Quod Erat Demonstrandum

2007/11/30

感想？我不敢想。

Filed under: Life — johnmayhk @ 6:33 下午

『不要問，只要信』，這年頭已用不著奢侈地去問為什麼，而是要做！如何做才是我們關心。果效比原則重要多。

2007/11/25

Hilbert’s space

Filed under: Family — johnmayhk @ 10:06 下午

More than just photos, they are the precious memories of my family. Just share with you my son’s recent photos.

http://www.flickr.com/photos/13918910@N02/sets/72157603291646571/show/

2007/11/24

[初中] Algebraic computation

Filed under: Junior Form Mathematics — johnmayhk @ 1:23 下午

After the uniform test, I’m really sad to see the poor performances made by my F.2 boys. We, teachers, keep on asking WHY, WHY, WHY: for certain kinds of questions, it is a MUST that students will make mistakes. During the discussion of paper, I said, ‘I knew many of you may make mistake here…’, then a student, Ho, replied,’ then why you set up this question?’ Oops.

What is the question in the discussion? It is a M.C. question.

$\frac{1}{x - y} - \frac{1}{x + y} =$

A. $\frac{-2}{x + y}$
B. 0
C. $\frac{2y}{x^2 - y^2}$
D. $\frac{2x}{(x - y)(x + y)}$

Yes, I intentionally set this question to test whether students could perform algebraic computation correctly. And we, teachers, knew that some students may make mistakes. Hence, I let students practice questions of similar types beforehand. But, according to student Ho’s comment, should I stop setting this kind of question instead?

Students asked what are techniques of doing mathematics? Teachers may reply that “practice more", but students just turn a deaf ear to this and want to have the painless short-cut. I try to clear up the concepts and show common techniques during lessons, but it is insufficent if students have no studying and drilling. Undoubtedly, we have tricks, but the effort takes the lead.

[AL][PM] Class discussions on complex numbers

Filed under: HKALE,Pure Mathematics — johnmayhk @ 1:21 下午
Tags:

We are learning the roots of unity in these F.7 pure math lessons. I randomly gave the following

Solve $x^5 - x^4 + x^3 - x^2 + x - 1 = 0$

Then I wrote

$x^5 - x^4 + x^3 - x^2 + x - 1 = 0$
$1 - x + x^2 - x^3 + x^4 - x^5 = 0$
$\frac{1 - (-x)^6}{1 - (-x)} = 0$ (for $x \neq -1$) – – – (1)

then I jumped to

$x^6 = 1$ – – – (*)

Then I stopped for a while because I had something in my mind:

started with another equation

$x^5 + x^4 + x^3 + x^2 + x + 1 = 0$

we can jump to the same equation $x^6 = 1$, why?

At that moment, Harris claimed that he had a quicker method: “just adding a negative sign".

Well, when solving (*), we have

$x = cis\frac{2k\pi}{6}$ (for $k = 0,1,2,4,5$ < – – – no ‘3’ since $x \neq -1$) is the roots to the original equation.

Then Harris wrote his method (H’s method) on the board as follows.

From (1), we have

$(-x)^6 = 1$
$-x = cis\frac{2k\pi}{6}$
$x = -cis\frac{2k\pi}{6}$ < – – – – this is so-called “adding a negative sign"

Well, it is OK with saying $k = 1,2,3,4,5$ < – – – – no ‘0’ since $x \neq -1$.

Is it a quicker way to obtain solutions, as claimed by Harris? Well, it depends.

But the most important thing is how to explain the solution given by Harris is the same as what I gave.

Of course, we can put different values of $k$ and check them one by one. However, it’s time consuming. We should have a general (or easier) way to show it up.

At that moment, some (or few) students tried to suggest something and think it in details. That is a wonderful moment because interaction, active learning began!

The first thing came to me is trying to see the geometric meaning. But how stupid that I was thinking about vertices of a regular pentagon!! I mistake that there are 5 roots so the roots represent the vertices of a regular pentagon!! No, $cis\frac{2k\pi}{6}$ are representing vertices of a regular hexagon!! A student, Lai, reminded me that there are 6 distinct values generated!!

Then the situation is easy now.

