFolderCalendar)

    For Each objAppointment In objFolder.Items
        strOutput = strOutput & "* " & objAppointment.Subject & vbCrLf
        strOutput = strOutput & "SCHEDULED: <" & Format(objAppointment.Start, "yyyy-MM-dd>") & vbCrLf
        strOutput = strOutput & "DESCRIPTION: " & objAppointment.Body & vbCrLf
        strOutput = strOutput & vbCrLf

        totalAppointments = totalAppointments + 1
    Next objAppointment

    Set objFile = CreateObject( "Scripting.FileSystemObject" ).CreateTextFile(strFilePath, True)
    objFile.Write strOutput
    objFile.Close

    MsgBox "Calendar exported to: " & strFilePath & vbCrLf & "Total appointments exported: " & totalAppointments, vbInformation
End Sub

Sub ShiftCalendarEvents()
    Dim calFolder As Object
    Dim calItems As Object
    Dim calItem As Object
    Dim startDate As Date
    Dim endDate As Date
    Dim shiftDays As Integer

    shiftDays = 7

    ' 9 represents olFolderCalendar in Outlook 2010
    Set calFolder = Application.Session.GetDefaultFolder(9)

    Set calItems = calFolder.Items

    For Each calItem In calItems
        If calItem.Class = 26 Then ' 26 represents olAppointment in Outlook 2010
            calItem.Start = DateAdd("d", shiftDays, calItem.Start)
            calItem.Duration = 60
            calItem.Save
        End If
    Next calItem

    MsgBox "Calendar events have been shifted by " & shiftDays & " days."
End Sub

— Me@2024-09-14 12:47:16 PM

.

.

2024.09.15 Sunday (c) All rights reserved by ACHK

Ex 1.31 Velocity transformation

Structure and Interpretation of Classical Mechanics

.

