Quod Erat Demonstrandum

2008/01/29

小報告:濫藥

Filed under: Report — johnmayhk @ 7:44 下午

主題:如何遏止青少年濫藥
日期:2008-01-23(FRI)
時間:1:45 p.m. ~ 5:30 p.m.
地點:Queen Elizabeth School

先是教育局的官員致辭,他多次高聲把『遏止』讀成『蠍子』,唉。

第一位講者,保安局禁毒處的代表,談的是青少年濫藥現況。他特別指出吸毒和藏毒兩者皆犯法。看他一個個的圖表,顯示了青少年(21歲以下者)濫藥情況嚴重。聽他把 powerpoint一張張讀出,不到十分鐘,我前後左右的同工,也難敵睡眼惺忪。

(more…)

廣告

2008/01/28

Hilbert’s work

Filed under: Family — johnmayhk @ 5:53 下午

下圖是兩歲多的兒子 Hilbert 的『攝影作品』。

圖中顯示的是某某星系嗎?不是啦,那只是Hilbert拿著數碼相機在我的手提電腦螢幕前隨意亂拍的『成果』。那一點點的不是星,而是螢幕上的塵。

讓兩歲小朋友拿相機拍攝?我和太太教子的方式不同。

若見兒子手拿剪刀,我和太太有不同的處理方式。
我:沒收剪刀,罵他危險不要再拿來玩。
太太:讓他繼續,並耐心教他如何安全地運用剪刀。

相機也是這個道理。

2008/01/27

Music appreciation: Vitas

Filed under: Fun — johnmayhk @ 1:37 上午

2008/01/25

Forum about Scilab

Filed under: Uncategorized — johnmayhk @ 11:07 下午

http://groups.google.com/group/ade-scilab/browse_thread/thread/1d62c7df46bc2b50/f8fdb984325a3b44#f8fdb984325a3b44

[CE][Math] Statistics: More dispersed?

Filed under: HKCEE — johnmayhk @ 2:52 下午

In CE mathematics, we know that there are at least 3 measurements for dispersion, namely, range, inter-quartile range and standard derivation.

Describing or comparing the dispersion of different sets may be a very important application.

About the standard derivation, you may read my old messages for your reference
http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1159940&t=1159940&v=t
http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=965435&t=959956

Given two numerical data sets A and B.

If we come across the following situations

range of A > range of B and
inter-quartile range of A < inter-quartile range of B

How to compare which set is more dispersed?

Well, you may say, just look at the standard derivation.

OK. If we have the following

range of A > range of B and
inter-quartile range of A > inter-quartile range of B and
standard derivation of A < standard derivation of B

Is it true to draw the conclusion that set B is more dispersed?

Further, how about when (set A \ne set B) such that

range of A = range of B and
inter-quartile range of A = inter-quartile range of B and
standard derivation of A = standard derivation of B

Urm, let me give the following as an example

Set A ={1,10,10,10}
Set B = (1,1,1,10}

Is it true to draw the conclusion that both sets have the same dispersion?

The problem may be something about the vagueness of the concept of “dispersion". How to resolve it?

2008/01/24

Pure Mathematics 補底

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 3:08 下午

更新日期:2008-02-28 (希望不斷更新。)
免插聲明:本文純粹為修純粹數學科的中七同學純粹補底之用,高手勿插,謝謝。

1.
If f(x) is a polynomial of degree 2, and f(1) = f(2) = 0, then it is NOT true to say
f(x) = (x-1)(x-2).
Instead, we should write
f(x) \equiv k(x-1)(x-2) for some non-zero constant k.
[Read : 2004-AL-pure math-paper I-Q.4]

2.
Let f(x) be a polynomial of degree not less than 3.
If f(x) is divided by (x-1)(x-2)(x-3), then it is NOT true to say
f(x) = (x-1)(x-2)(x-3)Q(x) + R
(where R is a constant, the remainder)
Instead, we should write
f(x) \equiv (x-1)(x-2)(x-3)Q(x) + ax^2 + bx + c
that is,
we should set the remainder as a polynomial of degree 2
(just less than the degree of the divisor (x-1)(x-2)(x-3) by 1)

3.
Resolve \frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} into partial fraction.
Don’t write the following
\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} = \frac{A}{x} + \frac{Bx + C}{x^2} + \frac{D}{x - 1} + \frac{Ex + F}{(x - 1)^2}.
Instead, we should write
\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} \equiv \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}
[Read : 2000-AL-pure math-paper I-Q.12]