Because the vertices of a regular hexagon is symmetrical about $O$,
that is,
if $cis\theta$ is a root
=> the point representing $cis\theta$ is a vertex P of the regular hexagon
since there is another vertex P’ of the hexagon such that P and P’ are symmetrical about O
and the corresponding complex number represented by P’ is $-cis\theta$
=> $-cis\theta$ is also a root

Don’t understand geometric meaning? Urm, try to use “pure algebra".

For $n$ being a positive even number,

$\cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}$
= $-\cos(\frac{2k\pi}{n} - \pi) - i\sin(\frac{2k\pi}{n} - \pi)$
= $-(\cos\frac{(2k - n)\pi}{n} + i\sin\frac{(2k - n)\pi}{n})$
= $-(\cos\frac{2(k - m)\pi}{n} + i\sin\frac{2(k - m)\pi}{n})$ where $n = 2m$ for some natural numbers $m$.
= $-(\cos\frac{2k'\pi}{n} + i\sin\frac{2k'\pi}{n})$ where $k' = k - m$

Back to our question, $n = 6$, hence $m = 3$; thus, by above,

$cis\frac{2(0)\pi}{6} = -cis\frac{2(0 - 3)\pi}{6} = -cis\frac{2(3)\pi}{6}$
$cis\frac{2(1)\pi}{6} = -cis\frac{2(1 - 3)\pi}{6} = -cis\frac{2(4)\pi}{6}$
$cis\frac{2(2)\pi}{6} = -cis\frac{2(2 - 3)\pi}{6} = -cis\frac{2(5)\pi}{6}$
$cis\frac{2(4)\pi}{6} = -cis\frac{2(4 - 3)\pi}{6} = -cis\frac{2(1)\pi}{6}$
$cis\frac{2(5)\pi}{6} = -cis\frac{2(5 - 3)\pi}{6} = -cis\frac{2(2)\pi}{6}$

For the last round of pure math teaching period, I’d come across other questions from F.7 students, just click here for your further reading.

I still want to say something more including a funny geometric application of “roots of unity" and replies to Justin. Just wait and see.

2007/11/19

[U] Solve sin(z) = 2

Filed under: University Mathematics — johnmayhk @ 3:21 下午
Tags:

A F.5 student, Yan, discussed with me something about “complex angles" in trigonometric formulae, in lunch hour, he asked me the way of solving

$\sin(\theta) = 2$

I just post it here. Starting with the Euler’s formula

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$

We can extend the angle $\theta$ to complex number by using the above formula. Replace $\theta$ by $-\theta$, we have

$e^{-i\theta} = \cos(\theta) - i\sin(\theta)$

Subtracting, yields

$e^{i\theta} - e^{-i\theta} = 2i\sin(\theta)$, thus

$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

Now, putting $\sin(\theta) = 2$, we have

$\frac{e^{i\theta} - e^{-i\theta}}{2i} = 2$
$u^2 - 4iu - 1 = 0$ where $u = e^{i\theta}$

$u = \frac{4i \pm \sqrt{(4i)^2 - 4(1)(-1)}}{2} = \frac{4i \pm \sqrt{-16 + 4}}{2} = (2 \pm \sqrt{3})i$

Thus,

$e^{i\theta} = (2 \pm \sqrt{3})i$
$e^{i(x + iy)} = (2 \pm \sqrt{3})e^{(2n\pi + \pi/2)i}$

Here, we put $\theta$ as a complex number $x + iy$ and try to find out values of $x$ and $y$. Also, by putting $\theta = 2n\pi + \frac{\pi}{2}$ into the Euler’s formula, we have $i = e^{(2n\pi + \pi/2)i}$. Continue,

$e^{-y}e^{ix} = (2 \pm \sqrt{3})e^{(2n\pi + \pi/2)i}$

$\therefore x = 2n\pi + \pi/2$
$e^{-y} = 2 \pm \sqrt{3}$, taking logarithm, yields
$\therefore y = -\ln(2 \pm \sqrt{3})$

Last time, I’d derived the formula of $\ln(z)$ to Yan, hence it is done.

2007/11/16

Filed under: Life — johnmayhk @ 5:00 下午

Just copy and paste (sorry) an article written by my secondary school English teacher Mr. James Hon (韓連山老師) in 教協報. Apart from daily heavy workload in secondary school, he committed himself in Hong Kong Professional Teachers’ Union. The following is a stereotype.

Tomorrow is another day

“Well, tomorrow is another day.”

Cheering team

Filed under: School Activities — johnmayhk @ 4:52 下午

[2007-11-15 morning]When taking about “green house”, people may think about “greenhouse effect”, however, the first concept turning up in Xavierians’ minds should be one of the four houses. In particular, the sense of belonging is overwhelming during the period of athletic meets. Cheering dancing on the track is one of the programs. Bystanders may have no idea on the enormous cost (I’m not professional and had no idea of what kind of movements should be good-looking and safe, I just randomly created something seemed to be so-called ‘dance’. Trial and errors, time consuming! Then followed by series of training and adjustments for 2 to 3 weeks) for the 3-miniute-only show. As the teacher-in-charge of the green house cheering team, I think our committee members and F.1 students were trying their best to practice and perform. Yellow house was the champion in the dancing part. Yes, as one of the judges, I voted for them. Although I believed that the dance steps in our team was OK, judges preferred varieties in overall pattern to details in dance steps.
Click to watch the practice
[2007-11-15 night] My mother-in-law’s birthday, my wife spent time on preparing the feast and made many surprises. But, the happiness was struck by one of the family members. Sigh.
[2007-11-16 morning] The pre-walk in Shing Mun Reservoir lasted about 2 hours something. I just remembered that I was attacked by a monkey because of my staring.

2007/11/07

[CE] A typical curve sketching question

Filed under: HKCEE — johnmayhk @ 9:40 下午

Just busy with the cheering team practice. For a break, write something quick…

This is a part of the question of 2000-CE-A.Math-Q.10, a typical question.Let $y = f(x) = \frac{7 - 4x}{x^2 + 2}$.
Show that the maximum and minimum values of $f(x)$ are $4$ and $-\frac{1}{2}$ respectively.

When it is given to F.5C students, some of them puzzled and asked, ‘is that something wrong in the question?’ They added, ‘$-\frac{1}{2}$ is maximum and $4$ is minimum.’ How come?

In fact, students’ concept is so vague to give that comment. Saying “$-\frac{1}{2}$ is maximum and $4$ is minimum." is not clear! They should say something like:

$y$ attains its maximum value when $x = -\frac{1}{2}$. Hence,
to find maximum value of $y$,
we put $x = -\frac{1}{2}$ into $y = \frac{7 - 4x}{x^2 + 2}$,
that is, the maximum value of $y = \frac{7 - 4(-\frac{1}{2})}{(-\frac{1}{2})^2 + 2} = 4$.

Also, they should say
$y$ attains its minimum value when $x = 4$. Hence,
the minimum value of $y = \frac{7 - 4(4)}{4^2 + 2} = -\frac{1}{2}$.

The designer of this question played little tricks on numbers. It may be quite interesting to set up similar question by choosing ‘suitable’ numbers $a$ and $b$ satisfying

$a < b$;
$f(a) = b =$ maximum value of $y$;
$f(b) = a =$ minimum value of $y$; and
It is not too easy (or too difficult) to find derivatives of $f(x)$.

Students try to find such function to help me set up A. Math Examination Paper ^_^.

[初中] Ratio

Filed under: Junior Form Mathematics — johnmayhk @ 9:32 下午

I gave F.2 boys the following simple question in a quiz.

Edmund has x stamps while Kelvin has 12 stamps more. Given that the ratio of the number of Edmund’s stamps to that of Kelvin’s is 2 : 3.
(a) Find the value of x.
(b) Kelvin then takes some stamps from Edmund such that the ratio of the number of Edmund’s stamps to that of Kelvin’s is changed to 7 : 23. How many stamps does Kelvin have finally?

Many students could give correct answer using something like

(a)
$x : (x + 12) = 2 : 3$
$\frac{x}{x + 12} = \frac{2}{3}$
$x = 24$

(b)
Let $y$ be the number of stamps taken by Kelvin.
$(24 - y) : (36 + y) = 7 : 23$
$\frac{24 - y}{36 + y} = \frac{7}{23}$
$y = 10$
Hence, Kelvin has $(36 + 10) = 46$ stamps in the end.

This method is converting ‘ratio problem’ into ‘solving equation problem’. Natural?

However, some students gave the following.

(a)
$x = 2 \times 12 = 24$.

(b)
Kelvin has $\frac{23}{7 + 23} \times 5(12) = 46$ stamps.

