Quod Erat Demonstrandum


[AL][PM] Sum an Infinite Series by Definite Integrals

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 6:39 下午

Using definite integrals to find the sum of an infinite series may be regarded as an application of definite integral. In the text book, we may find the following relation

\lim_{n\rightarrow \infty}\sum_{k = 1}^{n}f(\frac{k}{n})\frac{1}{n} = \int_0^1f(x)dx —— (*)

Actually, it is NOT the unique way of expressing \int_0^1f(x)dx. A more general way is something like

\int_0^1f(x)dx = \lim_{w \rightarrow 0}\sum_{i=1}^{n}f(w_i)\delta x_i

where {x_0 = 0,x_1,x_2,...,x_n = 1} is a partition of [0,1], \delta x_i = x_i - x_{i-1}, w_i \in [x_{i-1} , x_i] and w = \max_{1 \le i \le n}\delta x_i.

But this rigourous definition of Riemann Sum is not a good way of computation of the sum of infinite series, I need students to memorize (*). And I’d like to present the technique in applying (*).

e.g. 1 Evaluate \lim_{n \rightarrow \infty}(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + ... + \frac{1}{\sqrt{n^2 + 2n^2}})

STEP 1 : Taking out the common factor \frac{1}{n}
STEP 2 : Express in summation form (optional), try to write the expression involving the term \frac{k}{n}
STEP 3 : Convert the summation into definite integral by regarding

\sum as \int
\frac{1}{n} as dx
\frac{k}{n} as x
The lower limit of the integral = \lim_{n \rightarrow \infty} \frac{lower\:limit\:of\:k}{n}
The upper limit of the integral = \lim_{n \rightarrow \infty} \frac{upper\:limit\:of\:k}{n}

OK, let’s do e.g. 1

\lim_{n \rightarrow \infty}(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + ... + \frac{1}{\sqrt{n^2 + 2n^2}})
= \lim_{n \rightarrow \infty}\frac{1}{n}(\frac{1}{\sqrt{1 + \frac{1}{n}}} + \frac{1}{\sqrt{1 + \frac{2}{n}}} + ... + \frac{1}{\sqrt{1 + \frac{2n}{n}}}) [STEP 1 : taking out \frac{1}{n}]
= \lim_{n \rightarrow \infty}\sum_{k = 1}^{2n}(\frac{1}{\sqrt{1 + \frac{k}{n}}})\frac{1}{n} [STEP 2 : express in summation form, obtaining terms in \frac{k}{n}]
= \int_0^2\frac{1}{\sqrt{1 + x}}dx

This is the STEP 3.

\sum becomes \int
\frac{1}{n} becomes dx
\frac{k}{n} becomes x
The lower limit of k = 1, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{1}{n} = 0
The upper limit of k = 2n, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{2n}{n} = 2
Therefore, the lower and the upper limit of x are 0 and 2 respectively.
\lim_{n \rightarrow \infty}(\frac{1}{\sqrt{n^2 + n}} + \frac{1}{\sqrt{n^2 + 2n}} + ... + \frac{1}{\sqrt{n^2 + 2n^2}})
= \int_0^2\frac{1}{\sqrt{1 + x}}dx
= 2\sqrt{3} - 2

e.g. 2 Evaluate \lim_{n \rightarrow \infty}\sum_{k = 1}^{3n}\frac{1}{n}\sin(\pi + \frac{k\pi}{2n})

Here, STEP 1 & 2 are done. Go to STEP 3 directly.

\sum becomes \int
\frac{1}{n} becomes dx
\frac{k}{n} becomes x
The lower limit of k = 1, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{1}{n} = 0
The upper limit of k = 3n, hence \lim_{n \rightarrow \infty}\frac{k}{n} = \lim_{n \rightarrow \infty}\frac{3n}{n} = 3
Therefore, the lower and the upper limit of x are 0 and 3 respectively.