4.
If x > k, it is NOT true to draw the following conclusion
x^2 - x + 1 > k^2 - k + 1.
But, if we know x + k - 1 > 0, then it is true to have x^2 - x + 1 > k^2 - k + 1.
Proof: Multiply x + k - 1 > 0 and x - k > 0, we have (x^2 - k^2) - (x - k) > 0 and result follows.
[Read : 1999-AL-pure math-paper I-Q.2]

5.
Let {x_n} be a sequence of real numbers.
Knowing that x_n > n.
It is NOT true to say
{x_n} is increasing.
Just give an example.
x_1 = 10, x_2 = 9, x_3 = 8, x_4 = 7, x_5 = 6, ...
Obviously, x_n > n (for n = 1,2,3,4,5), but {x_n} is not increasing.
[Read 1999-AL-pure math-paper I-Q.2]

6.
Let {x_n} be a sequence of real numbers.
Knowing that x_n < 1 - \frac{1}{n}.
We cannot say {x_n} is bounded from above by 1 - \frac{1}{n}.
1 - \frac{1}{n} is NOT an upper bound of the sequence {x_n}.
Instead, with a small effort added, we have
x_n < 1 - \frac{1}{n} < 1 for all positive integers n.
And now, it is clear that, 1 is an upper bound of {x_n}.

7.
(a) Solve |x| < 3
Solution: -3 < x < 3
(b) Solve |x| > 3
Solution: x > 3 or x < -3

8.
Consider the quadratic equation
ax^2 + bx + c = 0
Given that the discriminant \Delta < 0.
Then it is NOT true to say that the quadratic equation has no real solution.
Just consider a quadratic equation with roots 0 and i.
That is
x(x - i) = 0
x^2 - ix = 0
Then \Delta = (-i)^2 - 4(1)(0) = -1 < 0.
But there is a real root (x = 0) for the quadratic equation.

9.
“Vector" is out of the syllabus now.
“Linearly independent", “linearly dependent" are out-of-syllabus concepts.

10.
Techniques in evaluating limits
(a) l’hôpital rule
(b) Special limit : \lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1
(c) Special limit : \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = e
(d) Special limit : \lim_{a \rightarrow 0}(1 + a)^{\frac{1}{a}} = e
(e) Boundedness : Knowing that {a_n} is bounded and \lim_{n \rightarrow \infty}b_n = 0, then \lim_{n \rightarrow \infty}a_nb_n = 0
(f) Squeezing (Sandwich) principle
(g) Taking limit on the recurrence relation : a_{n + 1} = 2a_{n} and suppose {a_n} converges \Rightarrow \lim_{n \rightarrow \infty}a_{n + 1} = \lim_{n \rightarrow \infty}(2a_{n}) \Rightarrow \lim_{n \rightarrow \infty}a_n = 2\lim_{n \rightarrow \infty}a_n \Rightarrow \lim_{n \rightarrow \infty}a_n = 0

11.
Symbols about left and right hand limits
x \rightarrow 3^+" does NOT mean “x \rightarrow +3"
x \rightarrow 3^-" does NOT mean “x \rightarrow -3"
See the difference? Try to obtain the following by sight.
(a) \lim_{x \rightarrow 3^+}\sqrt{x - 3} = 0
(b) \lim_{x \rightarrow 3^-}\sqrt{x - 3} does not exist.
(c) \lim_{x \rightarrow +3}\sqrt{x - 3} does not exist (since the left and right lmits are not the same)
(d) \lim_{x \rightarrow -3}\sqrt[3]{x + 3} = 0

12.
To evaluate \lim_{x \rightarrow \infty}[f(2004 + x) - f(x)], please do NOT split it as
\lim_{x \rightarrow \infty}f(2004 + x) - \lim_{x \rightarrow \infty}f(x)
No! Both limits may NOT exist!
You may try “rationalization", “sum to product", “MVT" etc.

13.
How to evaluate \lim_{x \rightarrow \infty}[f(\sqrt{2004 + x}) - f(\sqrt{x})] = ?
Try the Mean Value Theorem (of course, f should be differentiable at certain interval), that is
\lim_{x \rightarrow \infty}[f(\sqrt{2004 + x}) - f(\sqrt{x})]
= \lim_{x \rightarrow \infty}f'(c)(\sqrt{2004 + x} - \sqrt{x}) for some c \in [\sqrt{x} , \sqrt{2004 + x}]
= \lim_{x \rightarrow \infty}f'(c)[\frac{2004 + x - x}{\sqrt{2004 + x} + \sqrt{x}}] for some c \in [\sqrt{x} , \sqrt{2004 + x}]
= \lim_{x \rightarrow \infty}f'(c)[\frac{2004}{\sqrt{2004 + x} + \sqrt{x}}] for some c \in [\sqrt{x} , \sqrt{2004 + x}]
(if \lim_{x \rightarrow \infty}f'(c) is bounded, then the limit is zero, read 10.(e))

14.
If \lim_{x \rightarrow 0}\frac{g(x)}{x} exists, then \lim_{x \rightarrow 0}\frac{g(x)} = 0.
e.g. Given \lim_{x \rightarrow 0}\frac{f(x) - 1}{x} = 1, evaluate \lim_{x \rightarrow 0}f(x). (ANS: 1)
[Read : 2002-AL-pure math-paper II-Q.3]

15.
Be careful, the following may NOT be true!
\lim_{x \rightarrow a}(f(x) + g(x)) = \lim_{x \rightarrow a}(f(x) + \lim_{x \rightarrow a}g(x))
e.g. Given \lim_{x \rightarrow 0}\frac{f(x) - 1}{x} = 1, evaluate \lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}
It is WRONG to write
\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}
= \lim_{x \rightarrow 0}\frac{f(x) - 1}{x \times 1} (for \lim_{x \rightarrow 0}e^x = 1)
= 1
You should give somthing like
\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}
= \lim_{x \rightarrow 0}\frac{f(x) - 1 + 1 - e^x}{xe^x}
= \lim_{x \rightarrow 0}(\frac{f(x) - 1}{xe^x} + \frac{1 - e^x}{xe^x})
= \lim_{x \rightarrow 0}(\frac{f(x) - 1}{xe^x} + \frac{-e^x}{xe^x + e^x}) (l’hôpital rule)
= 0 (see, NOT 1)
[Read : 2002-AL-pure math-paper II-Q.3]

16.
Please please please DON’T make the following silly mistakes!
(a) \int\tan^n(x)dx \neq \frac{tan^{n + 1}}{n + 1} + C
(b) \int e^{x^2}dx \neq e^{x^2} + C
For (a), try the reduction formula.
For (b), there is no closed form, just stop there.

17.
The following definitions MUST be memorized!
(a) “f(x) is well-defined at x = a" means “f(a) can be found".
(b) “f(x) is continuous at x = a" means “\lim_{x \rightarrow a}f(x) = f(a)“.
(c) “f(x) is differentiable at x = a" means “\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h} exists".

18.
Q: At a look at the definitions above, why there is no mention of the left and right hand limits? When should we consider the left and right hand limits?
A: When f(x) has DIFFERENT expressions for x \ge a and x < a (say)
e.g.
(a) Let f(x) = x + 1 for x \ge 3 and f(x) = 2(x - 1) for x < 3. Then f(x) is continuous at x = 3. Proof: \lim_{x \rightarrow 3^+}f(x) = \lim_{x \rightarrow 3^+}(x + 1) = 4 while \lim_{x \rightarrow 3^-}f(x) = \lim_{x \rightarrow 3^-}2(x - 1) = 4 = f(3) Hence f(x) is continuous at x = 3.
(b) Let f(x) = x + 1 for x > 3 and f(x) = 2(x - 1) for x < 3 with f(3) = 5 Then f(x) is NOT continuous at x = 3. (why?)
(c) Let f(x) = x|x|. To prove f(x) is continuous at x = 0, it is NOT necessary to “divide cases", since both left and right hand limits are zero. i.e. \lim_{x \rightarrow 0^+}x = \lim_{x \rightarrow 0^-}x = 0, hence \lim_{x \rightarrow 0}x = 0 and therefore, \lim_{x \rightarrow 0}x|x| = 0 = f(0). Done.

19.
Given

f(x) = f_1(x) for x \ge a
f(x) = f_2(x) for x < a

(both f_1(x) and f_2(x) are ‘nice’ functions)
Then, to find f'(x), the problem only appears at x = a, hence DON’T write

f'(x) = f_1'(x) for x \ge a
f'(x) = f_2'(x) for x < a

Instead, you should write

f'(x) = f_1'(x) for x > a
f'(x) = f_2'(x) for x < a

And try to check the value of f'(x) at x = a by finding

f_{+}'(a) = \lim_{h \rightarrow 0^+}\frac{f_1(a + h) - f(a)}{h}
f_{-}'(a) = \lim_{h \rightarrow 0^-}\frac{f_2(a + h) - f(a)}{h}

and if f_{+}'(a) = f_{-}'(a), then f'(a) is the common limit.

20.
Techniques in integrations: substitution
(a) Involves \sqrt{a^2 - x^2} , substitute x = a\sin(\theta) or x = a\cos(\theta)
(b) Involves \sqrt{x^2 - a^2} , substitute x = a\sec(\theta)
(c) Involves x^2 + a^2 , substitute x = a\tan(\theta)

21.
Techniques in integrations: partial fractions
(a) Involves x^2 - a^2 , partial fractions may help
(b) Involves x^4 + a^4 , don’t forget
x^4 + a^4
= x^4 + 2x^2a^2 + a^4 - 2x^2a^2
= (x^2 + a^2)^2 - (\sqrt{2}ax)^2
= (x^2 + \sqrt{2}ax + a^2)(x^2 - \sqrt{2}ax + a^2)
and then partial fraction resolving

22.
Techniques in integrations: others
(a) Integration by parts
(b) Taking limits
e.g. \lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx = ?
Let I_n = \int_{0}^{1}\frac{x^{n}}{1 + x^2}dx, then
0 \le I_n = \int_{0}^{1}\frac{x^{n}}{1 + x^2}dx \ge \int_{0}^{1}\frac{x^{n + 1}}{1 + x^2}dx = I_{n + 1}
That is {I_n} is decreasing and bounded from below by zero.
Hence \lim_{n \rightarrow \infty}I_n exists, says L.
For sufficiently large n, write
\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx = \int_{0}^{1}\frac{x^{n - 2}(x^2 + 2x + 1 - 2x - 1)}{1 + x^2}dx = \int_{0}^{1}(x^{n - 2} -\frac{2x^{n - 1}}{1 + x^2} - \frac{x^n}{1 + x^2})dx = \frac{1}{n - 1} -2\int_{0}^{1}\frac{x^{n - 1}}{1 + x^2}dx - \int_{0}^{1}\frac{x^n}{1 + x^2}dx
Taking n \rightarrow \infty from both sides,
L = 0 - 2L - L
Hence, \lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx = L = 0
[Read : 1990-AL-pure math-paper II-Q.8(a)]

23.
Many questions involve the Mean Value Theorem in Cauchy form, that is
both f(x) and g(x) are continuous on [a , b] and differentiable on (a , b) with g(a) \neq g(b), then
then \frac{f(a) - f(b)}{g(a) - g(b)} = \frac{f'(c)}{g'(c)} for some c \in (a , b).
DON’T write the following as a so-called proof

f(a) - f(b) = f'(c)(a - b)
g(a) - g(b) = g'(c)(a - b)
\Rightarrow \frac{f(a) - f(b)}{g(a) - g(b)} = \frac{f'(c)}{g'(c)} for some c \in (a , b)

NO! We CANNOT ensure both c in f'(c) and g'(c) are the same!!! That is, all we have may be the following

f(a) - f(b) = f'(c_1)(a - b) for some c_1 \in (a , b)
g(a) - g(b) = g'(c_2)(a - b) for some c_2 \in (a , b)

and then we can do nothing at all…How to prove? Try the read
[Read 2006-AL-pure math-paper II-Q.11(a)]

24.
Get tired to give details, just give topics in major techniques:
(a) Integration by parts
(b) Fundamental theorem of integral calculus
(c) Inequalities
(d) Riemann sum

25.
Partial fractions : there is a short-cut for distinct linear factors, just in case you may forget…

\frac{6x^2 - 18x + 6}{x(x - 2)(x - 3)}
= \frac{6(0)^2 - 18(0) + 6}{x(0 - 2)(0 - 3)} + \frac{6(2)^2 - 18(2) + 6}{(2)(x - 2)(2 - 3)} + \frac{6(3)^2 - 18(3) + 6}{3(3 - 2)(x - 3)}
= \frac{1}{x} + \frac{3}{x - 2} + \frac{2}{x - 3}

2008/01/21

[報告] 某次崇拜的經驗

Filed under: Life,Teaching — johnmayhk @ 4:16 下午

星期天重返恩福堂,欲出席 10:30 a.m. 舉行的崇拜。我和太太遲了 5 分鐘,位於長沙灣道及大南西街交界處的恩福中心樓下已大排長龍,水泄不通!太太問隊中的某某,她說 10:30 a.m. 那埸已爆滿,現在排的是 11:00 a.m.!對於出身自蚊型教會的我,很難想像返教會崇拜好像往演唱會般,會『全場爆滿』,遲到者還被摒諸門外。(進天國的啟示??)我和太太現在還是『打游擊』信徒,於是『飛的』趕往九龍城靈糧堂。余牧師先播放『老媽子之歌』,再在講道中引『老師的啟示』一文。對那歌,我沒有什麼感覺,已聽過多次;但那文章,腦中除了『為何香水在多年後仍未揮發?』這類膚淺的問題外,心中確有所感。很贊同有人刻意分清『授課員』及『教師』這兩個稱謂。確實,『授課員』比真正育人的教師層次低很多,在這個教育的時空下,當一個稱職的授課員恐怕也不易,還可奢望達到那種層次(是政府宣傳的那種嗎?)?可以的,如果我們有『在人不能,在祢凡事都能』的信心。

下午忽聞舊同事陳維昌老師離世的消息,心中非常沉重。這位好好先生,享年六十三歲,懷念他在集隊時的宣佈,懷念和他一起上太極班的時光。

2008/01/16

[Fun] 濟記狂想 gag : 檢查頭髮

Filed under: Fun — johnmayhk @ 9:58 下午

礙於自己的學力和表達力,相信很多同學其實「唔知我寫咩」。好,今次讓我畫下「公仔」,大家應該看得明。下圖是昨天午飯時間草草而作:

jngag01sfxc.jpg

這是學生心態嗎?

[初中] Factorization, useful or not?

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 9:44 下午
Tags: ,

Is the topic “factorization" in junior form useful? Urm, apart from simplification, solving equation etc, will factorization play a critical role in advanced mathematics? No idea, but at least, I could tell you that we may generalize the idea of factorization in set theory. Take it easy. I just wanna say something in my daily teaching. This topic may have no application (?) in further study in mathematics, however, it may be useful as one of the tools of training of students’ mind. When letting my students play with some challenging (at least, not-straight-forward) problems, nealy all of them are willing to do and ask for help. Well, it is a good moment of promoting the beauty of mathematics!

The following are questions asked from one of my students, Lo, in class. Factorize

1. 2x^2 + 6x - 15z - 2xy + 5yz - 5xz
2. x^4 + 2x^3 + 3x^2 + 2x + 1

Questions above are not difficult, but at least, we could not just apply identities directly without doing some grouping beforehand. The art (art? yes, sometimes, it is more than ‘technique’, it may be an ‘art’) is: how?

For Q.1

2x^2 + 6x - 15z - 2xy + 5yz - 5xz
= 2x^2 - 2xy + 6x - 15z + 5yz - 5xz
= 2x(x - y) + 3(2x - 5z) + 5z(y - x)
= 2x(x - y) + 3(2x - 5z) - 5z(x - y)
= (x - y)(2x - 5z) + 3(2x - 5z)
= (2x - 5z)(x - y + 3)

Teachers can easily set up factorization problems by just expanding polynomials, i.e., starting from (2x - 5z)(x - y + 3), we can come up with 2x^2 + 6x - 15z - 2xy + 5yz - 5xz and ask students to factorize it back. However, if we change z (say) into number, the problem may not be easy to solve. That is, considering

(2x - 5)(x - y + 3)
= 2x^2 - 2xy + 6x - 5x + 5y - 15
= 2x^2 - 2xy + x + 5y - 15

Now, may be students find it difficult to factorize 2x^2 - 2xy + x + 5y - 15.

Further, if we consider the product of trinomials (three terms) with 3 variables x,y,z, it may already be a nightmare, say, could you factorize the following

3x^2 - 4y^2 - 3z^2 - 4xy + 8yz - 8zx ?

For Q.2

x^4 + 2x^3 + 3x^2 + 2x + 1
= x^4 + 2x^3 + (x^2 + 2x^2) + 2x + 1
= x^2(x^2 + 2x + 1) + 2x^2 + 2x + 1
= x^2(x + 1)^2 + 2x(x + 1) + 1
= (x(x + 1))^2 + 2x(x + 1) + 1
= (x(x + 1) + 1)^2
= (x^2 + x + 1)^2

Following this idea, we can create many, like

(a) x^8 + 2x^7 + 3x^6 + 4x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1
(b) x^{10} + 2x^9 + 3x^8 + 4x^7 + 5x^6 + 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1
(c) x^9 + 3x^8 + 6x^7 + 10x^6 + 12x^5 + 12x^4 + 10x^3 + 6x^2 + 3x + 1

It is extremely easy to set up questions above, while, it may not be easy to solve them immediately.

Just help you a bit, observe the following

111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321

Well, it is also an art to strike a balance: give some challenging questions to students without frightening them.

When I was in F.2, I found an old little mathematics book of only one single theme: factorization. I’d lost it long time ago. It taught me many techniques in factorization, as for example, I knew how to factorize something like a^3 + b^3 + c^3 - 3abc (read my old post if you want to). But, it is quite demanding to ask a F.2 student to factorize the following

x^{27} - 1

though it is just a piece of cake for F.6 or F.7 students to use the techniques in complex numbers and obtain

x^{27} - 1
= (x-1)(x^2 - 2x\cos(\frac{2\pi}{27}) + 1)(x^2 - 2x\cos(\frac{4\pi}{27}) + 1)(x^2 - 2x\cos(\frac{6\pi}{27}) + 1)\dots(x^2 - 2x\cos(\frac{26\pi}{27}) + 1)

 There was a funny problem called Tschebotarev problem concerning the factorization of x^n - 1, see if I can find more and share with you next time.

– – – – – –

Just wanna add something below.

When introducing the identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

in class, one student, Choi, puzzled that, ‘why can’t we write something like

a^3 - b^3 = ((a^{1.5})^2 - (b^{1.5})^2) = (a^{1.5} - b^{1.5})(a^{1.5} + b^{1.5})

Good question! When putting concrete numbers, it works fine, e.g.

16^3 - 9^3 = (16 ^{1.5})^2 - (9 ^{1.5})^2 = (16 ^{1.5} - 9 ^{1.5})(16 ^{1.5} + 9 ^{1.5})

But, what exactly is the question asking? Factorization of polynomials.

The mathematics object on the L.H.S., a^3 - b^3 is a polynomial in a, b; however, the thing on the R.H.S., like (a ^{1.5} - b ^{1.5}) is NOT a polynomial in a, b. It is because the indices involved are NOT non-negative integers. Hence, the suggestion by the student should not be acceptable.

This is one of the rules. The other is the kind of coefficients.

Also read

https://johnmayhk.wordpress.com/2007/12/02/%e5%88%9d%e4%b8%ad-factorization/
 

2008/01/14

[AL][PM] Hyperpower function

Filed under: HKALE,Pure Mathematics,University Mathematics — johnmayhk @ 4:39 下午

Justin asked me again about the hyperpower function. Sorry, I just re-post my old message in my old forum, which was prohibited by the adminstrators in the hkedcity. It is known that the following

converges if and only if

To know more about the hyperpower function, please visit the following

http://www.faculty.fairfield.edu/jmac/ther/tower.htm
http://mathworld.wolfram.com/PowerTower.html

We may create an AL pure mathematics question concerning hyperpower function like

Let
(a) Prove that {x_n} is increasing.
(b) By M.I. or otherwise, show that {x_n} is bounded above by 2.
(c) Hence solve the outdated problem:

2008/01/12

[軟件][教學] Scilab 在中學數學的應用初探

Filed under: HKALE,HKCEE,Junior Form Mathematics,Teaching — johnmayhk @ 6:01 下午

以往介紹過一些網上的數學工具,諸如 integrator 及 calc101 等。鑑於部分教師及同學沒有上網習慣,讓我介紹一個離線作業且完全免費的數學工具:Scilab,希望大家用得著。關於這個法國研發的開放軟件,只要大家 google 一下,不難找到更詳盡的資訊,在此不贅。感謝有關研究員之努力及慷慨!幾個月前,我才知這軟件存在,這裡我主要提及它在中學數學的一些應用(再進深的我不懂了):

大家先下載(現時)軟件的最先版本
http://www.scilab.org/download/4.1.2/scilab-4.1.2.exe

安裝後,執行便出現以下版面:

看到閃動著的游標(cursor)嗎?可以輸入東西了。為方便大家『唔駛打』,可下載以下的 text file,之後的例子都在這 file 中,大家可以 copy and paste 來試試例子的效果。

http://johnng.inscyber.net/mathgif2/Scilab-johnmayhk.txt

操作小提示:輸入後,按 Enter 顯示結果。打錯了或想修改某些之前輸入的東西,只要按向上箭咀鍵若干次,重新輸入便可。

數字(Numbers)

以下是一般基本計算機也可處理的運算工作。

這是一些初等函數(elementary functions)的運算,留意其輸入的方式。

更多的函數,比如 arcsin 是 asin(),可參考軟件中的 help file。

多項式(Polynomials)
這裡處理的是一元(single variable)的多項式。對多項式的輸入,我們不能直接輸入諸如 x+1 之類,因為電腦不知道何謂 x,我們要先定義清楚,才可繼續運算。這裡介紹 3 種定義方法:

(1) 直接輸入

(2) 透過系數(coefficients)
當多項式的項數多,直接輸入頗麻煩,我們可以透過系數定義,詳見如下:

留意,不一定用 x,其他字母作 variable 也可,但 coeff 這個就字不能修改。

(3) 透過根 (roots)

方便呀,forming equation 的題目,立即得到結果!

好了,定義了多項式,我們可以進行有關多項式的運算。

1. 基本運算

2. 有關除法

3. 解多項式方程 (solve polynomial equations)

4. 因式分解(factorization)

頗有用的功能!不過,對一些根為『不漂亮』的有理數之多項式,比如 – 7 + 26*x – 19*x^2 + 12*x^3,它的因式分解的答案也同樣『不漂亮』,大家不妨試試及想一想解決方法。(注,我知配合 Maxima.exe 或用 Mathematica 可處理多元的多項式。但不太懂,希望以後再研究一下。)

5. 求最大公因式(GCD)及解丟番圖方程(Diophant equation)
讀 AL Pure Mathematics 的同學注意了,這是 algebra 中輾轉相除法 (Euclidean algorithm) 的題目,感覺如何呢?

6. 補充一下:有關整數求 HCF 及 LCM 的方法

7. 有理函數化成部分分式(resolving into partial fractions)
Partial fractions 這類熱門的 pure mathematics 題目,也可以 Scilab 輕易處理。

顯示的三個ans就是答案了,即

矩陣(Matrices)
Scilab 最強大的功能是處理矩陣,諸君請看!開始時,當然要定義矩陣。方法是相當簡單。

好了,定義了矩陣,我們可以進行有關矩陣的運算。

注意:矩陣的 entries,可以是多項式,運算如舊!

這是有關矩陣『自身』的運算,修 Pure Mathematics 的同學,感覺良好嗎?

其他功能

Scilab 可做的遠超上述,除了有關統計數據資料之處理,還有下列一些。但因為別的,常用的軟件也有類似功能,我略舉以下數例作結,圖收拋磚引玉之效。

2008/01/09

yummy…

Filed under: Life — johnmayhk @ 7:33 下午

前天下課,午飯時間,中五的何同學問:

何:『阿 sir,你寧願食朱古力味的屎,定係屎味的朱古力?』
我:『吓,我可唔可以兩樣都唔揀?』
何:『唔得,一定要揀一樣!』
我:『咁就屎味的朱古力啦,朱古力冇害嘛。』
何:『我就寧願食朱古力味的屎,起碼好咪D。』
我:『Oops。』

原來只要『好味』,食 X 也無妨。這是否普遍人的心態?對話讓我想起電影《Matrix I》 內的某情節,也想到『果效與原則』的取捨,也想到我們的教育,然而,只是想一想而已,無力把思想沉澱。

[CE][AM] 錯在哪裡 – general solutions to trigonometric equations

Filed under: Additional / Applied Mathematics,HKCEE — johnmayhk @ 12:01 下午

Given that a,b are non-zero constants such that |a| < 1 and |b| < 1.

Solve the following simultaneous equations

\sin(x) = a – – – (1)
\cos(x) = b – – – (2)

Solution (so-called)

divide (1) by (2), yield

\tan(x) = \frac{a}{b}

Hence

x = 180^{o}n + \tan^{-1}(\frac{a}{b}) where n is any integer

上述的答案是錯的,為何?

2008/01/06

anniversary ceremony of Y.U.81

Filed under: Report,School Activities — johnmayhk @ 8:36 下午

Last night was the 30th anniversary ceremony of the establishment of Y.U.81, we had games and hot-potting, leaving school at 10 p.m. something. Although the activities were just similar to previous years, I think the change should be the quality of members. Senior members took care of the junior like brothers. (Esp. stayed with a crying member) I just recalled the days when chun tak was the HSL (or earlier), the dinner hours were a bit embarrassing for quite a lot of ‘men’s talk’ and ‘foul languages’ there. However, in recent years, members were becoming much more polite and respectful to teachers. As Mr. Ching always claims that the boys of SFXC are of very high quality in conduct. You know, my mother school was a band 5 school and I knew how trouble will be when students acting madly. One old boy, Mak, was there, he was good at solving 3*3*3 Rubik’s Cube and he is playing 5*5*5 Rubik’s Cube now. He said the principles were quite similar. How interesting to tell that two handsome young teachers, Mr. Rock and Mr. Zinc were enthusiastic about playing the 3*3*3 Rubik’s Cube these days in staff room. I think that it may be a good show if there are competitions or performances about playing Rubik’s Cube. A Talentine show? Apart from singing and musical performance, we may have other alternatives. Magic? Yoyo? One F.2 student in my class had a mini-show in the class party, just click the following and have a look!

http://www.youtube.com/v/Pot8GSTYJa4&rel=1

Occasionally, I found the following movie clip: a Japanese TV program.

http://www.youtube.com/v/cJolYpnLbCI&rel=1

Quite surprising that mathematics can be ‘used’ in a popular TV show! I’m afraid that we cannot watch similar programs in Hong Kong in the near future, even after the introduction of digital television broadcasting system with high resolution.

2008/01/05

[報告] 批判思考工作坊 (critical thinking workshop)

Filed under: Report,School Activities — johnmayhk @ 12:32 下午

中二學生被安排回校出席四次在星期六早上進行的『批判思考工作坊』,班主任(例如我)被安排做旁聽。

講者:Henry
時間:10:40 ~ 12:00

首先,講者指出,以前沒有這類課程,因為上一代的人怕教了學生批判思考,學生會『作反』。現在教這些課程是因為準備新高中。嗯…

跟著,講者要中二同學分辨,什麼是批評,什麼是批判。(Edmund 問:為何不是用英文教?)

好了,進入活動。

活動一:『心靈感應』其實是一個關於 9 的倍數的簡單數學遊戲,起碼在八九年前網上已玩過。講者說,批判思考去到最高層次就是涉及數字(?)

活動二:討論『有雞先還是有雞蛋先?』
看著中二同學討論得興高采烈,其中鄺同學認為先有蛋,因為上帝做蛋容易過做隻雞!也有麥同學和陳同學出來以『進化論』解釋他們的論點。接著是講者的解釋,他說,他是個科學家(!)。

以宗教角度:雞先(聖經記載上帝以生物形式做動物)
以哲學角度:看你如何定義
以科學角度:蛋先(始祖雞蛋突變生當代雞)

這個討論希望帶出『多角度』。

幾年前和蔡同學討論過類似問題,有興趣可看看吧:

http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=820806&t=817997

活動三:IQ 題
『今有兩人,阿甲只會在早上說謊,阿乙只會在晚上說謊,如何問兩個問題,可分辨誰是誰?』

(趙同學即問:如何定義早上和晚上?Oops)

講者提供的答案:
問:『你是誰?』
第一個人答:『我是甲。』
第二個人答:『我是甲。』

問:『現在是什麼時間?』
第一個人答:『早上八時。』
第二個人答:『晚上八時。』

好了,上述資料已經可以決定第一個及第二個人是誰。大家試試。這活動帶出何謂『思考』。

這個課程,學生每人給 $100(全級約有 200 學生,再加上一大部分由校方資助的費用)值得嗎?這又是一個可以『多角度思考』的問題:有趣的 metacognition。

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