Quod Erat Demonstrandum

2007/10/26

[Fun] Gag

Filed under: Fun — johnmayhk @ 4:24 下午

Whenever I’m in F.5C, I must hear someone making jokes. (Broken gag in Chinese!) They ask me some questions for making gags. OK, this time I give a question to revenge,

點解滔滔咁易肥? (猜一個護膚品牌子)

Think a bit and read the answer in the comment.

source : Facebook group “一日一爛gag, I am not a plastic bag."

廣告

Blood donation

Filed under: School Activities — johnmayhk @ 2:14 下午

The blood donation talk for form five students was held on this Monday. The purpose of the talk was for promotion, hoping more students would join the blood donation activity. The speaker was ms. MK. By delivering the talk through the PowerPoint, I wondered how many of students might be interested in the history of blood transfusion centers, the functions of different parts in blood cells, etc. When it closed to the lunch hour, a ‘short’ video was getting started! Students showed ‘ORZ’ when watching that 70s or 80s cartoon show with repeated information on the functions of blood cells and the importance of blood donation. The first blood donation day was held on 2007-10-25 (THU) and the number of blood donors was 58 only. I remembered that there were 108 blood donors once in a blood donation activity. It seems that the HSL was Tsang Chi Fung (?) that year. The main reason was not something about our students, but the lack of manpower from the blood transfusion centers. There were only 8 chairs! (Instead of 12 chairs in early time) Ms. MK asked me to extend the time of using the hall to 5 p.m.! Oh, it was impossible for the full booking. Before that, a red cross member’s mother called me and asked me that why her son was required to do the service in school hours! Oh, my.

2007/10/24

[AL][AM] Solving DE, sorry!

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 5:29 下午

It’s my fault to give the following D.E. to F.6 students for solving.

\frac{d^2y}{dx^2} + y = \tan(x)

They can’t use the method of undetermined coefficients as stated in the note to solve. The problem is what y_p we should take? They may use the method of Variation of Parameters by taking

y_p = L_1y_1 + L_2y_2

where L_1, L_2 are functions in x to be determined and y_1 and y_2 are two functions in the complementary solution y_c.

In this example, it’s easy to obtan y_c = A\sin(x) + B\cos(x) and we take y_1 = \sin(x) and y_2 = \cos(x).

To determine L_1 and L_2, we establish the following:

L_1'y_1 + L_2'y_2 = 0
L_1'y_1' + L_2'y_2' = \tan(x)

Solving the simultaneous equations above, we have

L_1' = \sin(x)
L_2' = \cos(x) - \frac{1}{\cos(x)}

Performing integration, omitting the arbitary constants, we have

L_1 = -\cos(x)
L_2 = \sin(x) - \ln|\tan(x) + \sec(x)|

Hence we obtain the y_p as

y_p = L_1y_1 + L_2y_2
= (-\cos(x))\sin(x) + (\sin(x) - \ln|\tan(x) + \sec(x)|)\cos(x)
= -\cos(x)\ln|\tan(x) + \sec(x)|

And the general solution to the D.E. is

y = A\sin(x) + B\cos(x) -\cos(x)\ln|\tan(x) + \sec(x)|

Just replace \tan(x) by f(x) for more general case.

Um, sorry to tell you that the method above is “out of syllabus", time wasting, right?

2007/10/23

[初中] Simple geometry problems

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 6:20 下午

Here’s just an assignment question given to F.2 students.
Refer to the figure, AB // CE. D lies on CE. BC cuts AD and AE at G and F respectively.

Show that \frac{AF \times EC}{EF} = \frac{AG \times DC}{DG}.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

In most of the cases, showing something about ratios in lengths, we look for similar triangles. It is easy to show that △ABF ~ △ECF and △ABG ~ △DCG. To help to visualize the situation, it may be better to draw similar triangles in “proper orientations”, that isFor △ABF ~ △ECF, we draw

Now, try to look at the conclusion we need to show, L.H.S. involves AF, EC, EF and all these lengths are shown in the figure above. By similarity of △ABF and △ECF, we have
\frac{AF}{EF} = \frac{AB}{EC}

\frac{AF \times EC}{EF} = AB — (1)

Now, we create the expression of L.H.S. now. Similarly, we look at another pair of similar triangles.

