Quod Erat Demonstrandum

2017/06/23

實數問題複數解決

Filed under: mathematics,NSS — johnmayhk @ 3:43 下午
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幾個月前的中學數學科諮詢文件見 M1,M2 外的 Further Mathematics 內容,重遇會考附加數學一些內容:

運用當中一個特性 z\overline{z}=|z|^2,輕易得出下式:

(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2

式子在說:平方和的積,仍是平方和。

現在的中學生,大部分不會知道甚麼是 (more…)

廣告

2015/09/13

一式過

Filed under: Pure Mathematics — johnmayhk @ 5:19 下午
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比如要寫出以下數列的通項

-1,1,-1,1,\dots

應該不難得出

(-1)^n

吧。(這樣假設上述數列的變化模式不變,一直都是「負 1 正 1」咁去。)

但還有沒有其他可能?

有的,比如 (more…)

2015/08/27

某 monic 多項式

Filed under: NSS,Pure Mathematics — johnmayhk @ 10:09 上午
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偶見高X論壇某題:

證明 x^4+x^3+x^2+x+1 > 0

回應者用了較麻煩的方法處理。

其實,當 x\ne 1(more…)

2014/02/20

平方和

Filed under: Fun — johnmayhk @ 11:44 下午
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隨便考慮 n 個正整數

a_1,a_2,a_3,\dots a_n

b=\sqrt{a_2^2+a_3^2+\dots +a_n^2}

(a_1+bi)^m=P+Qi

其中 m 是正整數,i^2=-1,那麼我們必有 (more…)

2012/05/01

無聊談通項

那天觀課,同事開始教等差數列(arithmetic sequence),(估計是隨便)問學生:

2,1,4,\frac{1}{2},8,\frac{1}{4},\dots

的通項(general term)是甚麼?

關於通項,之前也談過,除非先有「特殊規定」,否則所謂通項是無定義的。

談回上題,假設同事的「特殊規定」是把兩個等比數列(geometric sequence)併在一起。當然,學生在中一時接觸過數型(number pattern),但要寫出上述數列的通項,似乎不是一蹴而就的事。因那是引起動機的其中一例,同事沒有繼續討論,但卻引發我思考一些東西。

只要時間許可,相信學生不難得到: (more…)

2012/02/26

答網友:y=z^2

Filed under: HKALE,Pure Mathematics — johnmayhk @ 6:29 下午
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以下是網友在 2011-03-06 的提問:

By considering

(1+z)^8+(1-z)^8=0

show that

\displaystyle\sum_{k=0}^7 \tan^2\frac{(2k+1)\pi}{16}=56

不難 (more…)

2012/01/20

arctan,pi,complex numbers

Filed under: mathematics,Pure Mathematics — johnmayhk @ 12:02 下午
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式子

\pi=\tan^{-1}1+\tan^{-1}2+\tan^{-1}3

美麗嗎?我比較多見的形式是

\tan^{-1}(\frac{1}{2})+\tan^{-1}(\frac{1}{3})=\frac{\pi}{4}

(易知 \tan^{-1}(\frac{1}{n})+\tan^{-1}(n)=\frac{\pi}{2},故上述兩式等價。)

以圖來證明上式,可以考慮 (more…)

2011/11/26

利用複數尋寶

Filed under: Fun,Pure Mathematics — johnmayhk @ 4:51 下午
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此刻,純數還欠複數(complex numbers)這個課題未教完。當完成後,課題中精采之處,將長埋「新」高中之墓塚。

同事問:複數有沒有有趣的幾何應用?很多,這裡舉一個經典例子。

海盜留下藏寶位置之提示:

島上有一個絞刑架,一棵橡樹及一棵松樹。

由絞刑架出發前往橡樹, (more…)

2011/11/18

polar form

Filed under: Pure Mathematics — johnmayhk @ 11:46 上午
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基礎題:

Convert z=-\cos \theta -i\sin\theta into polar form.

問:「在何象限,cosine 和 sine 值也是負?」 (這是極不精確的說法,但學生又明我說甚麼。)

答:「第三。」 (more…)

2011/03/02

當 x 接近零,(1+1/x)^x 如何?

Filed under: Pure Mathematics — johnmayhk @ 5:07 下午
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課堂上,我通常會談「非例子」。

比如教(講)了

\displaystyle \lim_{x\rightarrow \infty}(1+\frac{1}{x})^{x} = e

隨即問

\displaystyle \lim_{x\rightarrow \infty}(2+\frac{1}{x})^{x} = ?