Use the procedure Gamma-bar to construct a procedure that transforms velocities given a coordinate transformation. Apply this procedure to the procedure p->r to deduce (again) equation (1.65).

~~~

(define ((Gamma-bar f-bar) path-q-local-tuple)
  (let* ((tqva path-q-local-tuple)
         (t (time tqva))         
         (O-tqva (osculating-path tqva)))   
    ((f-bar O-tqva) t)))

(define (F->C F)
  (define (f-bar q-prime)
    (define q
      (compose F (Gamma q-prime)))
    (Gamma q))
  (Gamma-bar f-bar))

(define (F->v F)
  (compose velocity (F->C F)))

(show-expression 
 ((F->v p->r)
  (->local 't
           (up 'r 'theta)
           (up 'rdot 'thetadot))))

— Me@2024-09-13 07:17:24 PM

.

.

2024.09.13 Friday (c) All rights reserved by ACHK

Euler problem 22.2

import Data.Char (ord)
import Data.List (sort)

score word
  = sum $ map (\char -> ord char - ord 'A' + 1) word

nameScore filename = do
  content <- readFile filename
  let    
    names = read $ "[" ++ content ++ "]"       
  return $
    sum $ zipWith (*) [1..] $ map score $ sort names

λ> nameScore "names.txt"
871198282
λ> 

— Me@2024-09-09 03:40:11 PM

.

.

2024.09.10 Tuesday (c) All rights reserved by ACHK

Euler problem 22.1

(ql:quickload "str")

(defun score (word)
  (loop :for char :across word
        sum (+ (- (char-code (char-upcase char))
                  (char-code #\A))
               1)))

(defun remove-quote (name)
  (remove #\" name))

(defmacro zipWith (f xs ys)
  `(mapcar ,f ,xs ,ys))

(defmacro range (max &key (min 0) (step 1))
  `(loop :for n :from ,min :below ,max :by ,step
        collect n))

(defun name-scores (filename)
  (with-open-file (stream filename)
    (let* ((names (read-line stream))
           (name-list-q (str:split #\, names))
           (name-list (mapcar #'remove-quote name-list-q))
           (sorted-name-list (sort name-list #'string<))
           (score-list (mapcar #'score sorted-name-list))
           (n-list (range (1+ (length score-list)) :min 1))
           (i-score-list (zipWith #'* n-list score-list)))
      (reduce #'+ i-score-list))))

(name-scores "names.txt")

; SLIME 2.28To load "str":
  Load 1 ASDF system:
    str
; Loading "str"
...

CL-USER> (name-scores "names.txt")
871198282
CL-USER> 

— Me@2024-09-04 10:43:19 AM

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2024.09.04 Wednesday (c) All rights reserved by ACHK

CSS, 4.2.2

htmlize, 2.2.2

.

<pre style="color: #333333;background-color: #f8f8f8;font-size: 16px">
...
</pre>
<pre style="color: #333333;background-color: #f8f8f8;font-size: 16px;font-family: DejaVu Sans Mono">
...
</pre>

— Me@2024-08-27 01:24:01 PM

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2024.09.01 Sunday (c) All rights reserved by ACHK

Posted in CSS

KDE Chinese, 2.2

<div style="color: #333333; font-size: 20px; font-family: AR PL UKai TW">
楷書測試
</div>
楷書測試

The above Chinese characters should be displayed in the regular script font (楷書). Otherwise, you need to install this package on your Linux distribution:

fonts-arphic-ukai

— Me@2024-08-13 01:41:14 PM

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2024.08.30 Friday (c) All rights reserved by ACHK

1.9 Abstraction of Path Functions, 3.2

Structure and Interpretation of Classical Mechanics

.

\displaystyle{\begin{aligned}  f (t, q(t), v(t), \cdots, q^{(n)}(t)) &= f(\Gamma[q])(t) \\  \bar \Gamma (\bar f) (t, q(t), v(t), \cdots, q^{(n)}(t)) &= \bar f [q](t) \\  \end{aligned}}

3. The key point is that since f is also a function with the path tuple as input, we can regard F as one possible f. This allows us to use \bar f (= f \circ \Gamma) to define F \to C.

\displaystyle{\begin{aligned}  q &= F \circ (\Gamma[q']) \\  \bar f [q'] &= \Gamma[q] \\  &= \Gamma[F \circ \Gamma[q']]  \end{aligned}}

Note that it is not the definition of \bar f[q']. Instead, it is just an instance of it.

(define (F->C F)
  (define (f-bar q-prime)
    (define q
      (compose F (Gamma q-prime)))
    (Gamma q))
  (Gamma-bar f-bar))

\displaystyle{\begin{aligned}  f &= \bar \Gamma (\bar f) \\  \end{aligned}}

4. However, in the original definition,

(define ((F->C F) local)
  (->local (time local)
           (F local)
           (+ (((partial 0) F) local)
              (* (((partial 1) F) local)
                 (velocity local)))))

there are some partial differentiation operators. Where are they in the new definition?

Those partial differentiation operators are within the definition of the function osculating-path.

(define ((Gamma-bar f-bar) path-q-local-tuple)
  (let* ((tqva path-q-local-tuple)
         (t (time tqva))         
         (O-tqva (osculating-path tqva)))   
    ((f-bar O-tqva) t)))

— Me@2024-06-13 03:47:47 PM

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2024.08.28 Wednesday (c) All rights reserved by ACHK

CSS, 4.3

Equation within a blockquote:

.

\displaystyle{E_{\rm {rel}}=\sqrt{(m_{0}c^{2})^{2}+(pc)^{2}}}

.

To improve it, replace the LaTeX command with this one:

$latex \displaystyle{E_{\rm {rel}}=\sqrt{(m_{0}c^{2})^{2}+(pc)^{2}}}&s=1&bg=EAEFF3&fg=15100c$

.

\displaystyle{E_{\rm {rel}}=\sqrt{(m_{0}c^{2})^{2}+(pc)^{2}}}

.

— Me@2023-12-27 12:33:01 PM

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2024.08.20 Tuesday (c) All rights reserved by ACHK