For △ABG ~ △DCG,

At a look at the conclusion, R.H.S. involves AG, DC and DG, and all these lengths are shown in the figure above. By similarity of △ABG and △DCG., we have,

\frac{AG}{DG} = \frac{AB}{DC}

\frac{AG \times DC}{DG} = AB — (2)

By comparing (1) and (2) , we can draw the conclusion that
\frac{AF \times EC}{EF} = \frac{AG \times DC}{DG}

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

Here is another textbook question, but I think not many students could manage to do it. Thus I should teach them here for saving my time. Refer to the following figure. △ACE and △CBF are equilateral. D lies on AB. \angleBCA and \angleCDB are right angles. Prove that △ADE ~ △CDF. [Please do it on you own first, then try to compare your method with mine.]

It is very easy to show \angle EAD = \angle FCD. Since \angle EAC = \angle FCB = 60^o and \angle CAD = \angle BCD just consider the right angles.

Then, what’s next? Can I find another pair of equal corresponding angles? It seems that it is not easy. Hence we need to switch to consider ratio of lengths.

If \frac{AD}{CD} = \frac{AE}{CF}, it is done by stating the reason that “ratio of 2 sides, inc. \angle”. But how to show \frac{AD}{CD} = \frac{AE}{CF}?

The ratio \frac{AD}{CD} makes me consider △ABC, we come across this type of setting many times in the topic of similar triangles. It is easy to find three triangles in △ABC such that they are similar to each other. Namely, △ADC ~ △CDB ~ △ABC.

Consider the similarity of △ADC and △ABC, it is easy to obtain

\frac{AD}{AC} = \frac{CD}{CB}
\frac{AD}{CD} = \frac{AC}{CB}

But AC = AE and CB = CF (look at equilateral triangles), hence we have

\frac{AD}{CD} = \frac{AE}{CF}

And the story ends.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

很多同學對學習幾何感到厭惡,因為課程使他們誤以為學習的重點是『背』reasons。多年前陳戈老師也提問過:寫 reasons 是否真的這樣重要?其實寫 reasons 為的是『方便』老師批改,希望看到同學在表達答案的理據是有效與否。參看上文,我沒有把每個 reasons 清楚交代,但如此表達答案相信(個人認為)已經足夠。在課本的幾何題目,只要一兩個 key steps,便已勾畫了解答,但要「鉅細無遺」地寫出每一個 reason,有時會把精妙的幾個步驟湮沒,解答變得冗長,嚇怕同學。不過,現實是…

2007/10/18

Happy blogging

Filed under: Life — johnmayhk @ 6:48 下午

1. The most frequently seen post in xanga or blog etc seems to be something like “long time no update la” (Um, I can see the date wor.)
2. Avoid saying something like “I’m busy” in blog, someone must give comment quickly that “then why you have time to write blog?” (People don’t believe that some blog writers just spend several minutes to finish writing)
3. Minimize the “copy and paste”, try to write something original.

[AL][CE][PM][AM] C + kL = 0

Filed under: Additional / Applied Mathematics,HKALE,HKCEE,Pure Mathematics — johnmayhk @ 6:46 下午


Suppose a line L cuts a circle C at two distinct points, then, the family of circles passing through the interesting points are given by C + kL = 0. (Um, we all know that C is a circle and L is a straight line, so it is not good to say C = 0, L = 0 are equations, just for convenience in presentation onwards.) How about when L touches C?

What is the meaning of C + kL = 0 in this case? This question and its further questions had been discussed some times ago. Suppose L touches C at P, it is not difficult to ‘guess’ that C + kL = 0 represents the family of circles touching C at the P. Well it’s true, but why?
In AL pure mathematics textbook, we have the following question.Let C: x^2 + y^2 + 2gx + 2fy + c = 0
Suppose P(x_1,y_1) lies on C.
Show that
(x^2 + y^2 + 2gx + 2fy + c) + k(x_1x + y_1y + g(x + x_1) + f(y + y_1) + c) = 0
represents a circle touches C at P for any real value of k.