\lim_{n \rightarrow \infty}\sum_{k = 1}^{3n}\frac{1}{n}\sin(\pi + \frac{k\pi}{2n})
= \int_0^3\sin(\pi + \frac{\pi x}{2})dx
= -\frac{2}{\pi}

After some practice, you may write down the integrals by using your naked eyes. That is, you should have no problem to obtain the following

\lim_{n \rightarrow \infty}\sum_{k = 1}^{2n}\frac{1}{n}\ln(1 + \frac{k^2}{n^2}) = \int_0^2\ln(1 + x^2)dx

Um…Enough? How to evaluate the following?

\lim_{n \rightarrow \infty}(\frac{1}{n + 1} + \frac{1}{n + 3} + \frac{1}{n + 5} + ... + \frac{1}{3n + 1})


[CE][AM][教學] Common tangents to two circles

Filed under: Additional / Applied Mathematics,HKCEE,Teaching — johnmayhk @ 4:24 下午

Reply to http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=1241115&t=1241115

Refer to the following figure, L is a common tangent to C_1 and C_2, find the equation of L.

There are, at least, two methods to cope with this question. Let me sketch one.

Let y = mx + c be the equation of L. Since L is a tangent to both circles, the distances between centres and L will be radii. Hence we have |\frac{c}{\sqrt{m^2 + 1}}| = 1 and |\frac{8m + c}{\sqrt{m^2 + 1}}| = 3

By squaring the above equations and eliminating c^2, we have 64m^2 + 16mc - 8(m^2 + 1) = 0 — (*) 

Then c = \frac{-7m^2 + 1}{2m}

Hence (*) can be converted into45m^4 - 18m^2 + 1 = 0

By solving and rejecting, we come up with the conclusion that y = -\frac{x}{\sqrt{15}} - \frac{4}{\sqrt{15}} is the equation of L.


Recalling before taking a rest

Filed under: Life,School Activities — johnmayhk @ 5:37 下午

I went to see the doctor in the morning. Before taking a rest, I just type something here.On 2007-09-15 (SAT), Kwong K.M. and Wong C.F. led us (Miss Ho and me) to Oil Man Estate to attend the annual U.O. meeting of the Red Cross. Actually, the meeting was informative and I did appreciate the work done by all the staff, especially those who work on the voluntary basis. During the sharing session, I’m glad to see Mr. Ho Chi Hang, steven, a sfxc alumnus was in charge of leading the discussion with us. There are 2 teams of Y.U. 126 and the U.O. was quite experienced. He suggested having more communications among U.O.s, well, it’s not bad.

I told F.2 students in my class that I spent 4 hours marking their homework (rate of tortoise \_/); therefore, if they just spend few minutes to copy others homework before the FT period, they should say sorry to me!

Green house meeting was held on 2007-09-21 (FRI), Suen Nam is the house caption this year (last year: Wong Kin Kueng). Mr. Ching’s speech emphasized the spirit of union of Green house, especially in cheering. He gave examples of old captions Lai Siu Cheong and Wu Ying. Yes, quite impressive. Yip Ming Kei, Chan Kin Lok, Ho Chun Yu…Once a caption, always…At the end, we (Mr. Ching, Suen Nam and I) were singing dancing (so-called) on the stage, happy ending? (I was ill and could not perform well at that moment) Sometimes I think, students are growing up and being “on the shore” sooner or later, but how about myself? I need to ‘teach’ (so-called) dancing (so-called) in the cheering team with my old bones years by years? May be, the most important element is not the age actually, it should the fire inside.

Many students were busy decorating boards for club promotion. Meanwhile, student Yan asked me some mathematics questions and I’m happy to share. I avoided saying something like, “um…you’ll learn it later”; instead, I tried to give basic explanations.


[AL][PM] Integration with absolute signs

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 11:09 上午

Learn from Mr. Nguyen’s one-minute talk, I present one-minute-silent-mathematics talk here. (i.e. spend one minute to read this post)

It’s a bit crazy to do complicated computations in so-called Pure Mathematics exercise. Well, take it easy man, mechanical computation is a cost of low-level thinking. Strike a balance, right.

