Apply Newton's Law of Cooling:
$latex dT/dt = k(T-T_s)$
$latex T_s =21$ (Room Temperature)
$latex dT/(T-T_s)= k.dt$
$latex \int \frac {dT}{T-T_s}=\int k.\mathrm{d}t$
$latex ln (T-T_s)=kt+C$
$latex e^{kt+C}= T-T_s$
$latex T= Ae^{kt} +T_s$
$latex A=e^{C}$
$latex T=Ae^{kt}+21$
At t=0, (normal body temperature)
$latex T_o= A + 21 =37$
=> A=16
Let t =x hour 1st temperature taken
$latex T(x)= 16e^{kx}+21=29$
$latex 16e^{kx} = 8$
$latex e^{kx}= \frac {1}{2}$ ...[1]
t = x+1 hour later
$latex T(x+1) =16e^{k(x+1)} + 21 = 27$
$latex e^{k(x+1)} = 3/8$
$latex e^{kx}e^{k}= 3/8$
$latex e^{k} = 3/4$
k = ln(3/4)
[1]: kx = ln(1/2)
x = ln(1/2) / ln(3/4)
x=2.41 hr = 2 hr 25m
Murder Time= 2am -2h25m =11:35 p.m. [QED]
Sunday, 31 March 2013
Sherlock Holmes & Math
Sherlock Holmes was called on the murder's scene at 2 am. He took the temperatures of the victim (29 ℃) & room (21 ℃).
One hour later the dead body's temperature dropped to 27 ℃.
He concluded the murder occurred 2.41 hrs ago at 11:35 p.m.
Why ?
[Assume normal person temp is 37 ℃ ]
One hour later the dead body's temperature dropped to 27 ℃.
He concluded the murder occurred 2.41 hrs ago at 11:35 p.m.
Why ?
[Assume normal person temp is 37 ℃ ]
Automorphic Number
Automorphic Number (n)
Automorphism φ
φ: n -> nxn
nxn = {...}n
Example: 1, 5, 6, 25 are Automorphic Numbers
5x5 =25 ={2}5
6x6 =36 ={3}6
25x25 =625 ={6}25
Automorphism φ
φ: n -> nxn
nxn = {...}n
Example: 1, 5, 6, 25 are Automorphic Numbers
5x5 =25 ={2}5
6x6 =36 ={3}6
25x25 =625 ={6}25
Egyptian Math
Egyptian Math In Pyramids
$latex \frac{3}{5} = \frac {1}{x} + \frac{1}{y}$
Find x, y?
Solution:
1) 2010 BCE Method
Cut 3 apples in half => 6 x ½
5 persons each takes ½ apple = 5x½
Left ½ apple, further cut into 5
= (½) x 1/5 = 1/10
Answer : 3/5 = 1/2 + 1/10
2) 2010 CE Method
3/5 = 0.6 = 0.5 + 0.1 = 1/2 + 1/10
$latex \frac{3}{5} = \frac {1}{x} + \frac{1}{y}$
Find x, y?
Solution:
1) 2010 BCE Method
Cut 3 apples in half => 6 x ½
5 persons each takes ½ apple = 5x½
Left ½ apple, further cut into 5
= (½) x 1/5 = 1/10
Answer : 3/5 = 1/2 + 1/10
2) 2010 CE Method
3/5 = 0.6 = 0.5 + 0.1 = 1/2 + 1/10
Mathematics and Physics
Mathematics and Physics are closely related as disciplines.
Their histories are interwined, sometimes in the person of a single figure such as Newton.
Their histories are interwined, sometimes in the person of a single figure such as Newton.
IMO Super-coach: Rukshin
Rukshin at 15 was a troubled russian kid with drink and violence, then a miracle happened: He fell in love with Math and turned all his creative, aggressive, and competitive energies toward it.
He tried to compete in Math olympiads, but outmatched by peers. Still he believed he knew how to win; he just could not do it himself.
He formed a team of schoolchildren a year younger than he and trained them.
At 19 he became an IMO coach who produced Perelman (Gold IMO & Fields/Clay Poincare Conjecture). In the decades since, his students took 70 IMO, include > 40 Golds.
Rukshin's thoughts on IMO:
1. IMO is more like a sport. It has its coaches, clubs, practice sessions, competitions.
2. Natural ability is necessary but NOT sufficient for success: The talented kid needs to have the right coach, the right team, the right kind of family support, and, most important, the WILL to win.
3. At the beginning, it is nearly impossible to tell the difference between future (Math) stars and those who will be good (at IMO) but never great (Mathematician).
He tried to compete in Math olympiads, but outmatched by peers. Still he believed he knew how to win; he just could not do it himself.
He formed a team of schoolchildren a year younger than he and trained them.
At 19 he became an IMO coach who produced Perelman (Gold IMO & Fields/Clay Poincare Conjecture). In the decades since, his students took 70 IMO, include > 40 Golds.
Rukshin's thoughts on IMO:
1. IMO is more like a sport. It has its coaches, clubs, practice sessions, competitions.
2. Natural ability is necessary but NOT sufficient for success: The talented kid needs to have the right coach, the right team, the right kind of family support, and, most important, the WILL to win.
3. At the beginning, it is nearly impossible to tell the difference between future (Math) stars and those who will be good (at IMO) but never great (Mathematician).
Differential Geometry & Yang-Mills Equation
Differential Geometry & Yang-Mills Equation
1. Characteristic class is quadratic.
=> Self-duality equation (Polyakov)
2. Possible to combine the classes with Yang-Mills functional.
[Hint by: S. Novikov, Fields Medal]
1. Characteristic class is quadratic.
=> Self-duality equation (Polyakov)
2. Possible to combine the classes with Yang-Mills functional.
[Hint by: S. Novikov, Fields Medal]
Physics Constants
n∈Z, denote $latex [n] =10^n$
$latex [8] = 10^8$
Physics Constants = 3R.[n]
Examples:
1. Light speed c = 3 [8]m/s
2. Gravitational ≈ 6.6 [-11] N.m² /kg²
3. Planck = 6.6 [-34] J.sec
4. Colomb = 9 [-9]
5. Electron weight = 9 [-28] gm
6. Earth rotate around Sun at speed 3 [4] m/sec.
7. g = 9.81 m/s²
Note: Speed of light 'c' from 'celeritas', in Latin (swiftness).
$latex [8] = 10^8$
Physics Constants = 3R.[n]
Examples:
1. Light speed c = 3 [8]m/s
2. Gravitational ≈ 6.6 [-11] N.m² /kg²
3. Planck = 6.6 [-34] J.sec
4. Colomb = 9 [-9]
5. Electron weight = 9 [-28] gm
6. Earth rotate around Sun at speed 3 [4] m/sec.
7. g = 9.81 m/s²
Note: Speed of light 'c' from 'celeritas', in Latin (swiftness).
Michael Bishop's Science
Michael Bishop (Nobel Medicine 1989)
School Science subjects should be taught in this sequence (unless in single 'Combined Science'):
1st Physics
2nd Chemistry
3rd Biology
School Science subjects should be taught in this sequence (unless in single 'Combined Science'):
1st Physics
2nd Chemistry
3rd Biology
ЛΟМΟΗΟСΟВ 罗蒙諾索夫
ЛΟМΟΗΟСΟВ 罗蒙諾索夫
(1711 ~1765)
“Chemistry is the hand of Physics; Math is the eye of Physics. ”
(1711 ~1765)
“Chemistry is the hand of Physics; Math is the eye of Physics. ”
Noether Theorem: Symmetry
Symmetry (hence Group) explains :
1. Conservation of Energy;
2. Conservation of Angular Momentum;
3. Periodic Table;
4. Laws of Thermodynamic.
Emmy Noether Theorem (1918): Conservation Laws owes to Symmetry :
1. In Linear motion
=> Conservation of Momentum
2. In Angular movement
=> Conservation of Angular Momentum
3. In Time
=> Conservation of Energy
1. Conservation of Energy;
2. Conservation of Angular Momentum;
3. Periodic Table;
4. Laws of Thermodynamic.
Emmy Noether Theorem (1918): Conservation Laws owes to Symmetry :
1. In Linear motion
=> Conservation of Momentum
2. In Angular movement
=> Conservation of Angular Momentum
3. In Time
=> Conservation of Energy
De Morgan's Imagination & Intuition
"The moving power of mathematical invention is not reasoning, but imagination."
“Mathematical proof is by logic; but Mathematical discovery is by intuition.”
“Mathematical proof is by logic; but Mathematical discovery is by intuition.”
Japanese Math 100
100 = 1+2+3+4+5+6+78+9
= 1+23-4+56+7+8+9
= 12+3-4+5+67+8+9
= 123+45-67+8-9
= 123-45-67+89
Or reverse
100 = 98+7+6-5-4-3+2-1
= 98-7+6+5+4-3-2-1
= 9+8+76+5-4+3+2+1
= 9-8+76-5+4+3+21
= 98-76+54+3+21
= 1+23-4+56+7+8+9
= 12+3-4+5+67+8+9
= 123+45-67+8-9
= 123-45-67+89
Or reverse
100 = 98+7+6-5-4-3+2-1
= 98-7+6+5+4-3-2-1
= 9+8+76+5-4+3+2+1
= 9-8+76-5+4+3+21
= 98-76+54+3+21
Babylon Math 60-base
Why Babylon used 60-base for time?