If you were a teacher, will you grant full marks to this kind of ‘one-line’ solution. For me, affirmative! That is mere RATIO. That is equation-free! How stupid that we had found $y = 10$ which is “useless" with respect to the method above. Yes, we emphasize always on the presentation of solution, writing appropriate steps. However, we may need to strike a balance for allowing delicate solution.

2007/11/06

[CE] Exploring on curve sketching

Filed under: HKCEE — johnmayhk @ 3:31 下午

1.

Could you guess that equation of the graph of the curve in red (as shown)? You may say it looks like $y = x^3$. Well, it is actually $y = x|x|$. Justin discovered that the graphs of $y = x^3$ and $y = x|x|$ are quite similar in shape. Further, the following shows the graph of $y = x^3|x|$ which looks similar to the graph of $y = x^5$.

Can we generalize the above as the graph of $y = x^{n-1}|x|$ looks quite ‘similar’ (quite a vague idea actually) to $y = x^{n+1}$ for $n > 1$? How to explain that finding? I just gave comment briefly that $y = x^{n + 1} = x^{n - 1}\times x^2$ ‘behaves similarly’ to $y = x^{n - 1}\times |x|$ in the way that $x^{n - 1}$ are multiplied by a non-negative factor. Could you give further comments and explanations?

2.

Many interesting matters have rooms for further discussions. Um, just give you something to ‘play’ with. Can you draw a square? What a silly question. Um, let me refine, could you use one single equation to plot a square with curve plotting software (e.g. Winplot.exe)? I had done one, see

Could you guess the equation of the graph above? Strictly speaking, the above graph is NOT a square (see the trouble corners). How to make the curve above? Just consider:

We all know that the equation of circle is something like

$x^2 + y^2 = 1$

$x^3 + y^3 = 1$ or
$x^4 + y^4 = 1$

You may use curve plotting software to see the difference of the graphs of the above curves (could you explain the difference?). When the power of $x$ and $y$ is getting larger and larger, what will be the changes? Try to use winplot.exe to check. In fact, the figure I posted is the graph of $x^{100} + y^{100} = 1$. It looks like a square except the situation at corners. Students, can you explain, without using any software, why the shape is something like a square? More, if we consider when the powers of $x$ and $y$ are getter smaller, what will be the changes of the shape of graph? Could you imagine the shape of the graphs of the following

$x^{\frac{1}{2}} + y^{\frac{1}{2}} = 1$
$x^{\frac{1}{3}} + y^{\frac{1}{3}} = 1$

Finally, when the powers are different, any further observation you may discover? As for example

$x^2 + y^3 = 1$

We still have a lot of elementary mathematics stuff for exploring, think more!

2007/11/03

Singing contest

Filed under: Life,School Activities — johnmayhk @ 5:11 下午

My wife was not at home and I’m happy to take care of my son one whole day. He’s sleeping sweet now. Just do some recalling.

The annual important event of the SFXC music club “$sing^3$ singing contest” was held yesterday. I’m not professional, but students asked me to be one of the judges (together with FC & SHY) of that event. Just like last year, singing while dancing is always impressive. The champion of the senior group event went to “4 sons” because we teachers enjoyed the happy atmosphere and their effort in preparing the dance.

When it comes to the score of ‘cooperation’, in my school days, my teachers would focus on the harmony of voices, like Simon and Garfunkel (Hello darkness my old friends…), but today, I seldom find that kind of harmonic performance, instead, the dancing or making jokes may be more dominant, part of the reasons should be the culture “fed” by similar Hong Kong TV programs.

It is very happy to see students on the stage and become ‘another one’, like some F.7 boys I know. Richard, Yip, Tsui, H, Wong, they had tried hard. A F.4 student, Chow, he was a naughty boy but I’m quite shocked to listen to his performance, his voice is clear with a bit sorrowful feeling. Actually, many senior students performing good, the major difference may be something about the “lasting power” (i.e. made no mistake even in the last note of the singing)

Actually, I prefer singing to listening, but I know my voice (together with my whole self) is “deteriorated”, but I think it’s enough to sing songs to my son (and wife sometimes :p).