Justin asked me the way to evaluate \int_{-1}^2|2x - |x| - 1|dx many days ago. We all know that getting rid of the absolute signs is our major concern. Yes, but how?


That is, we may start destroying the absolute signs from inside to outside. \int_{-1}^2|2x - |x| - 1|dx
= \int_{-1}^0|2x - (-x) - 1|dx + \int_0^2|2x - (x) - 1|dx
= \int_{-1}^0|3x - 1|dx + \int_0^2|x - 1|dx

See, the |x| inside is deleted (by divide cases of x\ge 0 and x\le 0). OK, go ahead. \int_{-1}^0|3x - 1|dx + \int_0^2|x - 1|dx
= \int_{-1}^0[-(3x - 1)]dx + \int_0^1[-(x - 1)]dx + \int_1^2(x - 1)dx
= \int_{-1}^0(-3x + 1)dx + \int_0^1(-x + 1)dx + \int_1^2(x - 1)dx
= \frac{7}{2}

Once upon a time, solving inequalities with absolute signs was a common practice in additional mathematics, but, may be the problem is shifted to AL students now, um, try to solve the following

|2x - |x| - 1| < 2


[AL][PM] Definite integrals — trouble? (3)

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 12:45 上午

Here is the theorem about the validity of the method of substitution in evaluating definite integrals.Let y = f(x) be a continuous function on [a,b]. Suppose that x = g(t) is a function satisfying g(\alpha) = a, g(\beta) = b and g'(t) remains the same sign on (\alpha , \beta). Then

\int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

Let me copy the proof from the textbook directly as follows.

Suppose F(x) is a primitive function of f(x), i.e.

\frac{d}{dx}F(x) = f(x)

and so,

\frac{d}{dx}F[g(t)] = f[g(t)]g'(t)

Hence, F[g(t)] is a primitive function of f[g(t)]g'(t). Then,

\int_a^bf(x)dx = F(b) - F(a) = F[g(\beta)] - F[g(\alpha)] = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

On reading the proof above, I puzzled that there is nothing to say about the condition that “the sign of g'(t) remains the same over certain range" in the proof. But after looking into the proof carefully, the condition is actually used. Where? It is something about the following implication

x = g(t)\:\Longrightarrow\:f(x) = f(g(t))

What? The above is a must, isn’t it? No, man. We must add the condition about g'(t) (as one of the ways) to ensure its validity. Caution: if x = g(t), we cannot come up with the conclusion that f(x) = f[g(t)] always. F.6 boys, when we are discussing function, don’t forget to consider the domain. f(x) and f(g(t)) may have different domains and hence they may represent different functions, and we cannot say immediately that f(x) = f[g(t)]!

I’d like to discuss this in details. Let me start with a simple example.

Let f:\mathbb{R}\rightarrow\mathbb{R} such that

f(x) = 2x + 1

Now, we use “substitution", let x = g(t) = \sin(t). How about the newly formed composite function f[g(t)] (i.e. f[\sin(t)] = 2\sin(t) + 1), can we say that f(x) = f[g(t)] now? No, of course. f(x) is very free to assume any real real value, but f[g(t)] is restricted on its domain [-1,1] (if t \in\mathbb{R}) and f can only assume values in [-1,3].

Ok, let’s get moving to the situation of integration. Refer to Fig. 1 & 2 below

x = g(t) is assumed to be differentiable on (\alpha,\beta), then is it true to say \int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt?

No! At least, f(x) is defined on [a,b] (see Fig. 1) while f[g(t)] is NOT defined for all t\in [\alpha,\beta], it is because, when t varies from \gamma to \beta, the value of x is greater than b (see Fig. 2) and f(x) is NOT defined for x > b (see Fig. 1 again). If the condition “g'(t) remains the same sign on (\alpha , \beta)" is added, as shown in Fig. 4 below,

it is certain that f(x) = f[g(t)] for t\in [\alpha,\beta] and hence \int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt.