Babylonians need to divide time by 1/2, 1/3, 1/4, 1/5.
LCM (2, 3, 4, 5) = 60
=> Time unit = 1/60 min (sec)
Babylonians need to divide time by 1/2, 1/3, 1/4, 1/5.
LCM (2, 3, 4, 5) = 60
=> Time unit = 1/60 min (sec)
Abstract Math discomforts
3 Wide Discomforts For Abstract Math Students
1. Group : Coset, Quotient group, morphism...
2. Limit ε-δ: Cauchy
3. Bourbaki Sets: Function f: A-> B is subset of Cartesian Product AxB.
Students should learn from their historical genesis rather than the formal abstract definitions
<a href="http://http://en.wikipedia.org/wiki/Wu_Wenjun">Wu Wenjun (吳文俊) on Learning Abstract Math
"...It is more important to understand the 'Principles' 原理 behind, à la Physics (eg. Newton's 3 Laws of Motion), and not blinded by its abstract 'Axioms' 公理."
Prof I.Herstein http://en.wikipedia.org/wiki/Israel_Nathan_Herstein
“... Seeing Abstract Math for the first time, there seems to be a common feeling of being adrift, of not having something solid to hang on to.”
“Do not be discouraged. Stick with it! The best road is to look at examples. Try to understand what a given concept says, most importantly, look at particular, concrete examples of the concept.”
"Abstract Math plays a dual role: that of unifying link between disparate parts of math and that of a research subject with a highly active life of its own. It plays an ever more important role in physics, chemistry, and computer science, etc."
1. Group : Coset, Quotient group, morphism...
2. Limit ε-δ: Cauchy
3. Bourbaki Sets: Function f: A-> B is subset of Cartesian Product AxB.
Students should learn from their historical genesis rather than the formal abstract definitions
<a href="http://http://en.wikipedia.org/wiki/Wu_Wenjun">Wu Wenjun (吳文俊) on Learning Abstract Math
"...It is more important to understand the 'Principles' 原理 behind, à la Physics (eg. Newton's 3 Laws of Motion), and not blinded by its abstract 'Axioms' 公理."
Prof I.Herstein http://en.wikipedia.org/wiki/Israel_Nathan_Herstein
“... Seeing Abstract Math for the first time, there seems to be a common feeling of being adrift, of not having something solid to hang on to.”
“Do not be discouraged. Stick with it! The best road is to look at examples. Try to understand what a given concept says, most importantly, look at particular, concrete examples of the concept.”
"Abstract Math plays a dual role: that of unifying link between disparate parts of math and that of a research subject with a highly active life of its own. It plays an ever more important role in physics, chemistry, and computer science, etc."
Weyl's Beauty
“My work always tried to unite the true with the beautiful; but when I had to choose one or the other, I usually chose the beautiful.”
Chen (1+ 2) Theorem
Chen Theorem 陈景润 @1966‘1+2’ 陈式定理
2n= p1 + p2 (Goldbach Conjecture "1+1")
or
2n = p1+ p2*p3 (Chen Theorem "1+2")
Examples:
62 = 43 + 19
62= 7 + 5*11
2n= p1 + p2 (Goldbach Conjecture "1+1")
or
2n = p1+ p2*p3 (Chen Theorem "1+2")
Examples:
62 = 43 + 19
62= 7 + 5*11
ST Yao丘成桐
丘成桐 (ST Yao 1949~) Fields Medal @1982 [33岁] proved Calabi Conjecture
1. 读私立 培正中学, 高中遇 好数学老师. @香港中文大学, 发觉 Math Beauty, '豁然开朗'.
2. Best Math student not necessary Mathematician, only sufficient!
3. 一名数学科学家 都应对 文学,哲学 这类 学科有基本的涉猎. 好的数学 使你体验到庄子讲的
"天地与我并生, 万物与我为一" 的境界
4. 成功 = 要有数学热情.
Strategy:
a. 深入思考
b. 在心中或纸上仔细研究
c. Find clues from book, till get answer.
d. 出题目给自己
1. 读私立 培正中学, 高中遇 好数学老师. @香港中文大学, 发觉 Math Beauty, '豁然开朗'.
2. Best Math student not necessary Mathematician, only sufficient!
3. 一名数学科学家 都应对 文学,哲学 这类 学科有基本的涉猎. 好的数学 使你体验到庄子讲的
"天地与我并生, 万物与我为一" 的境界
4. 成功 = 要有数学热情.
Strategy:
a. 深入思考
b. 在心中或纸上仔细研究
c. Find clues from book, till get answer.
d. 出题目给自己
4-level MathThinking
4 Levels:
L1. S&T (See & Touch) Concrete: 1 apple, 2 oranges...
e.g. Math Modeling: visualise the problem [Primary School]
L2. S~T (See, no Touch but can guess):
e.g. Guess x,y for 2x+3y=8 ? [Secondary School]
e.g. Chimpanzees can guess where you hide the banana.
L3. ~S~T&I (no See, no Touch but Imagine):
e.g. Complex i = [Junior College].
L4. ~S~T~I (no See, no Touch, no imagine)
e.g. Abstract Math: Galois Group, ε-δ Analysis, Ring, Field, etc. [University]
L1. S&T (See & Touch) Concrete: 1 apple, 2 oranges...
e.g. Math Modeling: visualise the problem [Primary School]
L2. S~T (See, no Touch but can guess):
e.g. Guess x,y for 2x+3y=8 ? [Secondary School]
e.g. Chimpanzees can guess where you hide the banana.
L3. ~S~T&I (no See, no Touch but Imagine):
e.g. Complex i = [Junior College].
L4. ~S~T~I (no See, no Touch, no imagine)
e.g. Abstract Math: Galois Group, ε-δ Analysis, Ring, Field, etc. [University]
Language & Math
1. Galois's mother home-schooling him Latin & other languages before entering Lycée Louis-Le-Grand.
2. William Hamilton: knew 15 languages include Chinese before discovered Quarternions (1,i,j,k) on Monday 16 Oct 1843 walking along Brougham Bridge, Ireland.
3. Pascal, Descartes are philosopher good in writing.
4. Gauss learnt even at old age Russian to read Lobatschefsky's Non-Euclidean Geometry
5. Cauchy's father heeded the advice of his neighbour Laplace to teach young Cauchy language before mathematics.
Prime Numbers * +
Prime Numbers
God multiplies Primes: Fundamental Law of Arithmetic (Euclid):
Any number (n) can be UNIQUELY Prime Factorized: n = P1 x P2 x ...Pn
e.g. 12= 2 x 2 x 3,
28=2 x 2 x7
Men add Primes: Goldbach Conjecture.
2n = P1+P2
e.g. 12= 5 + 7, 28 = 17 + 11, 30 = 11+19
God multiplies Primes: Fundamental Law of Arithmetic (Euclid):
Any number (n) can be UNIQUELY Prime Factorized: n = P1 x P2 x ...Pn
e.g. 12= 2 x 2 x 3,
28=2 x 2 x7
Men add Primes: Goldbach Conjecture.
2n = P1+P2
e.g. 12= 5 + 7, 28 = 17 + 11, 30 = 11+19
Butterfly Theorem
Butterfly Theorem
In a circle draw a chord PQ with mid-point M. Through M draw 2 chords AB, CD. Join AD, BC cut PQ at X, Y resp. (Butterfly M)
1. Prove: M = mid-point of XY
http://gogeometry.com/GeometryButterfly.html
2. If circle changed to ellipse, still true?
Yes. Affine transformation from circle elongated to ellipse, like distorted image through funny mirror => still MX = MY
[caption id="" align="alignright" width="300"]
Butterfly theorem (Pho[/caption]
In a circle draw a chord PQ with mid-point M. Through M draw 2 chords AB, CD. Join AD, BC cut PQ at X, Y resp. (Butterfly M)
1. Prove: M = mid-point of XY
http://gogeometry.com/GeometryButterfly.html
2. If circle changed to ellipse, still true?
Yes. Affine transformation from circle elongated to ellipse, like distorted image through funny mirror => still MX = MY
[caption id="" align="alignright" width="300"]
Butterfly theorem (Pho[/caption]
Good Mathematics
Good Mathematics [Lagrange's definition]
1. Easy to understand by people on the streets
2. Difficult to solve
Example: Fermat's Last Theorem
1. Easy to understand by people on the streets
2. Difficult to solve
Example: Fermat's Last Theorem
China 'Gauss': 秦九韶 Qin Jiushao
秦九韶 Qin Jiushao(1202-1261 AD): http://en.wikipedia.org/wiki/Qin_Jiushao
A Southern Song dynasty (南宋) officer. During his 3-yr leaves when his mother died, he generalised 孙子算经 (4th century)'s "Chinese Remainder Theorem" in '大衍求一术'. After leaves, he went back to chase money & women, produced no more Maths.
Note: '求一': solve a.X ≡ 1 (mod b); a < b
A Southern Song dynasty (南宋) officer. During his 3-yr leaves when his mother died, he generalised 孙子算经 (4th century)'s "Chinese Remainder Theorem" in '大衍求一术'. After leaves, he went back to chase money & women, produced no more Maths.
Note: '求一': solve a.X ≡ 1 (mod b); a < b
Topology
Topology (by Poincaré)
Moniker "Rubber-Sheet Geometry", compared with Geometry's 'rigid objects'.
[Greek]= τοΠοζ(Place) λΟγια(Study)
[Latin]= Analysis Situs (Situation)
1. Remove (invariants) of geometry:
2. Preserve 'Neighbourhood' (Nearness)
3. Elastic deformation (stretch, bend, twist)
Moniker "Rubber-Sheet Geometry", compared with Geometry's 'rigid objects'.
[Greek]= τοΠοζ(Place) λΟγια(Study)
[Latin]= Analysis Situs (Situation)
1. Remove (invariants) of geometry:
- a. Euclidean (distance)
- b. Affine (//, ratio)
- c. Projective (cross-ratio)
2. Preserve 'Neighbourhood' (Nearness)
- define 'Continuity' (Analysis)
3. Elastic deformation (stretch, bend, twist)
- a line is no longer a line.
153 St. Peter Fish
Bible Math: 153 St. Peter Fish
[John 21:3-11]
3 So they went out and got into the boat, but that night they caught nothing.
6 He said, "Throw your net on the right side of the boat and you will find some." When they did, they were unable to haul the net in because of the large number of fish... It was full of large fish, 153, but even with so many the net was not torn.
1) 153= 1! + 2! + 3! + 4! + 5! = (1+2+6+24+120)
2) $latex 1^3+5^3 +3^3$ = 1 + 125 + 27 = 153
3) Take any multiple of 3, e.g. 78
-> $latex 7^3+8^3$ = 855
-> $latex 8^3+5^3+5^3$ = 762
->567 ->684 ->792 ->1080 ->513 ->153 (Always!)
[John 21:3-11]
3 So they went out and got into the boat, but that night they caught nothing.
6 He said, "Throw your net on the right side of the boat and you will find some." When they did, they were unable to haul the net in because of the large number of fish... It was full of large fish, 153, but even with so many the net was not torn.
1) 153= 1! + 2! + 3! + 4! + 5! = (1+2+6+24+120)
2) $latex 1^3+5^3 +3^3$ = 1 + 125 + 27 = 153
3) Take any multiple of 3, e.g. 78
-> $latex 7^3+8^3$ = 855
-> $latex 8^3+5^3+5^3$ = 762
->567 ->684 ->792 ->1080 ->513 ->153 (Always!)
Logic: Pascal Wager
Pascal Wager:
1. We can choose to believe God exists, or we can choose not to so believe.
2. If we reject God and act accordingly, we risk everlasting agony and torment if He does exist (Type I error in Statistics lingo) but enjoy fleeting earthly delights if He doesn't exist.
3. If we accept God and act accordingly, we risk little if He doesn't exist (Type II error) but enjoy endless heavenly bliss if He does exist.
4. It's in our self-interest to accept God's existence.
5. Therefore God exists!
Mathematical Proof:
Pascal assumed
Probability of God exists = p
Probability God doesn't exist = 1-p
1. We can choose to believe God exists, or we can choose not to so believe.
2. If we reject God and act accordingly, we risk everlasting agony and torment if He does exist (Type I error in Statistics lingo) but enjoy fleeting earthly delights if He doesn't exist.
3. If we accept God and act accordingly, we risk little if He doesn't exist (Type II error) but enjoy endless heavenly bliss if He does exist.
4. It's in our self-interest to accept God's existence.
5. Therefore God exists!
Mathematical Proof:
Pascal assumed
Probability of God exists = p
Probability God doesn't exist = 1-p
You lead 2 lives, either Worldly (世俗) or Piously （虔,诚) , you get rewards X, Y, infinity or Z, as shown in table below.
In Worldly Life, the Expectation in probability is
Ew = p.X + (1-p).Y
In pious life, the Expectation is
Ep = p.∞ + (1-p).Z
Regardless how big the rewards in X (negative, punishment by God) ,Y, Z (self-sacrifice, negative reward)
Ep is infinitely bigger than Ew.
(Ew is good only if p= 0)
Conclusion: no matter what, pious life is better.
Einstein's day dream
1907- Einstein, a junior clerk recalled his 'Happiest thought of my life':-
"I was sitting in my Patent office chair when all of a sudden a thought occurred to me:
If a person falls freely he will not feel his own weight.
I was startled. It impelled me toward a theory of gravitation.”
=> The world changed by his 'day dream': Relativity Theory!
"I was sitting in my Patent office chair when all of a sudden a thought occurred to me:
If a person falls freely he will not feel his own weight.
I was startled. It impelled me toward a theory of gravitation.”
=> The world changed by his 'day dream': Relativity Theory!
The "Book"
Paul Erdös liked to talk about 'The Book', in which God maintains the perfect proofs for mathematical theorems (& he 'peeped' 1,000 from there).
Amateur vs Professional
Amateur versus Professional
1. Amateur is at liberty to study only those things he likes.
2. Professional must also study what he doesn't like.
3. Conclusion: Most famous theorems are found by Amateurs.
Examples:
Fermat = Judge (Number Theory, Probabilty),
Venn = Anglican Pastor (Venn Diagram),
Ramanujan = Railway clerk (Number Theory)
Cayley = Lawyer (Group),
Leibniz = Diplomat (Calculus, Binary 0 & 1)
1. Amateur is at liberty to study only those things he likes.
2. Professional must also study what he doesn't like.
3. Conclusion: Most famous theorems are found by Amateurs.
Examples:
Fermat = Judge (Number Theory, Probabilty),
Venn = Anglican Pastor (Venn Diagram),
Ramanujan = Railway clerk (Number Theory)
Cayley = Lawyer (Group),
Leibniz = Diplomat (Calculus, Binary 0 & 1)
Find Prime numbers
Primes (Landau's trick)
List down systematically the primes below n < 4,000
Start with 2
2x2= 4 => take 3 (<4)
< 2x3 =6 => 5
< 2x5 =10 =>7
< 2x7 =14 => 13
...
23, 43, 83, 163, 317, 631, 1259, 2503, (4001)
List down systematically the primes below n < 4,000
Start with 2
2x2= 4 => take 3 (<4)
< 2x3 =6 => 5
< 2x5 =10 =>7
< 2x7 =14 => 13
...
23, 43, 83, 163, 317, 631, 1259, 2503, (4001)
Learn with Example Space
Learn Math With Own Example Space
G. Polya / Paul Halmos advocate getting math students to construct not just one but classes of examples to:
1. Extend & enrich own Example Spaces;
2. Develop full appreciation of concepts, definitions, techniques that they are taught.
[Polya, Halmos, Feynman]: they collect and build a personal 'repertoire' of "Examples Space" (include counter-examples) for each abstract math idea, which they can relate to a concrete object.
Examples:
Group abelian = (Z,+)
Ring = Z
Principal Ideal = nZ
Equivalence Relation = mod (n)
Cosets = {3Z, 1+3Z, 2+3Z}
...
G. Polya / Paul Halmos advocate getting math students to construct not just one but classes of examples to:
1. Extend & enrich own Example Spaces;
2. Develop full appreciation of concepts, definitions, techniques that they are taught.
[Polya, Halmos, Feynman]: they collect and build a personal 'repertoire' of "Examples Space" (include counter-examples) for each abstract math idea, which they can relate to a concrete object.
Examples:
Group abelian = (Z,+)
Ring = Z
Principal Ideal = nZ
Equivalence Relation = mod (n)
Cosets = {3Z, 1+3Z, 2+3Z}
...
Triangle
Triangle
1. Not always true!
Sum of 3 internal angles = Π =180 degrees
depends on which geometry:
Euclidean / non-Euclidean / Riemann
2. True always!
Sum of 3 exterior angles = 2 Π = 360 degrees
1. Not always true!
Sum of 3 internal angles = Π =180 degrees
depends on which geometry:
Euclidean / non-Euclidean / Riemann
2. True always!
Sum of 3 exterior angles = 2 Π = 360 degrees
Prime and Perfect Square
For all primes p ≠2, (a,b ∈Z)
p= a² + b² <=> p ≡ 1 mod 4
(2=1² + 1²)
5= 1² + 2² = 1 + 4 ≡ 1 mod 4
13= 2² + 3² = 4 + 9 ≡ 1 mod 4
17=1² + 4² = 1 + 16 ≡ 1 mod 4
29= 2² + 5² = 4 + 25 ≡ 1 mod 4
37= 1² + 6² = 1 +36 ≡ 1 mod 4
Notes:
1) Perfect squares (4, 9, 16, 25... ) ≡ 0 or ≡ 1 mod 4
2) Prime (4n+1) = a² + b² (Euler took 7 yrs to prove)
3) Gauss expanded the proof to quadratic reciprocity (2 prime numbers p & q are linked by mod 4)
p= a² + b² <=> p ≡ 1 mod 4
(2=1² + 1²)
5= 1² + 2² = 1 + 4 ≡ 1 mod 4
13= 2² + 3² = 4 + 9 ≡ 1 mod 4
17=1² + 4² = 1 + 16 ≡ 1 mod 4
29= 2² + 5² = 4 + 25 ≡ 1 mod 4
37= 1² + 6² = 1 +36 ≡ 1 mod 4
Notes:
1) Perfect squares (4, 9, 16, 25... ) ≡ 0 or ≡ 1 mod 4
2) Prime (4n+1) = a² + b² (Euler took 7 yrs to prove)
3) Gauss expanded the proof to quadratic reciprocity (2 prime numbers p & q are linked by mod 4)
Bad mathematicians
Difference Between Good & Bad Mathematicians
S.S. Chern (陈省身): “The former gives many concrete examples, the latter has only abstract theories.”
S.S. Chern (陈省身): “The former gives many concrete examples, the latter has only abstract theories.”
Rigorous Calculus: ε-δ Analysis
Rigorous Analysis epsilon-delta (ε-δ)
Cauchy gave epsilon-delta the rigor to Analysis, Weierstrass 'arithmatized' it to become the standard language of modern analysis.
1) Limit was first defined by Cauchy in "Analyse Algébrique" (1821)
2) Cauchy repeatedly used 'Limit' in the book Chapter 3 "Résumé des Leçons sur le Calcul infinitésimal" (1823) for 'derivative' of f as the limit of
$latex \frac{f(x+i)-f(x)}{i}$ when i -> 0
3) He introduced ε-δ in Chapter 7 to prove 'Mean Value Theorem': Denote by (ε , δ) 2 small numbers, such that 0< i ≤ δ , and for all x between (x+i) and x,
f '(x)- ε < $latex \frac{f(x+i)-f(x)}{i}$ < f'(x)+ ε
4) These ε-δ Cauchy's proof method became the standard definition of Limit of Function in Analysis.
5) They are notorious for causing widespread discomfort among future math students. In fact, when it was first introduced by Cauchy in the Ecole Polytechnique Lecture, the French Napoleon top students booed at him and Cauchy received warning from the school.
Note 1: From the textbook 'Calculus' (1980, USA):
"If can't understand the 'ε-δ' definition, just memorize it like a poem - at least better than saying it wrongly."
E.g. “For all ε>0, there exists δ>0, …”
Note 2: George Polya: "The students are not trained in 'ε-δ', teaching them Calculus is like dropping these rules from the sky..."
Cauchy gave epsilon-delta the rigor to Analysis, Weierstrass 'arithmatized' it to become the standard language of modern analysis.
1) Limit was first defined by Cauchy in "Analyse Algébrique" (1821)
2) Cauchy repeatedly used 'Limit' in the book Chapter 3 "Résumé des Leçons sur le Calcul infinitésimal" (1823) for 'derivative' of f as the limit of
$latex \frac{f(x+i)-f(x)}{i}$ when i -> 0
3) He introduced ε-δ in Chapter 7 to prove 'Mean Value Theorem': Denote by (ε , δ) 2 small numbers, such that 0< i ≤ δ , and for all x between (x+i) and x,
f '(x)- ε < $latex \frac{f(x+i)-f(x)}{i}$ < f'(x)+ ε
4) These ε-δ Cauchy's proof method became the standard definition of Limit of Function in Analysis.
5) They are notorious for causing widespread discomfort among future math students. In fact, when it was first introduced by Cauchy in the Ecole Polytechnique Lecture, the French Napoleon top students booed at him and Cauchy received warning from the school.
Note 1: From the textbook 'Calculus' (1980, USA):
"If can't understand the 'ε-δ' definition, just memorize it like a poem - at least better than saying it wrongly."
E.g. “For all ε>0, there exists δ>0, …”
Note 2: George Polya: "The students are not trained in 'ε-δ', teaching them Calculus is like dropping these rules from the sky..."
Newtonian Calculus not rigorous !
Why Newton’s Calculus Not Rigorous?
$latex f(x ) = \frac {x(x^2+ 5)} { x}$ ...[1]
cancel x (≠0)from upper and below => $latex f(x )=x^2 +5 $
$latex \mathop {\lim }\limits_{x \to 0} f(x) =x^2 +5= L=5 $ ...[2]
In [1]: we assume x ≠ 0, so cancel upper & lower x
But In [2]: assume x=0 to get L=5
[1] (x ≠ 0) contradicts with [2] (x = 0)
This is the weakness of Newtonian Calculus, made rigorous later by Cauchy's ε-δ 'Analysis'.
$latex f(x ) = \frac {x(x^2+ 5)} { x}$ ...[1]
cancel x (≠0)from upper and below => $latex f(x )=x^2 +5 $
$latex \mathop {\lim }\limits_{x \to 0} f(x) =x^2 +5= L=5 $ ...[2]
In [1]: we assume x ≠ 0, so cancel upper & lower x
But In [2]: assume x=0 to get L=5
[1] (x ≠ 0) contradicts with [2] (x = 0)
This is the weakness of Newtonian Calculus, made rigorous later by Cauchy's ε-δ 'Analysis'.
Paul Dirac
Paul Ardrien Maurice Dirac [1902-1984] Quantum Physicist, Nobel Prize
- A natural introvert
- He self-paced Eddington's "Space, Time & Gravitation" to know Special & General Relativity inside out.
- 1925 encountered Quantum Mechanics: Heisenberg's 'lists'. Use Lie commutator AB-BA, not the product AB. Also Hamilton's formalism of mechanics Poisson bracket.
- Spin of electron: reverse, rotate 720 deg before the spin gets back to its original value.
For a ball, it is 360 degrees. Reason: Rotation of space = SO(3) but for quarternions & electrons = SU(2)
SU(2) is double in size of SO(3) -- "double cover" => 720=2x360
- Dirac didn't use group & quarternions
- 1927 X'mas 'spin matrices' (Spinors) which play the same role.
- Discovered Antimatter.
- 1956 Speech @ Moscow State University: 'A Physical law must possess mathematical beauty.'
- A natural introvert
- He self-paced Eddington's "Space, Time & Gravitation" to know Special & General Relativity inside out.
- 1925 encountered Quantum Mechanics: Heisenberg's 'lists'. Use Lie commutator AB-BA, not the product AB. Also Hamilton's formalism of mechanics Poisson bracket.
- Spin of electron: reverse, rotate 720 deg before the spin gets back to its original value.
For a ball, it is 360 degrees. Reason: Rotation of space = SO(3) but for quarternions & electrons = SU(2)
SU(2) is double in size of SO(3) -- "double cover" => 720=2x360
- Dirac didn't use group & quarternions
- 1927 X'mas 'spin matrices' (Spinors) which play the same role.
- Discovered Antimatter.
- 1956 Speech @ Moscow State University: 'A Physical law must possess mathematical beauty.'
Differential Theorem
Differential Theorem
If f differentiable at point a => f continuous at point a
Converse not true !
'Differentiability' stronger than 'Continuity'
Are all Continuous functions Differentiable ? False!
Counter-example (by Weierstrass):
$Latex f(x)=\Sigma{b^n}cos(a^n\pi x)$ n ∈[0,8], a= odd number, b∈[0,1], ab > 1+3Π/2
f(x) Continuous everywhere (cosine), but non-differentiable everywhere!
Note: Weierstrass Function is the first known fractal. (e.g. Snowflake Koch's curve).
[caption id="" align="alignright" width="256"]
Plot of Weierstrass Function (Photo credit: Wikipedia)[/caption]
Note: What it means a curve (function) is :
1. Continuous = not broken curve
2. Differentiable = no pointed 'V' or 'W' shape curve
If f differentiable at point a => f continuous at point a
Converse not true !
'Differentiability' stronger than 'Continuity'
Are all Continuous functions Differentiable ? False!
Counter-example (by Weierstrass):
$Latex f(x)=\Sigma{b^n}cos(a^n\pi x)$ n ∈[0,8], a= odd number, b∈[0,1], ab > 1+3Π/2
f(x) Continuous everywhere (cosine), but non-differentiable everywhere!
Note: Weierstrass Function is the first known fractal. (e.g. Snowflake Koch's curve).
[caption id="" align="alignright" width="256"]
Plot of Weierstrass Function (Photo credit: Wikipedia)[/caption]Note: What it means a curve (function) is :
1. Continuous = not broken curve
2. Differentiable = no pointed 'V' or 'W' shape curve
Prime Number
Prime Number
Prove we can always find an interval [p, q] wherein no prime exists, for any p, q ∈ N
Proof:
For any n ∈ N
Let H= (n+1)! = 1 x 2 x ... x(n+1)
Let G = {H+2, H+3,. . . , H+(n+1) }
=> G composites (trivial)
Choose [p, q] such that :
1. p the closest prime < H+2,
2. q the closest prime > H+(n+1)
=> no prime ∈ [p, q]
---n----------------n!-(H+1)-p[(H+2)---------H+(n+1)]q------------------
Prove we can always find an interval [p, q] wherein no prime exists, for any p, q ∈ N
Proof:
For any n ∈ N
Let H= (n+1)! = 1 x 2 x ... x(n+1)
Let G = {H+2, H+3,. . . , H+(n+1) }
=> G composites (trivial)
Choose [p, q] such that :
1. p the closest prime < H+2,
2. q the closest prime > H+(n+1)
=> no prime ∈ [p, q]
---n----------------n!-(H+1)-p[(H+2)---------H+(n+1)]q------------------
Perelman rejected Fields Medal & $1m Clay Prize
1. Perelman first published his Poincaré Conjecture proof at this site: http://arxiv.org/
2. After the USA study trip where he was attracted by Prof Hamilton's attempted proof with 'Ricci Flow', Perelman rejected a Harvard Professor job offer, returned to Russia to prove the Conjecture in isolation for 8 yrs as a low-pay free researcher.
“Absorbing the problem in its entirety and then boiling it down to an essence that proved simpler than everyone had assumed.”
3. In 2006 he rejected both Fields medal and US$1 m Clay Prize!
4. He won 1982 IMO Gold at 16 with full mark (42/42).
Perelman explained to the Math Olympiad jury his solution, who gave him full mark. Before the jury walked away, Perelman said: "Wait, I have 3 more solutions to this question!"
Saturday, 30 March 2013
God “The Super-Mathematician”
Dirac's God “The Super-Mathematician”
- Paul Ardrien Maurice Dirac [1902-1984] Quantum Physicist, Nobel Prize, a "belligerent atheist":
"It seems to be one of the fundamental features of nature that fundamental physical laws are described in terms of a mathematical theory of great beauty and power, needing quite a high standard of math for one to understand it. You may wonder: Why is nature constructed along these lines? One can only answer that our present knowledge seems to show that nature is so constructed. We simply have to accept it. One could perhaps describe the situation by saying that God is a mathematician of a very high order, and He used very advanced math in constructing the universe."
- Paul Ardrien Maurice Dirac [1902-1984] Quantum Physicist, Nobel Prize, a "belligerent atheist":
"It seems to be one of the fundamental features of nature that fundamental physical laws are described in terms of a mathematical theory of great beauty and power, needing quite a high standard of math for one to understand it. You may wonder: Why is nature constructed along these lines? One can only answer that our present knowledge seems to show that nature is so constructed. We simply have to accept it. One could perhaps describe the situation by saying that God is a mathematician of a very high order, and He used very advanced math in constructing the universe."
Fermat Little Theorem
Fermat ℓittle Theorem (FℓT)
∀ m ∈ N,
1) p prime => $Latex m^p$= m mod (p)
Note: Converse False
[Memorize Trick: military police = military in the mode of police]
Note:
p prime, ∀m,
if p | m,
=> m ≡ 0 mod (p) ...(1)
=> multiply p times:
$Latex m^p$≡ 0 mod (p) ...(2)
Substitute (1) to (2): m ≡ 0
=> p prime, ∀ m
$Latex m^p$≡ m mod (p)
=> No need (m, p) = 1 [co-prime]
E.g. (m, p) = (3, 2) =1
$Latex 3^2$= 9 ≡ 3 mod (2) ≡ 1 mod (2)
9 ≡ 1 mod (2)
E.g. (m, p) = (6, 2), p =2 |6
$Latex 6^2$= 36 ≡ 6 mod (2) ≡ 0 mod (2)
6 ≡ 0 mod (2)
WRONG EXAMPLE: p = 4 = non prime
$Latex 2^4$= 16 ≡ 2 mod (4)
but 16 ≡ 0 mod (4)
Equivalent:
p prime => $Latex m^{p-1}$= 1 mod (p)
by Contra-positive:
3) $Latex m^{p-1}$ ≠ 1 mod (p)
=> p non-prime
Apply: Prove 39 non-prime?
Take m =2, p= 39
Prove $Latex m^{p-1}$ ≠ 1 mod (p)?
a) $Latex 2^{38} =2^{36}.2^2$ .
b) $Latex 2^6$= 64 ≡ 25 mod (39)
c) $Latex 25^2$ = 625 ≡ 1 mod (39)
From (b) & (c):
$Latex (2^6)^6$ = $latex 2^{36}$ ≡ $latex 25^6$ mod(39)≡$latex (25^2)^3$ ≡$Latex 1^3$ ≡ 1
From (a): $Latex 2^{38}$ = $Latex 2^{36}.2^2$ ≡ 1x4 ≠ 1 mod(39)
=> 39 non-prime
∀ m ∈ N,
1) p prime => $Latex m^p$= m mod (p)
Note: Converse False
[Memorize Trick: military police = military in the mode of police]
Note:
p prime, ∀m,
if p | m,
=> m ≡ 0 mod (p) ...(1)
=> multiply p times:
$Latex m^p$≡ 0 mod (p) ...(2)
Substitute (1) to (2): m ≡ 0
=> p prime, ∀ m
$Latex m^p$≡ m mod (p)
=> No need (m, p) = 1 [co-prime]
E.g. (m, p) = (3, 2) =1
$Latex 3^2$= 9 ≡ 3 mod (2) ≡ 1 mod (2)
9 ≡ 1 mod (2)
E.g. (m, p) = (6, 2), p =2 |6
$Latex 6^2$= 36 ≡ 6 mod (2) ≡ 0 mod (2)
6 ≡ 0 mod (2)
WRONG EXAMPLE: p = 4 = non prime
$Latex 2^4$= 16 ≡ 2 mod (4)
but 16 ≡ 0 mod (4)
Equivalent:
p prime => $Latex m^{p-1}$= 1 mod (p)
by Contra-positive:
3) $Latex m^{p-1}$ ≠ 1 mod (p)
=> p non-prime
Apply: Prove 39 non-prime?
Take m =2, p= 39
Prove $Latex m^{p-1}$ ≠ 1 mod (p)?
a) $Latex 2^{38} =2^{36}.2^2$ .
b) $Latex 2^6$= 64 ≡ 25 mod (39)
c) $Latex 25^2$ = 625 ≡ 1 mod (39)
From (b) & (c):
$Latex (2^6)^6$ = $latex 2^{36}$ ≡ $latex 25^6$ mod(39)≡$latex (25^2)^3$ ≡$Latex 1^3$ ≡ 1
From (a): $Latex 2^{38}$ = $Latex 2^{36}.2^2$ ≡ 1x4 ≠ 1 mod(39)
=> 39 non-prime
Group: Proving 'Weapon' = Euclidean Division Algorithm
Euclidean Division Algorithm:
m= qn + r ; 0 ≤ r < n; (m, n, q, r) in Z
Apply:
Cyclic group of order n:
$Latex a^n = e$
Take any m, prove it is still within the cyclic group:
$Latex a^m = a^{qn+r} =a^{qn}.a^r =(a^n)^{q}.a^r = e^q.a^r = e.a^r =a^r $[QED]
m= qn + r ; 0 ≤ r < n; (m, n, q, r) in Z
Apply:
Cyclic group of order n:
$Latex a^n = e$
Take any m, prove it is still within the cyclic group:
$Latex a^m = a^{qn+r} =a^{qn}.a^r =(a^n)^{q}.a^r = e^q.a^r = e.a^r =a^r $[QED]
Hotel Bill Puzzle
A 3-bed hotel room costs $30 per nite. Three men each pays $10 to share the room. The hotel boss decides to refund $5 as discount. The hotel clerk pockets $2, return each man only $1.
Each man nett pays $10 - $1 =$9
The room costs the 3 men $9 x3 = $ 27
plus $ 2 pocketed by clerk
Total = $ 29
Where is the missing $ 1 (= $30 - $29 )?
Each man nett pays $10 - $1 =$9
The room costs the 3 men $9 x3 = $ 27
plus $ 2 pocketed by clerk
Total = $ 29
Where is the missing $ 1 (= $30 - $29 )?
Sociable Numbers
Sociable Numbers
Sum of all divisors = next chain number... -> back to 1st number.
12,496 -> 14,288 -> 15,472 -> 14,536 -> 14,264 -> 12,496
Sum of all divisors = next chain number... -> back to 1st number.
12,496 -> 14,288 -> 15,472 -> 14,536 -> 14,264 -> 12,496
Ring Memorize Trick
Memorize Trick For Ring:
1. Ring (R) is for Marriage with 2 operations: + (addition of kids), * (multiply asset).
2. (R , +) is Commutative Group
Analogy: your kids are also your wife’s kids, vice versa.
3. (R ,*) is Semi-Group (only closure & associative)
Analogy: your asset to multiply (*) is semi (50%) owned by your wife.
4. Fair distributive law for * with respect to +
=> distribute your asset (a) to your kids (k) & wife (w):
a*(k+w) = a*k + a*w
5. No division (/) operation => Ring can't be broken.
1. Ring (R) is for Marriage with 2 operations: + (addition of kids), * (multiply asset).
2. (R , +) is Commutative Group
Analogy: your kids are also your wife’s kids, vice versa.
3. (R ,*) is Semi-Group (only closure & associative)
Analogy: your asset to multiply (*) is semi (50%) owned by your wife.
4. Fair distributive law for * with respect to +
=> distribute your asset (a) to your kids (k) & wife (w):
a*(k+w) = a*k + a*w
5. No division (/) operation => Ring can't be broken.
Perfect Numbers
Perfect Numbers
6 = 1 x 2 x 3 = 1 + 2 + 3
1. Greek: 6, 28, 496, 8128
2. Found in Medieval manuscript: 33550336
3. To-date only 30 found.
4. Are they Infinite ?
6 = 1 x 2 x 3 = 1 + 2 + 3
1. Greek: 6, 28, 496, 8128
2. Found in Medieval manuscript: 33550336
3. To-date only 30 found.
4. Are they Infinite ?
Culture, Intellect & Attitude
Culture, Intellect & Attitude Most Important
Let f: {A,B...Y,Z} -> {1,2... 25,26}
f(AB}=f(A)+f(B)=1+2=3
=> f homomorphism
f(LUCK)= f(L)+f(U)+f(C)+f(K)= 12+21+3+11 =47
f(LOVE)=54
f(KNOWLEDGE)=96
f(LEADERSHIP)=97
f(HARDWORK)=98
f(CULTURE)=100
f(INTELLECT)=100
f(ATTITUDE)=100
=> Last 3 most important in life!
Let f: {A,B...Y,Z} -> {1,2... 25,26}
f(AB}=f(A)+f(B)=1+2=3
=> f homomorphism
f(LUCK)= f(L)+f(U)+f(C)+f(K)= 12+21+3+11 =47
f(LOVE)=54
f(KNOWLEDGE)=96
f(LEADERSHIP)=97
f(HARDWORK)=98
f(CULTURE)=100
f(INTELLECT)=100
f(ATTITUDE)=100
=> Last 3 most important in life!