It is given in the solution that,

since L : x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0 is the tangent to C at P, C + kL = 0 represents a circle touches C at P.

First of all, it’s unlikely to come across this type of question in the present public examination, so students, just be calm. For discussion purpose only, do you think that the one-sentence solution given is a proof? (“What is a proof?” or “What is a valid proof” may induce many further mathematical or philosophical discussions, I hope I have time to share it later.)

I think that we need to explain why C + kL = 0 touches C.

How to show two circles touching each other? There should be many ways. Intuitively, it is not difficult to see. Just think about “common tangent is the limiting case of common chord”. But is it a “proof”? Seems to be not satisfactory. One commonly used method in additional mathematics is to show the distance between centres of two circles is equal to either the sum or the difference of radii. But it’s a nightmare to apply this to the previous question. I tried another way in class and was buried in complicated algebraic operations very soon. Firstly, it’s easy to show P lies on C + kL = 0. Then I wrote down the tangent T to C + kL = 0 at P. Next, write down the tangent T_1 to C at P. Finally, I’d like to show T is the same as T_1 by showing the corresponding coefficients of x and y and the constant term are in proportion. But, it’s another nightmare.

Trying again, for P is a common point of C and C + kL = 0, I’d like to show that the slope of tangent to C + kL = 0 at P is exactly the slope of tangent to C at P. Then the amount of calculation is reduced. Here is the sketch.

The tangent to C at P is x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0 and hence the slope of tangent at P is -\frac{y_1 + f}{x_1 + g}. Besides, to find the slope of tangent to C + kL = 0 at P, we differentiate C + kL = 0 with respect to x, yield 2x + 2y\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} + k(x_1 + y_1 + g + f\frac{dy}{dx}) = 0, hence, putting (x,y) = (x_1,y_1), we have \frac{dy}{dx} = -\frac{(k + 2)(y_1 + f)}{(k + 2)(x_1 + g)} = -\frac{y_1 + f}{x_1 + g} which is the same as the slope of tangent to C at P.

Is it a proof? Or a good proof?

Then, I think another way just after finishing the above proof: method of contradiction. We know that P is a common point of C and C + kL = 0. Suppose C + kL = 0 does not touch C, then C + kL = 0 will cut C at another point Q (other than P) (See the figure below). Let L_1 be the line PQ, obviously, L_1 is NOT L. Now consider the family of circles passing through P and Q, i.e. C + k_1L_1 = 0. Note that C + kL = 0 is one of the member of that family, hence C + kL = C + k_1L_1 for certain k_1 (here, I must emphasize that C = 0 and L = 0 stand for equations of circle and line), resulting kL = k_1L_1 which is implying that the line L is the same as (or equivalent to) L_1, contradicting to the fact that L_1 is NOT L. In this proof, there’s nearly no calculation is involved.

Is it a proof? Or a good proof? Students, could you give another proof?

The remaining case should be, when there is no intersection between C and L, what is C + kL = 0? Together with the question asked by Zuson, “what is the meaning of x_1x + y_1y + g(x + x_1) + f(y + y_1) + c = 0 if (x_1,y_1) lies insides the circle?”, I need more time to study and present. Just end my utterance here.

2007/10/16

[初中] 對角線經過多少正方形?

Filed under: Junior Form Mathematics — johnmayhk @ 2:44 下午

蒙老師提出一個有趣的問題。

下圖顯示一個長闊分別是 7 單位及 4 單位的長方形。其對角線(diagonal)經過 10 個正方形(即圖中黃色的正方形)。那麼對於長闊分是 640 單位及 250 單位的長方形,其對角線會經過多少個正方形?