Fine? Discussion ends? Not yet. I’d to add more. When the condition “g'(t) remains the same sign on (\alpha , \beta)" is satisfied, the method of substitution is valid. How about when the condition is NOT satisfied, will the method of substitution be failed? Under basic training in simple logic, we CANNOT say so.

I think that the condition “g'(t) remains the same sign on (\alpha , \beta)" is too strong! To ensure the validity of the method of substitution, it is not necessary to have the condition above, let me suggest an example below (for my poor mathematical skills, I just use figures to illustrate without rigorous real analysis treatment, sorry.)

Fig. 6 shows the graph of x = g(t) which is assumed to be differentiable over (\alpha,\beta). Obviously, the so-called condition is not satisfied. However, in this case, we still have \int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt, can you see why?

Note that, in Fig. 6, g'(\mu) = g'(\lambda) = 0, hence we have g'(x) does not change signs over the intervals (\alpha,\gamma), (\gamma,\mu) and (\mu,\beta) respectively and hence the validity of the method of substituion is confirmed; therefore

\int_a^df(x)dx = \int_{\alpha}^{\lambda}f[g(t)]g'(t)dt
\int_d^cf(x)dx = \int_{\lambda}^{\mu}f[g(t)]g'(t)dt
\int_c^bf(x)dx = \int_{\mu}^{\beta}f[g(t)]g'(t)dt

Summing up the three equations above, yielding

\int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

Yes, the method of substitution is valid even by using “ugly" differentiable functions like Fig. 7 below

The discussions above drove me to re-think about the condition “g'(t) remains the same sign on (\alpha , \beta)“, I think it is not quite necessary now; instead, I prefer using “g(\alpha) = a, g(\beta) = b, g(t) is differentiable on (\alpha, \beta) and g([\alpha,\beta]) = [a,b]“. Do you agree?

Nelson gave me a note from mathematical analysis used in university, and I found that the way of presentation of the method of substitution is very clear (and clever), let me present here to round up my utterance:

Thereom (Change of Variable Formula): If \phi : [a,b] \longrightarrow \mathbb{R} is differentiable, \phi' is integrable on [a,b] and f is continuous on \phi([a,b]), then \int_{\phi(a)}^{\phi(b)}f(t)dt = \int_{a}^{b}f[\phi(x)]\phi'(x)dx.

Proof: Let g(x) = \int_{\phi(a)}^{\phi(b)}f(t)dt. Then g'(t) = f[\phi(x)]\phi'(x). So \int_a^bf[\phi(x)]\phi'(x)dx = \int_a^bg'(x)dx = g(b) - g(a) = \int_{\phi(a)}^{\phi(b)}f(t)dt.


[初中] Let me ‘teach’ you how to play Mark Six

Filed under: Junior Form Mathematics,Teaching — johnmayhk @ 10:34 下午

Reply to the question from 笨木:

“大學教授買六合彩獨中八千多萬" John Sir..我想問呢..你有冇睇過呢篇報導?mainly..個d大家教授話..買六合彩有方法..其中一個所謂的聰明..係話..買哂49個號碼..話d人買既只係49個數字既一部分 所以會有d數字唔中..我覺得呢個係白痴的..因為你買哂49個數字..抽出黎的一定係呢49個數字中間既數= =and..佢話賣既天書要188hkd一本..如果有傻仔買..當有100萬人買..(即6-7人中有一個人買)都有100萬x188..即係1億8千幾萬- -比起中六合彩仲要賺得多..John Sir你又點睇?By yahoo search, I come across the following

2006年10月,英國Bradford University 一班教授,合資買Lotto(六合彩),結果獨中頭獎5百多萬英磅,相等於港幣8千多萬"

The writer of the article claimed that the professors had a “smart" way of buying the Lotto. I’d like to say something about this.

