Mathematics: The Next Generation

Historical Backgroud:

Math evolves since antiquity, from Babylon, Egypt 5,000 years ago, through Greek, China, India 3,000 years ago, then the Arabs in the 10th century taught the Renaissance Europeans the Hindu-Arabic numerals and Algebra, Math progressed at a condensed rapid pace ever since: complex numbers to solve cubic equations in 16th century Italy, followed by the 17 CE French Cartersian Analytical Geometry, Fermat’s Number Theory,…, finally by the 19 CE to solve quintic equations of degree 5 and above, a new type of Abstract Math was created by a French genius 19-year-old Evariste Galois in “Group Theory”. The “Modern Math” was born since, it quickly develops into over 4,000 sub-branches of Math, but the origin of Math is still the same eternal truth.

Math Education Flaw: 本末倒置 Put the cart before the horse.

Math has been taught wrongly since young, either is boring, or scary, or mechanically (calculating).

This lecture by Queen Mary College (U. London) Prof Cameron is one of the rare Mathematician changing that pedagogy. Math is a “Universal Language of Truths” with unambiguous, logical syntax which transcends over eternity.

I like the brilliant idea of making the rigorous Math foundation compulsory for all S.T.E.M. (Science, Technology, Engineering, Math) undergraduate students. Prof S.S. Chern 陈省身 (Wolf Prize) after retirement in Nankai University (南开大学, 天津, China) also made basic “Abstract Algebra” course compulsory for all Chinese S.T.E.M. undergraduates in 2000s.

The foundations Prof Cameron teaches are centered around 4 Math Objects:

1. SET 集合
– Set is the founding block of the 20th century Modern Math, hitherto introduced into the world’s university textbooks by the French “Bourbaki” school (André Weil et al) after WW1.

Note: The last “Bourbaki” grand master Grothendieck proposed to replace Set by Category. That will be the next century Math for future Artificial Intelligence Era, aka “The 4th Human Revolution”.

2. FUNCTION 函数
– A vision first proposed by the German Gottingen School’s greatest Math Educator Felix Klein, who said Functions can be visualised in graphs, so it is the best tool to learn mathematical abstractness.

3. NUMBERS
– The German mathematician Leopold Kronecker, who once wrote that “God made the integers; all else is the work of man.”

– The universe is composed of numbers in “NZQRC” (ie Natural numbers, Integers, Rationals, Reals, Complex numbers). After C (Complex), no more further split of new numbers. Why?

4. Proofs
– reading and debugging proofs.

Example 1: Proof by Contradiction, aka Reductio ad Absurdum (Euclid’s Proof on Infinitely Many Prime Numbers)

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Challenge the proof: Why ?
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Induction intuitively by:
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Example 2: Proof by Logic
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[Hint:]
By Reasoning (which is unconscious), most would get “2 & A” (wrong answer)

By Logic (using consciousness), then you can proof …
Correct Answer: 2 and B
Test on all 3 Truth cases below in Truth Table:
p = front side
q = back side

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Recommended The Best Book on Abstract Algebra by Prof. Peter J. Cameron

French Curve

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 f and determine f'(x);
5. Find the limits of f within 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:

\displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}
Step 1: Determine the Domain of Definition D
D = R* = R – {0}

Step 2: There is no Periodicity and Symmetry of f
E = D = R*

[See Note below for Periodic and Symmetric example]

Step 3: Continuity of f
The function f is the quotient of 2 polynomial functions, therefore f is differentiable
=> f is continuous in ]-\infty,0[ \cup ]0,+\infty[
[See previous post CID Relation]

Step 4: Determine f’
\displaystyle\forall x \in R^{\star}, f'(x) = \frac{6x^{2}.2x^{2} - 4x (2x^{3}+27)}{4x^{4}} = \frac{4x^{4}-4.27x}{4x^{4}} = \frac{4x(x^{3}-27)}{4x^{4}}
\forall x \in R^{\star}, (x^{3} - 27 >0) \iff (x>3)
Therefore f’ has the same sign as x \mapsto x(x-3)

\begin{cases} \forall x \in ]-\infty,0[ \cup ]3,+\infty[, & f'(x)>0 \\  \forall x \in ]0,3[ , & f'(x)<0  \end{cases}

Step 5a: Limit at x=0

\displaystyle\lim_{x\to 0}(2x^{3}+27) = 27
\displaystyle\lim_{x\to 0} 2x^{2} = 0 , (\forall x \in R^{\star}, x^{2} >0)
Therefore, \displaystyle\lim_{x\to 0}f(x) = + \infty

Step 5b: Limit at x= + \infty
\displaystyle\lim_{x\to +\infty} f(x) =\lim_{x\to +\infty} \frac{2x^{3}+27}{2x^{2}}=\lim_{x\to +\infty} \frac{2x^{3}}{2x^{2}} = \lim_{x\to +\infty} x = +\infty

Step 5c: Limit at x= - \infty
Similarly,
\displaystyle\lim_{x\to -\infty} f(x) = \lim_{x\to -\infty} x = -\infty

Step 6: Construct the Table of Variation

\begin{array}{|l|l|l|}  \hline  x & - \infty \rightarrow \: \: \: \: 0 & 0 \:\:\:\:\: \rightarrow \:\:3 \rightarrow \:\:\: +\infty \\  \hline  f'(x) & \:\: \: \: \:\: \: + & \:\:\:\: - \:\:\:\:\:\:\:\:\: 0 \:\:\:\:\:\:\: + \\  \hline  f(x) & -\infty \nearrow +\infty & +\infty \searrow \: \frac{9}{2} \nearrow +\infty\\  \hline  \end{array}

Step 7: Study the infinite branches

7a) \displaystyle\lim_{x\to 0}f(x) = + \infty
=> y-axis is the asymptote

7b) \displaystyle\forall x \in R^{\star}, f(x) = \frac{2x^{3}+27}{2x^{2}}= x+\frac{27}{2x^{2}}
\displaystyle\lim_{x\to +\infty}\frac{27}{2x^{2}} = 0 , \displaystyle\lim_{x\to -\infty}\frac{27}{2x^{2}} = 0
=>
\displaystyle\lim_{x\to +\infty}f(x) = x , \displaystyle\lim_{x\to -\infty}f(x) = x
=> y= x is another asymptote
\forall x \in R^{\star}, \frac{27}{2x^{2}} >0
=> The curve C is above the asymptote y=x

Step 8: Study the remarkable points: intersection points with x-axis
\forall x \in R^{\star},(2x^{3}+27 =0)  \iff (x^{3}=-\frac{27}{2})  \iff (x=-\frac{3}{\sqrt[3]{2}}) = -2.38

Step 9: Draw the representative curve C of f.

french curve

french curve

Note:
\displaystyle\text{Let g: } x \mapsto \frac{sin x}{2- cos^{2}x}
D = R
g(x) is periodic of 2π => E = [0 , 2π]
\displaystyle\forall x \in R, g(-x)= \frac{sin (-x)}{2-cos^{2}(-x)}=-\frac{sin x}{2-cos^{2}x}=-g(x)
=> g(x) is symmetric with respect to the origin point O

We can restrict our study of g(x) in E = [0,π]

\displaystyle\forall x \in R, g(\pi-x)= \frac{sin (\pi-x)}{2-cos^{2}(\pi-x)}=\frac{sin x}{2-cos^{2}x}=g(x)
=> g(x) is symmetric w.r.t. to the equation x= π/2

Finally, we can further restrict our study of g(x) in E = [0, π/2]

g(x)_symmetric