我的解答

先分析一下 7*4 這個長方形。其對角線(除了端點)上一定沒有格點(Grid point,即 x 及 y 坐標為整數點),這是因為 7 和 4 是互素(relatively prime 即最大公因數是 1),可見對角線的斜率 4/7 是『約到最簡』,故對角線上(除了端點)再沒有格點,(假設對角線上存在某格點 (a,b),其中 0 < a < 7 及 0 < b < 4;那麼對角線的斜率 = b/a = 4/7,沒有可能。)

好了,要知道對角線經過多少正方形,只要知道對角線和格線(Grid lines)相交了多少點便可。參下圖:

那些藍色紅色的點就是對角線和格線相交的點。每一點,對應著一個正方形。由左下角的藍點數起,它對應左下角的淺藍色正方形。跟著第二個紅點,則對應著下面的粉紅色正方形,如此類推。圖中題示了 10 個藍紅點,即對應著 10 個正方形。亦即是說對角線經過 10 個正方形。

問題是,如何數算藍紅點的數目?心水清的讀者,可能知道我為何要用藍紅兩色來標示那些交點。

其實,藍點就是對角線和垂直格線(vertical grid lines)的交點,紅點就是對角線和水平格線(horizontal grid lines)的交點。即是就,只要我們知道有多少垂直及水平格線,便知道對角線和格線交點的數目。如本例,7*4 的長方形,不計 y-軸 和 x-軸,共有 7 條垂直格線和 4 條水平格線。那麼,對角線和格線的交點數目(即藍紅點的總數)是 7 + 4 – 1(『減 1』是因為對角線右上角的格點重覆數算了),即 10。

把這個結果一般化,對 a 單位乘 b 單位的長方形,如 a 和 b 互素,則對角線經過 (a + b – 1) 個正方形。

若 a 和 b 不互素如何?比如 21*12 的長方形,參考下圖:

我們可以把對角線沿格點[即 (7,4),(14,8),(21,12)]分成 3 等份,故對角線經過的正方形數目為 3 * 10 = 30。問題是,如何找出 3 這個數?心水清的讀者,你應該看到 3 正是 21 和 12 的最大公因數,即 3 = (21,12)。讓我又證明一下這樣的結果。

對 a 單位乘 b 單位的長方形,假設我們可以把對角線分 n 等分,使每段都是 c*d 長方形的對角線,其中 c | a,d | b 及 (c,d) = 1。我們有 a = nc 及 b = nd。要證明 n 是 (a,b)。假設 m(其中 m > n)是 a 和 b 的公因數,則 a = me,b = mf;必有 e < c 及 f < d。但 e/f = a/b = c/d,違反了 (c,d) = 1[即 c 和 d 已『約到最簡』]這個事實,故 a 和 b 的公因子 m,不能大於 n,故 n 是 (a,b) 是也。

好了,現在可以一拼寫出結果:對 a 單位乘 b 單位的長方形,其對角線會經過 (a,b)(\frac{a}{(a,b)}+\frac{b}{(a,b)}-1) 個正方形。現在,長闊分別是 640 及 250 單位,則 (640,250) = 10,故其對角線會經過 10(64 + 25 – 1) = 880 個正方形。

希望蒙老師不要怪我長氣啦!

2007/10/15

[AL][PM] Be careful with the fundamental theorem of integral calculus

Filed under: HKALE,Pure Mathematics — johnmayhk @ 5:50 下午

Fundamental theorem of integral calculus tells us that, for continuous function f(t), we have

\frac{d}{dx}\int_a^xf(t)dt = f(x)

Now, let’s consider the function f(t) = (x - t)^n, by using the above, we have

\frac{d}{dx}\int_0^x(x - t)^ndt = f(x) = (x - x)^n = 0

Fine? Um, it is not the correct answer. We could obtain the correct answer x^n by the following way:

Let u = x - t, du = -dt
\frac{d}{dx}\int_0^x(x - t)^ndt = \frac{d}{dx}\int_x^0u^n(-du) = \frac{d}{dx}\int_0^xu^ndu = x^n.

