Landau’s Beautiful Proofs

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Landau’s beautiful proofs:
1= cos 0 = cos (x-x)

Opening cos (x-x):
1 = cos x.cos (-x) – sin x.sin (-x)
=> 1= cos² x + sin² x
[QED]

Let cos x= b/c, sin x = a/c
1= (b/c)² + (a/c)²
c² = b² + a²
=> Pythagoras Theorem
[QED]

Landau (1877-1938) was the successor of Minkowski at the Gottingen University (Math) before WW II.

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Open Cubic Root

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Mental Trick

It was discovered by the Martial Art writer Liang Yusheng 武侠小说家 梁羽生 (《白发魔女传》作者), who met Hua Luogeng (华罗庚) @1979 in England:
2³= [8]
8³= 51[2]
3³= 2[7]
7³= 34[3]

The last digit pairs :
[2 <->8] , [3 <-> 7]
Others unchanged.

Example:

$latex \sqrt[3]{658503} = N$
Last three digits 503 <-> …[7]
First three digits 658:
 (8³ =512)< 658 < (729 = 9³)
=>  8
Answer : $latex \sqrt[3]{658503} = N$= 87
Note: Similar trick for opening $latex \sqrt[23] {200 digits}$ by an indian lady Ms Shakuntala (83) dubbed “Human computer”.

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Solution 2 (Eigenvalue): Monkeys & Coconuts

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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

Since each iteration of the transformation caused the coconut status ‘unchanged’, which means λ = 1 (see remark below)

$Latex \frac{4}{5}(x-1)=x$
We get
x = – 4

Also by recursive, after the fifth monkey: $Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x-1)- (\frac{4}{5})^4-(\frac{4}{5})^3- (\frac{4}{5})^2- \frac{4}{5}$

$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x) – (\frac{4}{5})^5 – (\frac{4}{5})^4 – (\frac{4}{5})^3+(\frac{4}{5})^2 – \frac{4}{5}$

 

$Latex 5^5$ divides (x)

Minimum positive x= – 4 mod ($Latex 5^{5}$ )= $Latex 5^{5} – 4$= 3,121 [QED]

 

Note: The meaning of eigenvalue  λ in linear transformation is the change  by a scalar of λ factor (lengthening or shortening by λ) after the transformation. Here

λ = 1 because “before” and “after” (transformation A)  is the SAME status (“divide coconuts by 5 and left 1”).

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Solution 1 (Sequence): Monkeys & Coconuts

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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]

Note: The solution was given by Paul Richard Halmos (March 3, 1916 – October 2, 2006)

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Monkeys & Coconuts Problem

Math Online Tom Circle

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).

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French Curve

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The French method of drawing curves is very systematic:

“Pratique de l’etude d’une fonction”

Let f be the function represented by the curve C

Steps:

1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of fand determine f'(x);
5. Find the limits of fwithin the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.

Example:

$latex \displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}$
Step 1: Determine the Domain of Definition D
D = R* = R –…

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Cut a cake 1/5

Math Online Tom Circle

Visually cut a cake 1/5 portions of equal size:

1) divide into half:

20130513-111010.jpg

2) divide 1/5 of the right half:

20130513-133441.jpg

3) divide half, obtain 1/5 = right of (3)

$latex \frac{1}{5}= \frac{1}{2} (\frac{1}{2}(1- \frac{1}{5}))= \frac{1}{2} (\frac{1}{2} (\frac{4}{5}))=\frac{1}{2}(\frac{2}{5})$

20130513-171052.jpg

4) By symmetry another 1/5 at (2)=(4)

20130513-174541.jpg

5) divide left into 3 portions, each 1/5

$latex \frac{1}{5}= \frac{1}{3}(\frac{1}{2}+ \frac{1}{2}.\frac{1}{5}) = \frac{1}{3}.\frac{6}{10}$

20130513-174742.jpg

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Tuition That We May Have To Believe In

This insightful article makes a really good read.

Quotes from the article:

To be honest, the amount to be learnt at each level of education is constantly increasing, and tuition could just help you get that edge over others. After all, it was meant to be supplementary in nature.

The toughest part at the end of the day however, is probably this: getting the right tutor.

guanyinmiao's musings (Archived: July 2009 to July 2019)

This commentary, “Tuition That We May Have To Believe In”, is a reply to a previous article on tuition by Howard Chiu (Mr.), “Tuition We Don’t Have To Believe In” (Read).

I must say Howard’s article had me on his side for a moment. He appealed to me emotively. Nothing like a mental picture of some kid attending hours and hours of tuition immediately after school when he could well be enjoying himself thoroughly with… an iPhone or iPad (I highly doubt kids these days still indulge their time at playgrounds). But the second time I read his article, I silenced the part of my brain which still prays the best for children, so do pardon me if I sound a tad too pragmatic at times.

The overarching assertion that Howard projects his points from is that there is “huge over consumption of this good”. Firstly, private tutoring…

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Generalized Analytic Geometry

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Generalized Analytic Geometry

Find the equation of the circle which cuts the tangent 2x-y=0 at M(1,4), passing thru point A(4,-1).

Solution:

1st generalization:
Let the point circle be:
(x-1)² + (y-4)² =0

2nd generalization:
It cuts the tangent 2x-y=0
(x-1)² + (y-4)² +k(2x-y) =0 …(C)

Pass thru A(4,-1)
x=4, y= -1
=> k= -2
(C): (x-3)² + (y-1)² =…
[QED]

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Math Chants

Math Online Tom Circle

Math Chants make learning Math formulas or Math properties fun and easy for memory . Some of them we learned in secondary school stay in the brain for whole life, even after leaving schools for decades.

Math chant is particularly easy in Chinese language because of its single syllable sound with 4 musical tones (like do-rei-mi-fa) – which may explain why Chinese students are good in Math, as shown in the International Math Olympiad championships frequently won by China and Singapore school students.

1. A crude example is the quadratic formula which people may remember as a little chant:
ex equals minus bee plus or minus the square root of bee squared minus four ay see all over two ay.”

$latex \boxed{
x = \frac{-b \pm \sqrt{b^{2}-4ac}}
{2a}
}$

2. $latex \mathbb{NZQRC}$
Nine Zulu Queens Rule China

3. $latex \boxed {\cos 3A = 4\cos^{3}…

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What is “sin A”

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What is “sin A” concretely ?

1. Draw a circle (diameter 1)
2. Connect any 3 points on the circle to form a triangle of angles A, B, C.
3. The length of sides opposite A, B, C are sin A, sin B, sin C, respectively.

Proof:
By Sine Rule:

$latex \frac{a}{sin A} = \frac{b}{sin B} =\frac{c}{sin C} = 2R = 1$
where sides a,b,c opposite angles A, B, C respectively.
a = sin A
b = sin B
c = sin C

20130421-193110.jpg

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