2007/11/02

[Fun] Gag

Filed under: Fun — johnmayhk @ 12:06 下午

[AL][PM] Asymptote

Filed under: HKALE,Pure Mathematics — johnmayhk @ 12:05 下午

A livejournal writer, Mr. Ng, felt surprising that many Hong Kong secondary school mathematics teachers have no idea in the definition of asymptote. He quoted that “$\lim_{x\rightarrow \infty} f(x) = 0 \Rightarrow x$-axis is a horizontal asymptote of $y = f(x)$" is completely wrong! He then gave a counter-example $y = \frac{\sin(x)}{x}$. He says that even in MathWorld, the definition that “An asymptote is a line or curve that approaches a given curve arbitrarily closely." is 一塌糊塗。Some times ago, I found the following in the internet.

From a page of an article “The Asymptotes of Plane Curves" written by H. G. Green, published in “The Mathematical Gazette" (Vol. 13, No. 185 (Dec., 1926), pp. 232-235), we see a definition of asymptote as follows.

An asymptote of a curve of the $n^{th}$ degree is the limit of a continuous series of parallel lines which cut the curve in less than $n$ points and whose intersections therewith, as the lines approach the limit, become and remain farther from the origin than any given distance, however great."

As in Mr. Ng’s further discussion, there are MANY definitions for asymptote. I think it is a good topic to study further.

2007/11/01

[CE] Linear programming – system of inequalities

Filed under: HKCEE — johnmayhk @ 3:58 下午

1.

Just start teaching the solving of system of 2-variable linear inequalities. One F.5C student, Yeung, told me that “no solution for parallel lines", Oops, his comment was too brief but I understood him immediately. Let me refine his observation, as follows.

Consider a system of two 2-variable linear inequalities (x and y are real with no extra restriction), the system MAY have no solution only when two different corresponding straght lines are parallel to each other.

As for example,

$x + 2y < 1$
$2x + 4y > 4$

The two corresponding lines $x + 2y = 1$ and $2x + 4y = 4$ are parallel, and the system above has no solution.

In other words, the (two 2-variable linear inequalities) system MUST be solvable if the system is ‘corresponding to’ two non-parallel lines. Although it’s not something very striking, I appreciate the discovery, because it is totally orignated from a student himself. Now, could you show me the proof of Yeung’s observation?

[But, most of F.5C students were out of mood of attending the lesson that day, may be they thought that the content is TOO EASY.]

2.

To solve $Ax + By + C < 0$,
we draw $L: Ax + By + C = 0$, and
$L$ divides the x-y plane into 2 regions, namely $R_1$ and $R_2$.
Knowing that $P(x_1,y_1)$ lies on $R_1$ and
$P$ satisfies $Ax + By + C < 0$, that is
$Ax_1 + By_1 + C < 0$, then
$R_1$ is the required region.
That is, $R_1$ represents the solution to the inequality $Ax + By + C < 0$.

The question is: why we just try one point? Should we try another? Is it possible that there is another point $Q$ lying on $R_1$ but $Q$ does not satisfy the inequality $Ax + By + C < 0$? How to prove that “one point is enough for representing the whole region"?

Well, the first thing in my mind is the signed distance (有向距離) from a point to a line.

Additional Mathematics tells us that the distance between a point $P(x_1,y_1)$ and a line $L : Ax + By + C = 0$ is $|\frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}}|$. The absolute sign is for ensuring non-negative distance. How about getting rid of the absolute sign? The numerator $Ax_1 + By_1 + C$ may be positive, zero or negative. What is the significant geometric meaning of the positive and negative values obtained?

Just tell you the following.

If $Ax_1 + By_1 + C$ is positve, then $P$ and $O$ are at the different sides of $L$ (as shown)

If $Ax_1 + By_1 + C$ is negative, then $P$ and $O$ are at the same sides of $L$ (as shown)

Back to our questions, if $O$ does not lie on $L$, $O$ must lie on one of the regions $R_1$ and $R_2$. Suppose $O$ lies on $R_1$, then all the points on $R_1$ and $O$ are at the same side of $L$. Hence any point $Q(a,b)$ lying on $R_1$ will lead to the result $A(a) + B(b) + C < 0$. Hence one point is enough! It is impossible to have another point $T$ lying on $R_1$ such that $T$ does not satisfy $Ax + By + C < 0$.

Please complete the discussion when $O$ lies on $L$.

3.

F.5C students asked, ‘how about the case for 3 or more variables?’ To the best of my memory, we may use simplex method.