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”)

    Ref:

    http://www.leparisien.fr/politique/emmanuel-macron-etait-un-etudiant-exceptionnel-selon-ses-profs-de-sciences-po-09-05-2017-6932907.php

    http://www.sciencespo.fr/actualites/actualit%C3%A9s/emmanuel-macron-promotion-2001/2998

    Summary of Persistent Homology

    Singapore Maths Tuition

    We summarize the work so far and relate it to previous results. Our input is a filtered complex $latex K$ and we wish to find its $latex k$th homology $latex H_k$. In each dimension the homology of complex $latex K^i$ becomes a vector space over a field, described fully by its rank $latex beta_k^i$. (Over a field $latex F$, $latex H_k$ is a $latex F$-module which is a vector space.)

    We need to choose compatible bases across the filtration (compatible bases for $latex H_k^i$ and $latex H_k^{i+p}$) in order to compute persistent homology for the entire filtration. Hence, we form the persistence module $latex mathscr{M}$ corresponding to $latex K$, which is a direct sum of these vector spaces ($latex alpha(mathscr{M})=bigoplus M^i$). By the structure theorem, a basis exists for this module that provides compatible bases for all the vector spaces.

    Specifically, each $latex mathcal{P}$-interval $latex (i,j)$ describes a basis element…

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    Inspirational Scientist: Dan Shechtman

    Singapore Maths Tuition

    Source: https://www.theguardian.com/science/2013/jan/06/dan-shechtman-nobel-prize-chemistry-interview

    To stand your ground in the face of relentless criticism from a double Nobel prize-winning scientist takes a lot of guts. For engineer and materials scientist Dan Shechtman, however, years of self-belief in the face of the eminent Linus Pauling‘s criticisms led him to the ultimate accolade: his own Nobel prize.

    The atoms in a solid material are arranged in an orderly fashion and that order is usually periodic and will have a particular rotational symmetry. A square arrangement, for example, has four-fold rotational symmetry – turn the atoms through 90 degrees and it will look the same. Do this four times and you get back to its start point. Three-fold symmetry means an arrangement can be turned through 120 degrees and it will look the same. There is also one-fold symmetry (turn through 360 degrees), two-fold (turn through 180 degrees) and six-fold symmetry (turn through 60…

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    2016 Nobel-Prize Winning Physics Explained Through Pastry 

    2016 Nobel Prize Physics is Mathematics (Topology) applied in Superconductor and Superfluid to explain the Phase Transitions and Phases of Matter. 

    Phases of Matter: Solid, Liquid, Gas

    Phase Transitions: Solid ->  Liquid -> Gas

    Superconductor below Tc (critical temperature) : zero resistance.

    Superfluid below Tc : zero viscosity.

    Reason explained by Mathematics : Topological invariant (拓扑不变量) increased step-wise.

    Eg. Disk (0 hole), Circle (1 hole), Donut (2 holes), Coffee Cup (2 holes)… XYZ (n holes). [n increased by steps from 0, 1, 2, 3… ]

    We say donut and coffee cup are homeomorphic (同胚) because they have the same topological invariant (2 holes).

    Why call “Affine” Geometry ?

    The term “Affine” was coined by Euler  (1748b), motivated by the idea that images related by affine transformation  have an affinity with one another.

    Note 1: The term “Affine Geometry” is never used in GCE A-level Math, but commonly taught in French Baccalaureate.

    Note 2:  “Affinity” 亲和力 => 模仿
    Affine transformation => 仿射 变化

    In geometry, an affine transformationaffine map[1] or an affinity (from the Latin, affinis, “connected with”) is a function between affine spaces which preserves points, straight lines and planes. Also, sets of parallel lines remain parallel after an affine transformation. An affine transformation does not necessarily preserve angles between lines or distances between points, though it does preserve ratios of distances between points lying on a straight line.

    Examples of affine transformations include translationscalinghomothetysimilarity transformationreflectionrotationshear mapping, and compositions of them in any combination and sequence.

    Chinese Remainder Theorem (韩信点兵) “The Problem of 6 Professors”

    How to formulate this problem in CRT (Chinese Remainder Theorem, aka “韩信点兵” dated since 3rd century in China.) ?

    Let d = week days {1, 2, 3, 4, 5, 6, 7} for {Monday (Prof M), tuesday (Prof t), Wednesday (Prof W), Thursday (Prof T), Friday (Prof F), saturday (Prof s), Sunday (Prof S)}

    d : 1  2   3   4 5  6 [7] 1  2 3 4 5 6  [7] 1 2
    M: m 0 m 0 m 0 [m]  ==> fell on 1st sunday
    t:    –  t  0  0  t  0  [0 ] t 0 0 t 0 0 [t ]  ==> fell on 2nd sunday
    W: –  – w  0  0  0 [w] 0  0 0 w 0 0 [0] ==> fell on 1st sunday
    T:  –  –  –  T  T  T  [T]  ==> fell on 1st sunday (TRIVIAL CASE!)
    F:  –  –  –  –  f   0  [0] 0 0 0 f 0 0 [0] 0 0 f 0 0 0 [0] 0 f 0 0 0 0 0 f [0] 0 0 0 0 f 0 [0] 0 0 0 f 0 0 [0] 0 0 f  0 0 0 [0] 0 f 0 0 0 0 [0] f 0 0 0 0 0 [f] ==> fell on 9th sunday
    s:  –  –   –  – –  s   [0] 0 0 0 s 0 0 [0] 0 s 0 0 0 0 [s] ==> fell on 3rd sunday

    The solution X (number of days counted from the first announcement by Monday professor “M”.will be the first time all 6 professors simultaneously fall on the same Nth sunday, when ALL are compelled to omit a lecture.

