Inspirational Scientist: Dan Shechtman

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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:

Kids with tuition fare worse?

Singapore Maths Tuition

Article: http://www.straitstimes.com/opinion/kids-with-tuition-fare-worse

Those who read the news, either online or in print, would probably have seen this article: “Kids with tuition fare worse”.

In the article, it is claimed that: “In fact, children who received tuition actually scored about 0.256 standard deviations lower on their tests than those who did not (standard deviation is a measure of how spread out test scores are from the average).”

The headline is actually quite misleading, causing people to think that tuition causes worse performance. One needs to read the final part of the article: “The first is that students who receive tuition choose to receive it precisely because they are not doing well in school. In other words, weak performance may be what is driving students to enrol for tuition.”

The correct way to measure the effect of tuition is via a “before and after” experiment. Scores of students before and…

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