Or, more elementary,

\frac{d}{dx}\int_0^x(x - t)^ndt
=\frac{d}{dx}\int_0^x\sum_{r=0}^nC_r^nx^{n-r}(-1)^rt^rdt
=\frac{d}{dx}\sum_{r=0}^nC_r^n(-1)^rx^{n-r}\int_0^xt^rdt
=\frac{d}{dx}\sum_{r=0}^nC_r^n(-1)^rx^{n-r}(\frac{x^{r+1}}{r+1})
=\frac{d}{dx}\sum_{r=0}^n\frac{C_r^n(-1)^r}{r+1}x^{n+1}
= (n+1)\sum_{r=0}^n\frac{C_r^n(-1)^r}{r+1}x^n
= x^n
(Note: F.6 students, try to verify the last equality)

Then why we obtain the incorrect answer in using the first method?

2007/10/05

[初中] Correct answer is the most important?

Filed under: Junior Form Mathematics — johnmayhk @ 5:42 下午

I gave the following question to my F.2 class.
Refer to the following figure.

Given that AB = CD and \angle A = \angle D = 36^o, find the size of \angle B.
Many of them could obtain correct answer 144^o, but only one student could give correct method.

Well, it’s not bad to give 144^o, right? Could I appreciate their “number sense" or “spatial sense" by the way? Or, next time, I should draw the figures ugly enough to remind students that “seeing is believing" may not come first, at least, the proofs are at the higher priority in mathematics.

As you may guess, many students claimed that \angle B = \angle C or AD // BC without any reason.

My suggested solution is extending AB and CD, and suppose two lines intersect each other at a point E. (Well, how to ensure the existence of E anyway?) Then, we try to show that EB = EC and hence \angle EBC = \angle ECB; therefore we have \angle ABC = \angle DCB.

Many students hate to do problems in geometry, because it requires clear and logical presentation, sometimes, the process is more important than the conclusions (answers); however, people tend to pay attention to the conclusion (outcome) only without enjoying or appreciating the process. Well, it may be my fault to set the expectation too high. I need adjustment.

[AL][PM] Past Paper 1998 Paper II Q.11

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 5:01 下午

(a)

Let f be a non-negative continuous function on [a , b]. Define F(x) = \int_a^xf(t)dt for all x\in [a,b].

Show that F is an increasing function on [a , b]. Hence deduce that if \int_a^bf(t)dt = 0, then f(x)\:=\:0 for all x\in [a,b].

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

這裡說 f 是連續,目的為了使 f 可積(Riemann integrable)而已。由 F 的定義,運用積分基本定理,立知 F'(x) = f(x)。故由 f 非負,得 F'(x) \ge 0,即:F(x) is increasing on [a , b].

第二部分,學生很直觀地得到結論,因為 f 是非負的,而 \int_a^bf(t)dt 就是 f 圍出來的面積。現在 \int_a^bf(t)dt 是零,面積亦即是零,故 f 應該也是零。然而,pure mathematics 不太接受這個『直觀』的解,學生要『數學化』地表達答案。那麼我們要運用剛才的結果:F is increasing.

F 的遞增性,得 F(a)\:\le\:F(x)\:\le\:F(b),即 \int_a^af(t)dt \le F(x) \le \int_a^bf(t)dt,亦即 0\le F(x)\le 0,故 F(x) = 0 on [a , b],意思是 F(x) 在 [a , b] 區間內恆等於零,即 F(x) 是 constant function 是也。那麼,對 F(x) 求導,得 F'(x) = f(x) = 0,證畢。

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

(b)

Let g be a continuous function on [a , b]. If \int_a^bg(x)u(x)dx = 0 for any continuous function u on [a , b], show that g(x) = 0 for all x \in [a,b].