    [Hint]: Prof M starts on Monday (“décalage” / shift by  1 day in a week cycle), …Prof W on Wednesday (shift by 3 days in a week cycle) …  a week has 7 days.

    Prof M (1st day in week) repeats every 2 days : (mod 2)
    d(M) – 1 = 0 (mod 2)
    => d(M) = 1 (mod 2)

    Similarly, Prof W (3rd day in week) repeats at interval of 4 days : (mod 4)
    d(W) – 3 = 0 (mod 4)
    => d(W) = 3 (mod 4)

    Additionally, a “virtual” Prof S of Sunday (7th day in week) repeats at interval of 7 days: (mod 7)
    d(S) – 7 = 0 (mod 7)
    => d(S) = 7 (mod 7) = 0 (mod 7)

    The problem is reduced to solving the following 7 modular equations : 

    X = 1 (mod 2) … [1]
    X = 2 (mod 3) … [2]
    X = 3 (mod 4) … [3]
    X = 4 (mod 1) = 4 ×1  = 0 (mod 1) … [4]
    X = 5 (mod 6) … [5]
    X = 6 (mod 5) = 1 (mod 5)  … [6]
    X = 0 (mod 7) … [7]

    We can simplify them into only 3 equations (with co-prime mods) by elimination and combination rules.

    Eliminate [4] which is trivial (useless) for any equation (mod 1).

    X =3 (mod 4) .. (3) can be simplified :
    since 2 | 4, 
    so X= 3 (mod 4÷2) = 3 (mod 2)
    => X = 1 (mod 2) which is same as (1).
    => eliminate (3), keep (1)

    Similarly, equation (5):
    X = 5 (mod 6) = 5 (mod 6 ÷2) = 5 (mod 3)
    X = 2 (mod 3) which is same as (2).
    => eliminate (5), keep (2)

    X = 1 (mod 2) … (1)
    X = 6 (mod 5) = 1 (mod 5) … (6)
    We can combine them by Easy CRT (since both same remainder 1): 
    X = 1 (mod LCM (2,5)) = 1 (mod 10) … (8)

    Solve :
    X = 2 (mod 3) …(2)
    X = 1 (mod 10) … (8)
    X = 0 (mod 7) … (7)

    Check CRT Condition: 3, 10, 7 are co-primes pairwise, ie the pairs (3,10) (3,7) (10,7) are co-primes, thus we can apply CRT to solve [2], [8] & [7]:
    Applying the CRT,
    Mod: 3 10 7
    Rem: 2  1 0
    X = (10.7).u1 + (3.7).u2 +(3.10).u3 (mod 3.10.7)

    (10.7).u1 = 70.u1 = (69+1).u1 = 2 (mod 3)
    => 1.u1= 2 (mod 3) => u1 =2
    (3.7).u2 = 21.u2 = (20+1).u2 = 1 (mod 10)
    => 1.u2 = 1 (mod 10) => u2 = 1
    (3.10).u3 = 30.u3 = (28 + 2).u3 = 0 (mod 7)
    => 2.u3 = 0 (mod 7) => u3=7
    X = 10.7.2  + 3.7.1 + 3.10.7 = 140 + 21 + 210
    X = 371 (mod 3.10.7) = 371 (mod 210) =161 (mod 210)

    \boxed {X = 161}

    That is: On the  (161 / 7 = ) 23rd Sunday when all 6 professors are simultaneously compelled to omit a lecture.

    –2nd Method : —

    If choose different [3] & [5] instead of [1] & [2]:

    X = 3 (mod 4) … [3]
    X = 5 (mod 6) … [5]
    X = 6 (mod 5) = 1 (mod 5)  … [6]
    X = 0 (mod 7) … [7]
    can be re-rewritten as:
    X = -1 (mod 4) … [3]
    X = -1 (mod 6) … [5]
    X =   1 (mod 5)  … [6]
    X = 0 (mod 7) … [7]

    [3] & [5] : LCM (4,6) = 12
    By Easy CRT: X = -1 (mod 12 ) 
    Finally, solve:
    X = -1 (mod 12) … [9]
    X = 1 (mod 5) … [6]
    X = 0 (mod 7)… [7]
    Apply the same CRT as above : 
    X = 371 (mod 12.5.7) = 371 (mod 420)
    Notice 2 | 420,
    we can simplify the solution:
    X = 371 (mod 420 ÷2) = 371 (mod 210)
    X = 161 (mod 210)

    (Source : Excellent Video on CRT Techniques)

    Easy CRT :

    (Reference only: Revision)

    CRT Definition: