Franco-Anglo Mathematics Research Cooperation

French people are rational (think before action), English people are imperative (action before think).

17CE French Mathematician & Philosopher René Descarte : “I think therefore I am” 我思故我在 。He invented Analytical Geometry (xyz cartesian geometry).

De Moivre Theorem:
(cos \: {x} + i.sin \: {x})^{n} = cos \: {nx} + i.sin \: {nx}


French youngest President Emmanuel Macron and his Education 

Emmanuel Macron is the youngest French President (39) since Napoleon Bonaparte (40).

A brilliant student since young, he impressed his secondary school Drama teacher 24 years older, finally married her.

Like any genius (Einstein, Galois, Edison, …) who doesn’t adapt well in the traditional education system, Macron entered the prestigious and highly competitive Classe Préparatoire (Art Stream) Lycée Henri IV in Paris to prepare for the “Concours” (法国抄袭自中国的)”科举” Entrance Exams in France’s top Ecole Normale Supérieure (ENS). Like the 19CE Math genius Evariste Galois who failed the Ecole Polytechnique Concours twice in 2 consecutive years, Macron also failed ENS “Concours” in 2 consecutive years. 

He revealed recently,  “The truth was I didn’t play the game. I was too much in love (with my former teacher) for seriously preparing the Concours …”

Note: French traditional  name for the elitist tertiary education (first 2 or 3 years if repeat last year):  “Khâgne” is the name of Classe Préparatoire for Art Stream. The Classe Préparatoire for Science Stream is called “Taupe” (Mole 鼹鼠 ). 

Also some cute  names for:

  • Art students:
  1. 1st year: hypo-khâgne
  2. 2nd year: khâgne / carré  (square)
  3. 3rd year (repeat): cube (cubic)
  • Science students:
  1. 1st year: “1/2” (Mathématiques Supérieures)
  2. 2nd year: “3/2” (Mathématiques Spéciales)
  3. 3rd year (repeat): “5/2”

    His party “En Marche” did a survey on French Education: “The elitist national education system for the elites & rich families. ”  

    Emmanuel <=> Contract with God 上帝与他同在

    — ” 谋事在人, 成事在天 “

    ( “Man proposes, God disposes”)


    Rigorous Prépa Math Pedagogy

    The Classe Prépa Math for Grandes Écoles is uniquely French pedagogy – very rigorous based on solid abstract theories.

    In this lecture the young French professor demonstrates how to teach students the rigorous Math à la Française:

    \displaystyle {\lim_{n\to\infty} \bigl( 1 + \frac{1}{n} \bigr)^{n} = e} 

    1st Mistake:
    \displaystyle { \bigl( 1 + \frac{1}{n} \bigr) \xrightarrow [n\to\infty] {} 1 \implies \boxed {{\bigl( 1 + \frac{1}{n} \bigr)}^{n} \xrightarrow [n\to\infty] {} 1^{n} =1}} (WRONG!!)

    THEOREM 1 :
    \boxed { \text {If } {U}_{m} \xrightarrow [n\to\infty] {} \alpha \text{  and  } f(x) \xrightarrow [x \to\alpha] {} \ell \text { then } f (U_{m}) \xrightarrow [n\to\infty] {} \ell}

    Note: The ‘x’ in f (x) is {U}_{m} hence f is ‘fixed’ by a value \alpha

    In the above mistake:
    {U}_{m} = 1 + \frac{1}{n}
    f _{n} (1 + \frac{1}{n} ) \text { where } f_{n} : x \mapsto x^{n}
    {f}_{n} is not fixed, but depends on ‘n’. It is wrong to apply Theorem 1.

    2nd Mistake:
    \forall n \in \mathbb {N}^{*}, \bigl( 1 + \frac{1}{n} \bigr) > 1

    THEOREM 2:
    \boxed { \forall q > 1, q^{n} \xrightarrow [n\to +\infty] {} +\infty}
    Note: q is a fixed value.

    \text {Let  } q_{n} = 1 + \frac{1}{n}
    \boxed {{\bigl(1 + \frac{1}{n} \bigr)}^{n} \xrightarrow [n\to\infty] {} +\infty } (WRONG!!)

    Reason: q_{n} = \bigl( 1 + \frac{1}{n} \bigr) is not fixed value but depends on variable ‘n’. It is wrong to apply Theorem 2.

    3rd Mistake: Binomial

    \boxed { \displaystyle \forall n \in \mathbb {N}^{*}, {\bigl( 1 + \frac{1}{n} \bigr)}^{n} = \sum_{k=0}^{n} \binom {n}{k} {(\frac {1}{n})}^{k} . (1)^{n-k} = \sum_{k=0}^{n} \binom {n}{k} {(\frac {1}{n})}^{k} } 

    {\bigg[ \bigl( 1 + \frac{1}{n} \bigr)}^{n} = \binom {n}{0} {(\frac {1}{n})}^{0} + \binom {n}{1} {(\frac {1}{n})}^{1} + ... = 1 + 1 + ... > 2 \bigg]
    Expanding the binomial,
    \displaystyle \binom {n}{k} = \frac {n. (n-1). (n-2)... (n-k+1)}{k!} \sim_{n\to +\infty} \frac {n^k}{k!}

    \boxed {\displaystyle\binom {n}{k}.{(\frac {1}{n})}^{k} \sim_{n\to +\infty}\frac {n^k}{k!}. {(\frac {1}{n})}^{k} =\frac {1}{k!}} Note: valid if k is fixed value.

    \boxed {\displaystyle {\bigl( 1 + \frac{1}{n} \bigr)}^{n} = \sum_{k=0}^{n} \binom {n}{k} {(\frac {1}{n})}^{k} \sim_{n\to +\infty} \sum_{k=0}^{n}\frac {1}{k!}} (WRONG !!)
    Reason: k in the \displaystyle \sum_{k=0}^{n} is not fixed, it varies from k = 0 to n

    THEOREM 3:
    \displaystyle\boxed { {U}_{n} \sim_{n\to +\infty}  {V}_{n}, \forall p \ge 1,  ({U}_{n})^{p} \sim_{n\to +\infty} ({V}_{n})^{p} }
    Note: p is fixed value.
    \boxed {\displaystyle ({U}_{n})^{n} \sim_{n\to +\infty} ({V}_{n})^{n} } (WRONG!!)
    Reason: n is variable to infinity.

    \text {If  } {U}_{n}\xrightarrow [n\to\infty] {} 0, \text {then } {(1+{U}_{n})}^{\alpha} \xrightarrow [n\to\infty] {} \alpha.{U}_{n}
    Is below correct ?
    {(1+\frac{1}{n})}^{n} \xrightarrow [n\to\infty] {??} n. \frac{1}{n} = 1
    [Hint] n is variable, \alpha is fixed value.

    Final Solution: Exponential

    THEOREM 4:
    \boxed {\forall (a,b) \in \mathbb {R}_{+}^{*} \text { x }\mathbb {R}, a^{b} = e^{b.\ln {a}}}

    \boxed { \displaystyle \forall n \in \mathbb {N}^{*}, {\bigl( 1 + \frac{1}{n} \bigr)}^{n} = e^{n. {\ln (1+ \frac {1}{n})}} } … [*]

    THEOREM 5:
    \boxed {\text {If  } {U}_{n}\xrightarrow [n\to\infty] {} 0, \text {then } \ln (1+{U}_{n}) \sim_{n\to\infty} {} {U}_{n} }

    \frac{1}{n} \xrightarrow [n\to\infty] {} 0, \text {then } \ln (1+ \frac{1}{n}) \sim_{n\to\infty} {}  \frac{1}{n}
    n.\ln (1+ \frac{1}{n}) \sim_{n\to\infty} {}1

    THEOREM 6: From equivalence (\sim_{n\to\infty} {}) to find Limit (\xrightarrow [n\to\infty] {}), and vice-versa
    \boxed { \text {If  } {U}_{n} \sim_{n\to\infty} {} \ell \in \mathbb {R},  \text {then } {U}_{n}   \xrightarrow [n\to\infty] {} \ell }

    Converse is true only if \ell \neq 0
    eg. \ell = (\frac {1}{n})_{n \in {N}^{*}} \neq 0
    because (\frac {1}{n}) \xrightarrow [n\to\infty] {} 0 but \nsim_{n\to\infty} {} 0

    Therefore (from Theorem 6):
    \displaystyle \lim_{n\to\infty}  n.\ln (1+ \frac{1}{n})= 1

    Since exponential function is continuous at 1 (why must state the condition of Continuity? )
    hence, from [*], we have:
    \boxed { \displaystyle {\lim_{n\to\infty} \bigl( 1 + \frac{1}{n} \bigr)^{n} = e}}

    Espaces Vectoriels

    Cours math sup, math spé, BCPST.

    The French University (engineering) 1st & 2nd year Prépa Math: “Vector Space” (向量空间), aka Linear Algebra (线性代数), used in Google Search Engine. The French treats the subject abstractly, very theoretical, while the USA and UK (except Math majors) are more applied (directly using matrices).

    Note: First year French (Engineering) University “Classe Prépa”: Math Sup (superior); 2nd year Math Spé (special).

    Part 2:

    Applications Lineaires (Linear Algebra):