With basic training in applied mathematics, to ensure “sure-win" in playing Mark Six, we need to buy C_6^{49} = 13983816 different tickets. Cost? Sorry to tell you that I had never bought any of Mark Six ticket before and had no idea of how much it costs. The article claimed that, instead of  13983816, it is required only 17 lottery tickets. It is of course questionable. The meaning of “all numbers appearing in that 17 tickets" means we can find a “lucky number" (out of 6 “lucky numbers") in one of the 17 tickets. But, there is no guarantee to have ALL 6 lucky numbers appearing in 1 ticket! In fact, it is extremely simple to ensure ONE lucky number in the tickets (no need 17 tickets), we need at most 9, the arrangement is as follows.


However, the following question may be meaningful and mathematical.

How many tickets we need to buy to ensure TWO lucky numbers appearing in one of the tickets? Or, what is the minimum number of tickets we need to ensure there must be TWO lucky numbers in one ticket and how to buy?

Consider the following 5 number sets.

A_1 = {01,02,03,04,05,06,07,08,09,10,11,12}
A_2 = {13,14,15,16,17,18,19,20,21,22,23,24}
A_3 = {25,26,27,28,29,30,31,32,33}
A_4 = {34,35,36,37,38,39,40,41,42,43}
A_5 = {44,45,46,47,48,49}

Now, no matter which 6 lucky numbers are, TWO of them must belong to one of the 5 number sets above. (Why? Pigeon hole principle)

OK, suppose the TWO lucky numbers are in A_1 which is a set of 12 numbers.

Then, all we need is 6 tickets to ‘capture’ those TWO lucky numbers. The arrangement is


Follow the pattern above, we can consider the cases when the TWO lucky numbers appearing in the number set A_2. Hence, up to now, we have to buy 6\times 2 = 12 tickets.

Now, suppose the TWO lucky numbers are in A_3 which is a set of 9 numbers.

Then, all we need is 3 tickets to ‘capture’ those TWO lucky numbers. The arrangement is


OK, up to now, there are 12 + 3 = 15 tickets required. Go ahead.

Suppose the TWO lucky numbers are in A_4 which is a set of 10 numbers.

Then, we need 4 tickets to ‘capture’ those TWO lucky numbers. The arrangement is


OK, up to now, there are 15 + 4 = 19 tickets. And finally, if the TWO lucky numbers are in A_5, we need to buy one more ticket


Here, we are certain that TWO lucky numbers can be found in those 20 tickets mentioned above.

But, is 20 the minimum number required? Can we serve the same purpose by buying less than 20 tickets? My mathematics knowledge is so poor to come up with any further conclusion. Please share your idea or finding if you have time. This is actually a good topic to go further studies. As for example, how to ensure THREE lucky numbers there? Mark Six is more than just gambling or probabilities. Many years ago, I hope to have a topic of “mathematics in gambling" as a project for F.3 students, but I blamed myself immediately for this silly thought.

Looking back to that $188 book, I think I am able to write similar book by presenting pages of numbers (actually rubbish, non-sense, like what I usually do in forum and blog).



Filed under: Life — johnmayhk @ 1:08 上午

Recently, I’d tried to use Windows Vista and Office 2007. I found it difficult to operate (at least, I cannot use the Chinese input method invented by one old boy Mr. Lee C.K.) and started complaining. Someone told me that there still many bugs in the system of Windows Vista, OMG! Yes, people, like me, are not willing to leave their “comfort zone” and accept new changes. It is believed that the “new version” is better than the old one. Hence I tend to believe that I need to change myself to fit the new “arrangement”. However, it is only a hypothesis, are the new ones always better than the old ones? Or, it is just a matter of habits? I’m afraid to change for the sake of change. Anyway, playing with new computer techniques should be enjoyable; I just feel that I’m too old.


[AL][PM] Definite integrals — trouble? (2)

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 11:20 下午

Here is a standard question. Evaluate


It is easy to use the substitution x = \pi - u to convert the integral into the following


Knowing that there is an even power of sine function, we may substitute t = \tan(x) and the integrand will be a rational function in t, i.e. \frac{1}{4 + 5t^2}. However, when considering the changes in the upper and lower limits, we will find that when x = 0 or x = \pi, we have t = 0. That is both the upper and lower limits are zero! Does it mean that the definite integral \int_0^{\pi}\frac{dx}{4\:+\:\sin^2(x)} = \int_0^0\frac{dt}{4 + 5t^2} = 0? No, the integrand is positive over the interval [0,\pi], how come the integral being zero?

Actually, the substitution of t = \tan(x) is invalid.

The validity of the method of substitution is based on the following theorem. (Direct quotation)

Let y = f(x) be a continuous function on [a,b]. Suppose that x = g(t) is a function satisfying g(\alpha) = a, g(\beta) = b and g'(t) remains the same sign on (\alpha , \beta). Then

\int_a^bf(x)dx = \int_{\alpha}^{\beta}f[g(t)]g'(t)dt

The condition that “g'(t) remains the same sign on (\alpha , \beta)" implies g(t) (and hence g^{-1}(t)) is bijective between (a,b) and (\alpha, \beta).

Now, looking back to the example above, the substitution of t = \tan(x) (or x = \tan^{-1}(t)) is problematic.

Is t = \tan(x) bijective on (0,\pi)? Definitely not, \tan(x) is undefined at x = \frac{\pi}{2}, or, when t = 0, x is not single-valued.

However, we still may take t = \tan(x) on certain interval such that the function is really bijective over that range. As for example, t = \tan(x) is bijective on (0,\frac{\pi}{2}). Hence we may obtain correct answer by writing

\int_0^{\pi}\frac{dx}{4\:+\:\sin^2(x)} = \int_0^{\frac{\pi}{2}}\frac{dx}{4\:+\:\sin^2(x)} + \int_{\frac{\pi}{2}}^{\pi}\frac{dx}{4\:+\:\sin^2(x)}

Using the substitution of u = \pi - x, it is easy to know that \int_0^{\frac{\pi}{2}}\frac{dx}{4\:+\:\sin^2(x)} = \int_{\frac{\pi}{2}}^{\pi}\frac{dx}{4\:+\:\sin^2(x)}, hence \int_0^{\pi}\frac{dx}{4\:+\:\sin^2(x)} = 2\int_0^{\frac{\pi}{2}}\frac{dx}{4\:+\:\sin^2(x)}.

Now, we are confident to substitute t = \tan(x) and obtain \int_0^{\frac{\pi}{2}}\frac{dx}{4\:+\:\sin^2(x)} = \int_0^{\infty}\frac{dt}{4+5t^2}

By further substitution of t = \frac{2}{\sqrt{5}}\tan(\theta), yields, \int_0^{\infty}\frac{dt}{4+5t^2} = \frac{1}{2\sqrt{5}}\int_0^\frac{\pi}{2}d\theta = \frac{\pi}{4\sqrt{5}}, hence

\int_0^{\pi}\frac{xdx}{4\:+\:\sin^2(x)} = \frac{\pi}{2}\times 2(\frac{\pi}{4\sqrt{5}}) = \frac{\pi^2}{4\sqrt{5}}

Satisfactory? Um, you may wonder why the condition “g'(t) remains the same sign on (\alpha , \beta)" is a must for the validity of the method of substitution? Try to read the proof in the textbook and it seems that there is nothing to say about the condition of “g'(t) being same sign" over the range. Then I tried to read different textbooks and notes, and I found that some writers said “the proof is out of scope".

Well, the reason for ensuring the substitution function being single-valued should be quite obvious. I’ll prove the validity of the method of substitution in an elementary way (i.e. by the definition) so as to see the importance of the single-valuedness. Just wait.


Definite integrals — trouble? (1)

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 9:54 下午

Recently, I’d started teaching the topic of “definite integrals”. After some discussions with Nelson, I’d like to write something here.

e.g. 1 \int_{-1}^1\frac{dx}{x^2} = [-\frac{1}{x}]^1_{-1} = -2

Do you notice that there’s something wrong with the above result. Since the integrand \frac{1}{x^2} must be positive and the meaning of \int_{-1}^1\frac{dx}{x^2} is the area under the curve, and hence the answer must be positive! How come we have -2? What’s wrong?


In e.g. 1, actually, we are using the Newton-Leibniz’s formula \int_a^b\:f(x)dx = [F(x)]^b_a. Before using this formula, we must ensure that F(x) is really a primitive of f(x), i.e. F'(x) = f(x) on [a,b]. In other words, F(x) is differentiable at every x in the interval [a,b]. But, in this example, F(x) = -\frac{1}{x} which is undefined at x = 0 and hence F(x) is not differentiable at x = 0. Thus, we cannot apply the Newton-Leibniz’s formula. As a matter of fact, the integrand f(x) = \frac{1}{x^2} is unbounded on [-1,1] and hence it is not integrable over the interval.

For the following example, I must say sorry to Nelson for my wrong calculation. Actually, the operation of the following is fine.

\int_{-1}^1\:e^{|x|}dx = \int_{-1}^0\:e^{-x}dx + \int_0^1\:e^{x}dx = [-e^{-x}]^0_{-1} + [e^x]^1_0 = 2(e - 1) Problem appears only when we takeF(x) =\{\begin{array}{lr}e^x&x\ge 0\\-e^{-x}&x < 0\end{array}

as a primitive of f(x) = e^{|x|} and obtaining something like

e.g. 2 \int_{-1}^1\:e^{|x|}dx = [F(x)]_{-1}^1 = e - (-e) = 2e


In e.g. 2, we see the same problem as e.g. 1, the F(x) is not a primitive of f(x) = e^{|x|} because F(x) is not differentiable at x = 0. Instead,

G(x) =\{\begin{array}{lr}e^x& x\ge 0\\2 - e^{-x}& x < 0\end{array}

is a primitive of f(x) = e^{|x|} on [-1,1] and hence

\int_{-1}^1\:e^{|x|}dx = [G(x)]_{-1}^1 = e - (2 - e) = 2(e - 1)


[U] Nowhere differentiable continuous functions (1)

Filed under: University Mathematics — johnmayhk @ 6:27 下午

Here is an email from one of my students, Justin.
“ 早前閱讀了有關三次數學危機的書籍 ( 只用 $20 購買的大陸書籍 ) 名稱是數學悖論與三次數學危機 , 韓雪濤著 ,John Sir 有沒有看過 ? 讀到有關微積分的時候 , 看到 : 維爾斯特拉斯 ( 是誰 ?) 構造了一個處處連續但處處不可微的 function. 這條 function 是什麼 ? 還有有關無窮小量的問題 , 它的定義是什麼 , 而且是代表著什麼意思 , 它當初定義的模糊衍生了第二次數學危機 , 就現有知識大概是 delta x tends to 0 之類的東西。”

I’m so glad to know students would spend time on reading popular mathematics books. I read that book mentioned last year and found it interesting, especially the information about the Third Crisis near the end of the book. Though the contents of the book are old stuff, I regard it as a revision of history of mathematics.

Nowhere differentiable continuous functions are very strange and interesting. In secondary school, we learn that a differentiable function must be continuous (on its domain), however, for continuous function, it may not be differentiable, just like y = |x|. It is not differentiable at x = 0, but apart from x = 0, the function is very nice on \mathbb{R}. This function is differentiable a.e. (almost everywhere). Can we imagine a “bad” function such that it is not differentiable at every x?

Karl Theodor Wilhelm Weierstrass (維爾斯特拉斯) is a Germany mathematician, a great teacher, according to a small Chinese popular mathematics book “世界數學名題欣賞 – 無處可微的連續函數”, there are many examples of “nowhere differentiable continuous functions” and the example given by Weierstrass is as follows.

 f(x) = \sum_{n=0}^{\infty}a^ncos(b^n\pi x) where b is an odd natural number, 0 < a < 1, ab > 1 + \frac{3\pi}{2}. This example was found in 1860. Secondary school students should understand the proof of non-differentiablity of the function, but I need time to type. Many years ago, I copied an easier example, just read http://www.hkedcity.net/ihouse_tools/forum/read.phtml?forum_id=27877¤t_page=&i=716330&t=716330

The problem of the meaning of “tending to zero” was resolved and developed in the path of mathematical analysis. The vague meaning of limit was refined by introducing the \epsilon - \delta language. It is an elementary mathematics concept.


Hilbert space

Filed under: Family — johnmayhk @ 7:04 下午

Hilbert, a famous mathematician, is one of my idols. That’s why I named my son Hilbert. I don’t really know whether he is gifted or not, but it is certain that he is a gift from heaven. Just share with you some of his photos.





Filed under: Life — johnmayhk @ 3:57 下午

論調一:『九歲入大學 = 沒有童年』



不要忽略『少時天才,大時更厲害』大不乏人。遠的不說,近的有 Terence Tao(看看他的 video)更多的應是『少時的天才很早已被埋沒』。腦中忽爾迴盪著挪威天才數學家阿貝爾,叫人感慨的是,天才,往往有著悲劇般的人生。



[初中][教學] Mathematics is much easier than English

Filed under: Junior Form Mathematics — johnmayhk @ 3:06 下午

What is the focus of secondary mathematics training in Hong Kong? Students may have acquired mastery of calculations. However, when it comes to presentation, students are not well-prepared for giving mathematics ideas or presenting mathematical discoveries both in written or orally. After taking the DOLACEE course, we need to do more in learning and making meaning through English language. I tried to give my F.2 boys to do the following question:

1 = 1
3 + 5 = 8
7 + 9 + 11 = 27
13 + 15 + 17 + 19 = 64
21 + 23 + 25 + 27 + 29 = 125

Observe the above pattern.
(a) Follow the same pattern, write two more rows.
(b) Describe the pattern you discover.

Most of them could finish (a), it requires only a little amount of number sense. However, there were only 3 students tried to do (b). The answers given by those three students were quite different. Because I just set this question in hurry, I did not think it carefully and design the assessing criteria in details. May be later, after the “standards referenced” mode for assessment being implemented exclusively, teachers may put much afford it setting this kind of criteria. Look back to the question above, three students gave answers as follows:

S1. It is a triangle.
S2. Each number is 2 more than the first number.
S3. I find that the no. of terms will elect (affect?) the answer that means the answer will be the cube of the no. of terms.

It may be clear that the answer given by S3 seems to be the best for “better” mathematical sense. We cannot say S1 gave wrong answer, though the message was not clear enough. S2 told something but seems to unfold just a very small amount of mathematics content. Well, they were only 3 students and I can anticipate that if all 39 students gave me answers, the varieties and the complexities of setting criteria are troublesome. Of course, if students could give, not only the descriptions, but also the reasons or proof behind, he may score the highest marks! Students, will you try?


Formulae in life

Filed under: Life — johnmayhk @ 10:25 下午

Life phenomena are not a must, but after listening to some sharing, I know that some phenomena are predictable, or precisely, explainable. Sometimes, we cannot but admit that something has to be happening. Murphy’s Laws? (i.e. If anything can go wrong, it will.) I will choose to hide up because I am not brave enough to fit into the formula! Saluting to courageous free men… There are many unfinished tasks, too sad. Pray for my uncle in Guangzhou, he is still staying in hospital after the operation for the Colorectal cancer.

Form teacher of 2V

Filed under: School Activities — johnmayhk @ 10:24 下午

I was assigned to be a form teacher of F.3C 2 years ago. May be my performance was not satisfactory or something, I had not been form teacher in the following two years after that. This year, the two-form-teacher policy is going to carry out in junior forms, I am given a chance to be a form teacher (I prefer using ‘class teacher’ instead) this year. Together with Ms. Wong (another form teacher of F.2V), I do hope to enjoy this academic year with students in my class. I knew some pupils coming from F.1J last year, I wish some of them may change their impressions on me (i.e. I am not kind and mild).

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