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

Part (a) 的結論是 f(x) = 0,part (b) 的結論是 g(x) = 0,我們自然想到 f(x)g(x) 有沒有什麼關聯?Part (a) 中,f(x) 是一個非負函數;Part (b) 中,g(x) 不一定是非負函數,所以我們斷不能直接 put g(x) = f(x) 的。看看 g 的條件:對任何連續函數 u(x),恒有 \int_a^bg(x)u(x)dx = 0。Pure Mathematics 的題目常有這樣的技巧:既然對任意的 u 也成立,那麼我們可否代入某個特別的 u,使我們做出一些結果?如果要運用 part (a),我們要代入一個特別的 u,以產生一個非負函數。不妨設 u(x) = g(x)(這不是唯一選擇,g^{2n-1}(x) 也可),則,由 g(x) 的條件,有 \int_a^bg(x)g(x)dx = 0\int_a^bg^2(x)dx = 0,看,我們製作了一個非負函數 g^2(x),我們便可運用 part (a),設所謂的 f(x)g^2(x),得 g^2(x) = 0,即 g(x) = 0.

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

(c)

Let h be a continuous function on [a , b]. Define A = \frac{1}{b-a}\int_a^bh(t)dt.

(i) If v(x) = h(x) - A for all x \in [a,b], show that \int_a^bv(t)dt = 0.

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

這部分最簡單,直接驗算便可。

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

(ii) If \int_a^bh(x)w(x)dx = 0 for any continuous function w on [a , b] satisfying \int_a^bw(x)dx = 0, show that h(x) = A for all x \in [a,b].

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

此部分,學生常有誤解。他們以為直接用 part (b),設 g(x) = h(x) 便可完工。問題是,h 是『沒有資格』做 g 的。請再看清楚 gh 的條件:

1. \int_a^bg(x)u(x)dx = 0 for any continuous function u on [a , b].

2. \int_a^bh(x)w(x)dx = 0 for any continuous function w on [a , b] satisfying \int_a^bw(x)dx = 0.

可以這樣說,g 是『強勁』得多,因為只要 u 是連續函數,進行這個積分運算 \int_a^bg(x)u(x)dx,則答案就是零。但 h 就沒有這樣的『強勁』,不是所有連續函數 w(x) 可以產生 \int_a^bh(x)w(x)dx = 0 這個效果,只有一部分的連續函數 w(x),就是滿足 \int_a^bw(x)dx = 0 這個額外條件的連續函數,才可以產生 \int_a^bh(x)w(x)dx = 0 這個效果。

要證明 h(x) = A for all x \in [a,b],可證明 h(x) - A = 0 for all x \in [a,b]。設 v(x)\:=\:h(x)\:-\:A,即設法證明 v(x)\:=\:0 for all x \in [a,b].

這個 v(x) 有什麼特別?剛巧,part (c)(i) 告訴我們,這樣的 v(x),其積分的結果是零,即 \int_a^bv(x)dx = 0. 而 \int_a^bv(x)dx = 0 這個結果,正正告訴我們,v 『有資格』做 part (c)(ii) 提及的所謂 w。即是我們有以下結果:

\int_a^bh(x)v(x)dx = 0

我們嘗試由這個結果出發,希望最終推導 v(x) = 0,因為這個目標,自然地我們要上式變成涉及 v(x) 的東西,即

\int_a^b(A + v(x))v(x)dx = 0

從而

\int_a^bAv(x)dx + \int_a^bv(x)v(x)dx = 0

\int_a^bv^2(x)dx = 0 [因為 \int_a^bv(x)dx = 0]

好了,我們又產生一個非負函數 v^2(x),由 part (a),我們得 v^2(x) = 0 for all x \in [a,b],即是說 h(x) = A for all x \in [a,b]. 證畢。

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後話一:有學生誤以為 \int_a^bh(x)dx = \int_a^bAdx\:\:\:\Rightarrow\:\:\:h(x) = A 這是錯的。我們不能如此『消滅』積分符號,正如就算有 \sin(x) = \sin(y),我們不一定有 x = y。我們不能貿然『消滅』\sin 的。不同的函數 h(x)g(x),也可有相同的定積分數值,則 \int_a^bh(x)dx = \int_a^bg(x)dx. 理由是函數縱然不同,但其圖像圍出來的面積也可以一樣。

後話二:我很喜歡這類題目,這才是『純數的感覺』(沒有背誦的包袱,純粹玩弄定義),更有大學數學基礎課 real analysis 一些題目的影子。

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