Automorphism = Symmetry
Automorphism of a Set is an expression of its SYMMETRY.
1. Geometry figure (e.g. triangle) under certain transformations (reflection, rotation, ...), it is mapped upon itself, certain properties (distance, angle, relative location) are preserved.
=> the figure admits certain automorphism relative to its properties.
2. Automorphism of an arbitrary Set (with arbitrary relations between its elements) form an Automorphism Group of the set.
1. Geometry figure (e.g. triangle) under certain transformations (reflection, rotation, ...), it is mapped upon itself, certain properties (distance, angle, relative location) are preserved.
=> the figure admits certain automorphism relative to its properties.
2. Automorphism of an arbitrary Set (with arbitrary relations between its elements) form an Automorphism Group of the set.
Golden Ratio Φ
A---------------C----------B
$Latex \frac {AB}{AC} = \frac{AC}{CB}$
= 1.61803... = Φ
= $Latex \frac {1+ \sqrt{5}} {2}$
$Latex \frac {6}{5} \Phi^2$
= ∏ = 3.14159...
Donald Knuth (Great Computer Mathematician, Stanford University, LaTex inventor) noted the Bible uses a phrase like:
"as my Father is to me, I am to you"
=> F= Father = line AB
I (or me) = AC
U = You = CB
=> F/I = I/U = Φ
Note: Φ = 1.61803 = - 2 sin 666°
$Latex \frac {AB}{AC} = \frac{AC}{CB}$
= 1.61803... = Φ
= $Latex \frac {1+ \sqrt{5}} {2}$
$Latex \frac {6}{5} \Phi^2$
= ∏ = 3.14159...
Donald Knuth (Great Computer Mathematician, Stanford University, LaTex inventor) noted the Bible uses a phrase like:
"as my Father is to me, I am to you"
=> F= Father = line AB
I (or me) = AC
U = You = CB
=> F/I = I/U = Φ
Note: Φ = 1.61803 = - 2 sin 666°
Baccalaureat Origin
1. Arabic Origin: Latin translation of Arabic word 'Bi-haqq al-ruwayeh'. Arabs learnt from China.
2. Chinese Origin: Chinese ‘秀才' (xiu-cai) scholar system (equivalent to Cambridge GCE A level).
3. First appeared as 'Baccalareus' in the University of Paris (1232 AD) by Pope Gregory IX, means 'The right (license) to teach on the authority of another'.
2. Chinese Origin: Chinese ‘秀才' (xiu-cai) scholar system (equivalent to Cambridge GCE A level).
3. First appeared as 'Baccalareus' in the University of Paris (1232 AD) by Pope Gregory IX, means 'The right (license) to teach on the authority of another'.
Amicable Number (220, 284)
Genesis 32:14 "Jacob gives Esau 200 she-goats and 20 he-goats, 200 ewes & 20 rams..."
In the Bible story, Jacob cheated his elder brother Esau for the inheritance from their father Issac, Esau was angry. To show friendship later, Jacob gave him 220 (=200+20) goats, 220 (=200+20) ewes & rams.
Amicable Number = (220, 284): Sum of all factors of 220 = 284, vice versa.
220 = 1+2+4+5+10+11+20+22+44+55+110 = 284
284 =1+2+4+71+142= 220
Give Amicable 220 => to exchange friendship (284)
Discovered by Arab Thabit bin Qurra (836 AD)
1)al-Farisis (13 AD): 17,296 & 18,416
(before Euler in 18 CE)
2) Muhammed Baqir Yazdin (17 CE): 9,363,584 & 9,437,056
Other Amicable numbers:
1184 & 1210
2620 & 2924
5020 & 5564
6232 & 6368
10744 & 10856
In the Bible story, Jacob cheated his elder brother Esau for the inheritance from their father Issac, Esau was angry. To show friendship later, Jacob gave him 220 (=200+20) goats, 220 (=200+20) ewes & rams.
Amicable Number = (220, 284): Sum of all factors of 220 = 284, vice versa.
220 = 1+2+4+5+10+11+20+22+44+55+110 = 284
284 =1+2+4+71+142= 220
Give Amicable 220 => to exchange friendship (284)
Discovered by Arab Thabit bin Qurra (836 AD)
1)al-Farisis (13 AD): 17,296 & 18,416
(before Euler in 18 CE)
2) Muhammed Baqir Yazdin (17 CE): 9,363,584 & 9,437,056
Other Amicable numbers:
1184 & 1210
2620 & 2924
5020 & 5564
6232 & 6368
10744 & 10856
Chinese Remainder Theorem 中国剩余定理
中国剩余定理CRT (Chinese Remainder Theorem)
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
Solve X?
明. 程大位 "算法统宗" (1593)
3人同行70稀
5树梅花21支
7子团圆半个月(15)
除百零五(105)便得知
Let remainders:
$Latex r_3=2, r_5=3, r_7=2$
$Latex r= r_3.70+ r_5.21 + r_7.15 (mod 3.5.7)$
r= 2.70 +3.21 +2.15 (mod 105)
r= 140 +63 +30 (mod 105)
r= 233 (mod 105)
$latex r= 23 = x_{min}$
or X= 23 +105Z (23 + multiples of 105)
-------------------------------------------
CRT: Why 3:70, 5:21, 7:15
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
1) Find A such that
A ≡ 1 (mod 3)
A ≡ 0 (mod 5)
A ≡ 0 (mod 7)
=> 5|A, 7|A => 35 |A
----------------Ring Theory -------------------------------
Commutative Ring CRT by Ring/ Ideal Theory:
If R is a commutative ring & A1,...An are pairwise coprime ideals,
Prove that if
$Latex r_1, r_2, \cdots, r_n$ belong to R
Then there exists 'a' belongs to R with
$Latex a + A_j= r_j + A_j$ ; j ∈[1,n]
Interpretation:
Let R= Z ring
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
Or
X ≡ rj (mod mj)
Ideal = (m) = mZ
$Latex A_3=(3), A_5=(5), A_7=(7)$
$latex r_3=2, r_5=3, r_7=2$
There exists
a=23
$Latex a +A_3 = r_3 + A_3$
23+(3)= 2+ (3)
23 + 1x3 =2 + 8x3 =26
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
Solve X?
明. 程大位 "算法统宗" (1593)
3人同行70稀
5树梅花21支
7子团圆半个月(15)
除百零五(105)便得知
Let remainders:
$Latex r_3=2, r_5=3, r_7=2$
$Latex r= r_3.70+ r_5.21 + r_7.15 (mod 3.5.7)$
r= 2.70 +3.21 +2.15 (mod 105)
r= 140 +63 +30 (mod 105)
r= 233 (mod 105)
$latex r= 23 = x_{min}$
or X= 23 +105Z (23 + multiples of 105)
-------------------------------------------
CRT: Why 3:70, 5:21, 7:15
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
1) Find A such that
A ≡ 1 (mod 3)
A ≡ 0 (mod 5)
A ≡ 0 (mod 7)
=> 5|A, 7|A => 35 |A
A=35, 70 ...
70 ≡ 1 (mod 3)
=> 70x2 ≡ 2 (mod 3)
2) Find B s.t.:
B ≡ 0 (mod 3)
B ≡ 1 (mod 5)
B ≡ 0 (mod 7)
3|B, 7|B => 21|B
21 ≡ 1 (mod 5)
=> 21x3 ≡ 3 (mod 5)
3) Find C s.t. :
C ≡ 0 (mod 3)
C ≡ 0 (mod 5)
C ≡ 1 (mod 7)
=> 3|C, 5|C => 15|C
=> C=15≡ 1 (mod 7)
15x2 ≡ 2 (mod 7)
4)
X ≡ 70x2 +21x3+ 15x2 (mod 3x5x7)
X≡ 233 (mod 105)
X≡ 23 (mod 105)
X= 23+105Z
----------------Ring Theory -------------------------------
Commutative Ring CRT by Ring/ Ideal Theory:
If R is a commutative ring & A1,...An are pairwise coprime ideals,
Prove that if
$Latex r_1, r_2, \cdots, r_n$ belong to R
Then there exists 'a' belongs to R with
$Latex a + A_j= r_j + A_j$ ; j ∈[1,n]
Interpretation:
Let R= Z ring
X ≡ 2 (mod 3)
X ≡ 3 (mod 5)
X ≡ 2 (mod 7)
Or
X ≡ rj (mod mj)
Ideal = (m) = mZ
$Latex A_3=(3), A_5=(5), A_7=(7)$
$latex r_3=2, r_5=3, r_7=2$
There exists
a=23
$Latex a +A_3 = r_3 + A_3$
23+(3)= 2+ (3)
23 + 1x3 =2 + 8x3 =26
Mental Calculation: Hourly Rate
Mental quick calculation of hourly rate:
Example
Estimate
Basic Monthly salary $ 3,258
Remove last two digits = 32
Divide by 2 = $ 16
Actual:
3,258 x 12 / (52 x 40) = $18.8
Estimate + 10% ~ Actual
Example
Estimate
Basic Monthly salary $ 3,258
Remove last two digits = 32
Divide by 2 = $ 16
Actual:
3,258 x 12 / (52 x 40) = $18.8
Estimate + 10% ~ Actual
Solution 3 (Modelling): Monkeys & Coconuts
Let x the min number of coconuts initially.
1st monkey took "a" coconuts away, 2nd monkey "b" coconuts....5th monkey took "e" coconuts.
[a][a][a][a][a] + 1 =x
Loan 4 coconuts to the initial pool of x coconuts to divide by 5 evenly at each monkey.
[a][a][a][a][a] + 1 + 4 = [a][a][a][a][a] + 5 = x+4 = X (inflated x by 4 )
1st Monkey: [a’][a’][a’][a’][a’] = X
a’ = $Latex \frac {1}{5} X$
... Left 4a’= $Latex \frac {4}{5} X$
[a’][a’][a’][a’] => [b][b][b][b][b]
b= $latex \frac{1}{5} $ .4a’ = $latex \frac{4}{25}X$
... Left 4b= $Latex \frac {16}{25}X$
[b][b][b][b] => [c][c][c][c][c]
c=$Latex \frac {1}{5}$ .4b= $Latex \frac {16}{125}X$
... Left 4c= $Latex \frac {64}{125}X$
[c][c][c][c] => [d][d][d]d][d]
d= $Latex \frac {1}{5}$.4c= $Latex \frac {64}{625}X$
... Left 4d= $Latex \frac {256}{625}X$
[d][d][d]d] => [e][e][e][e][e]
e= $Latex \frac {1}{5}$.4d= 256.(X/3125)
Since e is integer
=> X = 3125 or multiples of 3125
Minimum X=3125
x+4 = 3125
x= 3121 = minimum Coconut initially.
Note: This solution used the Singapore Modelling Math taught in all Primary Schools for 11-year-old pupils.
1st monkey took "a" coconuts away, 2nd monkey "b" coconuts....5th monkey took "e" coconuts.
[a][a][a][a][a] + 1 =x
Loan 4 coconuts to the initial pool of x coconuts to divide by 5 evenly at each monkey.
[a][a][a][a][a] + 1 + 4 = [a][a][a][a][a] + 5 = x+4 = X (inflated x by 4 )
1st Monkey: [a’][a’][a’][a’][a’] = X
a’ = $Latex \frac {1}{5} X$
... Left 4a’= $Latex \frac {4}{5} X$
[a’][a’][a’][a’] => [b][b][b][b][b]
b= $latex \frac{1}{5} $ .4a’ = $latex \frac{4}{25}X$
... Left 4b= $Latex \frac {16}{25}X$
[b][b][b][b] => [c][c][c][c][c]
c=$Latex \frac {1}{5}$ .4b= $Latex \frac {16}{125}X$
... Left 4c= $Latex \frac {64}{125}X$
[c][c][c][c] => [d][d][d]d][d]
d= $Latex \frac {1}{5}$.4c= $Latex \frac {64}{625}X$
... Left 4d= $Latex \frac {256}{625}X$
[d][d][d]d] => [e][e][e][e][e]
e= $Latex \frac {1}{5}$.4d= 256.(X/3125)
Since e is integer
=> X = 3125 or multiples of 3125
Minimum X=3125
x+4 = 3125
x= 3121 = minimum Coconut initially.
Note: This solution used the Singapore Modelling Math taught in all Primary Schools for 11-year-old pupils.
Group Definition
Memorize Trick for Group Definition
C.A.N. I. ?
C= Closure
A= Associative: (ab)c = a(bc)
N= Neutral element (e): ae=ea=a
I= Inverse: a $Latex ^ {-1} $ = e
If only 50% (C.A.)=> Semi-Group
If Semi-Group + Neutral = (C.A.N.) = MoNoid
Note: No Inverse => not a Group
Arthur Cayley (UK, 1821-1895): first gave an abstract definition of Group @1854 while being a lawyer for 14 yrs, couldn't find a teaching job. His definition was ignored for 25 years until 1882 by Walter Van Dyck who gave the final Axiomatic definition of Abstract Group. ie above [C.A.N.I.]
C.A.N. I. ?
C= Closure
A= Associative: (ab)c = a(bc)
N= Neutral element (e): ae=ea=a
I= Inverse: a $Latex ^ {-1} $ = e
If only 50% (C.A.)=> Semi-Group
If Semi-Group + Neutral = (C.A.N.) = MoNoid
Note: No Inverse => not a Group
Arthur Cayley (UK, 1821-1895): first gave an abstract definition of Group @1854 while being a lawyer for 14 yrs, couldn't find a teaching job. His definition was ignored for 25 years until 1882 by Walter Van Dyck who gave the final Axiomatic definition of Abstract Group. ie above [C.A.N.I.]
'Zero' (Love) in Tennis Game
Why they say 'Love' instead of ‘Zero’?
'Love' is actually French's l'oeuf (egg), sound alike.
'Love' is actually French's l'oeuf (egg), sound alike.
Kolmogorov Teaching Method
Andrei Nikolaevich Kolmogorov (1903-1987)
Russian great mathematician with unique teaching style:
1. He didn't lecture, just ask questions, waiting for remarkable answer.
2. He introduced literature, music, hikes, excursions, expeditions.
3. He didn't mind if students didn't become mathematicians, as long as they had broad outlook and unsatisfied curiosity.
Russian great mathematician with unique teaching style:
1. He didn't lecture, just ask questions, waiting for remarkable answer.
2. He introduced literature, music, hikes, excursions, expeditions.
3. He didn't mind if students didn't become mathematicians, as long as they had broad outlook and unsatisfied curiosity.
Sum of cubes = Square of sum
The sum of cubes will always be a square – Beautiful !
$Latex 1^3+2^3+3^3+\cdots+n^3=(1 + 2 + 3 +\cdots+ n)^2$
$Latex 1^3+2^3+3^3+\cdots+n^3=(1 + 2 + 3 +\cdots+ n)^2$
π
Memorize: 1 3 5 => 113355
=> ∏ = 355/113
Note: Chinese mathematician Zhu Chongzhi (祖冲之 429–500 AD) computed ∏ up to 7 decimals, accuracy 1,000 yrs earlier than the West.
=> ∏ = 355/113
Note: Chinese mathematician Zhu Chongzhi (祖冲之 429–500 AD) computed ∏ up to 7 decimals, accuracy 1,000 yrs earlier than the West.
Solution 2 (Eigenvalue): Monkeys & Coconuts
Solution 2: Use Linear Algebra
Eigenvalue equation: A.X = λ.X
A =S(x)= $Latex \frac{4}{5}(x-1)$ where x = coconuts
S(x)=λx
Set λ = 1
$Latex \frac{4}{5}(x-1)=x$
x =-4
Also fifth monkey:
$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x-1)+(\frac{4}{5})^5+(\frac{4}{5})^4+\cdots+\frac{4}{5}$
$Latex 5^5$ divides (x-1)
Minimum positive x= -4 mod ($Latex 5^5$ )= $Latex 5^5$-4= 3121 [QED]
Note: λ=1 because "before" and "after" (transformation A) is the same.
Eigenvalue equation: A.X = λ.X
A =S(x)= $Latex \frac{4}{5}(x-1)$ where x = coconuts
S(x)=λx
Set λ = 1
$Latex \frac{4}{5}(x-1)=x$
x =-4
Also fifth monkey:
$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x-1)+(\frac{4}{5})^5+(\frac{4}{5})^4+\cdots+\frac{4}{5}$
$Latex 5^5$ divides (x-1)
Minimum positive x= -4 mod ($Latex 5^5$ )= $Latex 5^5$-4= 3121 [QED]
Note: λ=1 because "before" and "after" (transformation A) is the same.
Solution 1 (Sequence): Monkeys & Coconuts
Monkeys & Coconuts Problem
Solution 1 : iteration problem => Use sequence
$Latex U_{j} =\frac {4}{5} U_{j- 1} -1 $
(initial coconuts)
$Latex U_0 =k$
Let
$Latex f(x)=\frac{4}{5}(x-1)=\frac{4}{5}(x+4)-4$
$Latex U_1 =f(U_0)=f(k)= \frac{4}{5}(k+4)-4$
$Latex U_2 =f(U_1)=f(\frac{4}{5}(k+4)-4)= \frac{4}{5}((\frac{4}{5}(k+4)-4+4)-4$
$Latex U_2=(\frac{4}{5})^2 (k+4)-4$
$Latex U_3=(\frac{4}{5})^3 (k+4)-4$
$Latex U_4=(\frac{4}{5})^4 (k+4)-4$
$Latex U_5=(\frac{4}{5})^5 (k+4)-4$
Since
$Latex U_5$ is integer ,
$Latex 5^5 divides (k+4)$
k+4 ≡ 0 mod($Latex 5^5$)
k≡-4 mod($Latex 5^5$)
Minimum {k} = $Latex 5^5 -4$= 3121 [QED]
Solution 1 : iteration problem => Use sequence
$Latex U_{j} =\frac {4}{5} U_{j- 1} -1 $
(initial coconuts)
$Latex U_0 =k$
Let
$Latex f(x)=\frac{4}{5}(x-1)=\frac{4}{5}(x+4)-4$
$Latex U_1 =f(U_0)=f(k)= \frac{4}{5}(k+4)-4$
$Latex U_2 =f(U_1)=f(\frac{4}{5}(k+4)-4)= \frac{4}{5}((\frac{4}{5}(k+4)-4+4)-4$
$Latex U_2=(\frac{4}{5})^2 (k+4)-4$
$Latex U_3=(\frac{4}{5})^3 (k+4)-4$
$Latex U_4=(\frac{4}{5})^4 (k+4)-4$
$Latex U_5=(\frac{4}{5})^5 (k+4)-4$
Since
$Latex U_5$ is integer ,
$Latex 5^5 divides (k+4)$
k+4 ≡ 0 mod($Latex 5^5$)
k≡-4 mod($Latex 5^5$)
Minimum {k} = $Latex 5^5 -4$= 3121 [QED]
Note: The solution was given by Paul Richard Halmos (March 3, 1916 – October 2, 2006)
Monkeys & Coconuts Problem
5 monkeys found some coconuts at the beach.
1st monkey came, divided the coconuts into 5 groups, left 1 coconut which it threw to the sea, and took away 1 group of coconuts.
2nd monkey came, divided the remaining coconuts into 5 groups, left 1 coconut again thrown to the sea, and took away 1 group.
Same for 3rd , 4th and 5th monkeys.
Find: how many coconuts are there initially?
Note: This problem was created by Nobel Physicist Prof Paul Dirac (8 August 1902 – 20 October 1984). Prof Tsung-Dao Lee (李政道) (1926 ~) , Nobel Physicist, set it as a test for the young gifted students in the Chinese university of Science and Technology （中国科技大学－天才儿童班）.
1st monkey came, divided the coconuts into 5 groups, left 1 coconut which it threw to the sea, and took away 1 group of coconuts.
2nd monkey came, divided the remaining coconuts into 5 groups, left 1 coconut again thrown to the sea, and took away 1 group.
Same for 3rd , 4th and 5th monkeys.
Find: how many coconuts are there initially?
Note: This problem was created by Nobel Physicist Prof Paul Dirac (8 August 1902 – 20 October 1984). Prof Tsung-Dao Lee (李政道) (1926 ~) , Nobel Physicist, set it as a test for the young gifted students in the Chinese university of Science and Technology （中国科技大学－天才儿童班）.
Math is like Music
Prof Alice Chang Sun-Yung (University of Princeton, Geometric Analysis)
"Math is a language like music. To learn it systematically, it is necessary to master small pieces & gradually add another piece & then another... In a sense, Math is like the classical Chinese language - very polished, very elegant. Sitting in a good math lecture is like sitting in a good opera. Everything comes together."
"Math is a language like music. To learn it systematically, it is necessary to master small pieces & gradually add another piece & then another... In a sense, Math is like the classical Chinese language - very polished, very elegant. Sitting in a good math lecture is like sitting in a good opera. Everything comes together."
International Math Awards
1. Abel Prize: each yr, 1 or few share US$1m.
2. Wolf Prize: US$100K.
3. Shaw Prize (by Hong Kong Run-Run Shaw): US$ 1m, from 2002, every yr.
4. Clay Prize: 7 x US$ 1 m conjectures.
5. Beal Prize: US$100K
Prove Beal Conjecture: x^n + y^n = z^n if x, y, z have a common factor for n>2.
6. Fields Medal (equivalent to Nobel Prize): US$13,500, every 4 yrs, < 40 yr-old.
2. Wolf Prize: US$100K.
3. Shaw Prize (by Hong Kong Run-Run Shaw): US$ 1m, from 2002, every yr.
4. Clay Prize: 7 x US$ 1 m conjectures.
5. Beal Prize: US$100K
Prove Beal Conjecture: x^n + y^n = z^n if x, y, z have a common factor for n>2.
6. Fields Medal (equivalent to Nobel Prize): US$13,500, every 4 yrs, < 40 yr-old.
Catalan Conjecture
Catalan Conjecture (1 = 9 - 8 = 3² - 2³)
No more other 2 'consecutive' numbers (exclude 0, 1) like (8, 9), each of them is a pure power.
[2004 proved by Romanian Preda Mihailescu]
No more other 2 'consecutive' numbers (exclude 0, 1) like (8, 9), each of them is a pure power.
[2004 proved by Romanian Preda Mihailescu]
Calendar Math
Mental quick calendar days:
Reference Date: last day in Feb (doomsday) 28/2/2013= Thursday
The following days are all Thursday: 4/4, 6/6, 8/8, 10/10, 12/12
The odd Thursday:
5/9, 9/5 (9 am to 5 pm work)
7/11, 11/7 (Seven-eleven Shop)
7/3 (7+3 = 10)
Quick applications:
Singapore National day 9/8/2013 = Fri (8/8 + 1)
Xmas 26/12/2013 = Thurs (12/12 +2*7)
Reference Date: last day in Feb (doomsday) 28/2/2013= Thursday
The following days are all Thursday: 4/4, 6/6, 8/8, 10/10, 12/12
The odd Thursday:
5/9, 9/5 (9 am to 5 pm work)
7/11, 11/7 (Seven-eleven Shop)
7/3 (7+3 = 10)
Quick applications:
Induction in Geometry
Given: a unit length.
Use only a straightedge (ruler without markings) & a compass.
Prove: we can construct a line segment of √n for all n ∈N.
Proof:
1) n=1 (given).
2) Assume true for n, i.e. can construct √n
3) Construct a right-angled triangle with height = 1, base= √n
=> hypotenuse = $latex \sqrt {n+1} $
=> True for n+1
Therefore true for all n ∈N [QED]
Use only a straightedge (ruler without markings) & a compass.
Prove: we can construct a line segment of √n for all n ∈N.
Proof:
1) n=1 (given).
2) Assume true for n, i.e. can construct √n
3) Construct a right-angled triangle with height = 1, base= √n
=> hypotenuse = $latex \sqrt {n+1} $
=> True for n+1
Therefore true for all n ∈N [QED]
Singapore Modelling Math
Modelling Math versus Usual Method
Marie had 20 m of cloth. She used 3/5 of it to make some dresses. How many meters of cloth did she use?
1. Model method:
【】【】【】〖〗〖〗
5 units = 20 m
1 unit = 20 / 5= 4 m
3 units= 3 x 4 = 12 m
2. Usual method:
3/5 x 20 = 12 m
The 2nd method is too abstract for 8 or 9 yr-old kids who will resort to learn fraction “x” or “/” formula by rote.
Models appear 'silly' to adults who prefer abstract concept '3/5 x 20';
Note 1: Singapore Primary School Modelling Math
Adopted since 1990 from Polya’s “How to Solve It” (1945) 4-Step Problem Solving Process
1. Understand the problem in English Word Math.
2. Devise a plan: draw models
3. Carry out the plan: solve
4. Look back: verify
The Concrete-Pictorial-Abstract Math approach is developed by Dr. Kho Tek Hong (retired in 2007) & his MOE team @ 1980s.
Note 2: Models can also help visualize & 'concretize' Abstract Algebra (Set, Group, Coset, subgroup, kernel, isomorphism ...Galois theory). Feynman invented 'model' (Feynman diagram) to visualize quantum theory.
Marie had 20 m of cloth. She used 3/5 of it to make some dresses. How many meters of cloth did she use?
1. Model method:
【】【】【】〖〗〖〗
5 units = 20 m
1 unit = 20 / 5= 4 m
3 units= 3 x 4 = 12 m
2. Usual method:
3/5 x 20 = 12 m
The 2nd method is too abstract for 8 or 9 yr-old kids who will resort to learn fraction “x” or “/” formula by rote.
Models appear 'silly' to adults who prefer abstract concept '3/5 x 20';
Note 1: Singapore Primary School Modelling Math
Adopted since 1990 from Polya’s “How to Solve It” (1945) 4-Step Problem Solving Process
1. Understand the problem in English Word Math.
2. Devise a plan: draw models
3. Carry out the plan: solve
4. Look back: verify
The Concrete-Pictorial-Abstract Math approach is developed by Dr. Kho Tek Hong (retired in 2007) & his MOE team @ 1980s.
Note 2: Models can also help visualize & 'concretize' Abstract Algebra (Set, Group, Coset, subgroup, kernel, isomorphism ...Galois theory). Feynman invented 'model' (Feynman diagram) to visualize quantum theory.
Paul Erdös
[caption id="" align="alignright" width="246"]
head of Paul Erdös, Budapest fall 1992 (Photo credit: Wikipedia)[/caption]
Paul Erdös
1. The mathematician could carry his equations in his head, like a chess master playing blindfolded.
2. "A mathematician is a machine for turning coffee into theorems."
head of Paul Erdös, Budapest fall 1992 (Photo credit: Wikipedia)[/caption]Paul Erdös
1. The mathematician could carry his equations in his head, like a chess master playing blindfolded.
2. "A mathematician is a machine for turning coffee into theorems."
Mathematical Induction
1. First used @1575 by Francisco Maurolycus (Galileo's teacher) to prove 1 + 3 +5+...+ (2n-1) = n²
2. Pascal proved Pascal Triangle's Binomial Coefficient Theorem.
3. 'Induction' coined by Augustus De Morgan.
Ignorant Joke
A mathematician was interviewed by a TV reporter on Euclid who said 2,500 yrs ago that there are infinitely many primes.
The ignorant reporter asked immediately:
"Are there still infinitely many primes now?"
The ignorant reporter asked immediately:
"Are there still infinitely many primes now?"
Fields Medal Inscription
"Transire suum pectus mundoque potiri"
[To transcend human limitations & to master the universe]
[To transcend human limitations & to master the universe]
Friday, 29 March 2013
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