這個易,再問

\displaystyle \lim_{x\rightarrow 0}(1+\frac{1}{x})^{x} = ?

學生 (more…)

2009/10/26

利用圖像尋找非實根

Filed under: HKALE,NSS,Pure Mathematics — johnmayhk @ 8:24 下午
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在高中一 NSS 數學課,我開始教二次圖像和二次方程之根(roots)的關係。現在課程涉及複數,卻沒有教如何利用圖像尋找複根(complex roots)或非實根(unreal roots),我在此補充一下。

以下是在下用極速粗製濫造的 ETV,同學先看看:

解說: (more…)

2009/09/29

今天雜記

Filed under: Life,mathematics,NSS,School Activities — johnmayhk @ 11:02 下午
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難以轉化

科主任擬了兩班高中一 M2 的聯測卷,其中有一道題目只有三名學生懂得處理:

Given that x and y are purely imaginary numbers. If (x + y) + (2x - y)i = 12 - i, find (the values of) x and y. (more…)

2008/04/21

Complex trigonometric functions

Filed under: University Mathematics — johnmayhk @ 1:03 下午
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F.5 student Hoover found it interesting to know some properties of complex trigonometric functions. He asked if the following is still true?

\sin^2(z) + \cos^2(z) = 1

for any complex number z. (more…)

2008/01/16

[初中] Factorization, useful or not?

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

2007/12/18

[AL][PM] Applications of roots of unity

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 11:42 上午
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Oh, it’s the time for switching the channel into English. Actually, I’m afraid to write in English because I’m blamed not to use English to write blog too often, it may do harm to students. (My English is poor ma, right!) But, firstly, there are less than 10 students reading this blog, so it may be harmless relatively. And next, because of the beautiful fonts, I try to use English in this blog, and that is the only reason ^_^!

F.7C students were just given a quiz, here are some old stuff for your (so-called) enrichment.

Q.1 [Remainder theorem?]

Show that x^{1987} + x^{1997} + x^{2007} is divisible by x^{1987} + x^{1988} + x^{1989}.

Q.2 [Binomial theorem?]

Evaluate C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007}.

Q.3 [Geometry problem?]

Let P_1, P_2, \dots P_{2007} be the vertices of a regular 2007-gon inscribed in a unit circle. Evaluate P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2007}.

The questions above may look different to each other, however, by using the roots of unity, we can solve them.

Solution to Q.1

Let \omega be a complex cube root of unity, i.e. \omega^3 = 1 and \omega \neq 1.

If P(x) is a polynomial (over \mathbb{R}) such that P(\omega) = P(\omega^2) = 0, then (x - \omega)(x - \omega^2) is a factor of P(x). Hence P(x) is divisible by 1 + x + x^2.

Now, let P(x) = 1 + x^{10} + x^{20}

It is easy to check

1 + \omega^{10} + \omega^{20} = 1 + \omega^{20} + \omega^{40} = 0 (Why?)

Hence P(\omega) = P(\omega^2) = 0; thus P(x) is divisible by 1 + x + x^2. Or

1 + x^{10} + x^{20} = (1 + x + x^2)Q(x) for some polynomial Q(x)

Now, x^{1987} + x^{1997} + x^{2007}
= x^{1987}(1 + x^{10} + x^{20})
= x^{1987}(1 + x + x^2)Q(x)
= (x^{1987} + x^{1988} + x^{1989})Q(x)

Result follows.

Solution to Q.2

Let \omega be a complex forth root of unity, i.e. \omega^4 = 1 and \omega \neq 1.

The following results are trivial.

1 + \omega^k + \omega^{2k} + \omega^{3k} = 4 if k is a multiple of 4
1 + \omega^k + \omega^{2k} + \omega^{3k} = 0 if k is not a multiple of 4

Now

(1 + x)^{n} = C_{0}^{n} + C_{1}^{n}x + C_{2}^{n}x^2 + \dots + C_{n}^{n}x^n (where n = 2007)

Put x = 1, \omega, \omega^2, \omega^3 into the above accordingly. We have

2^{n} = C_{0}^{n} + C_{1}^{n} + C_{2}^{n} + \dots + C_{n}^{n}
(1 + \omega)^{n} = C_{0}^{n} + C_{1}^{n}\omega + C_{2}^{n}\omega^2 + \dots + C_{n}^{n}\omega^n
(1 + \omega^2)^{n} = C_{0}^{n} + C_{1}^{n}\omega^2 + C_{2}^{n}\omega^4 + \dots + C_{n}^{n}\omega^{2n}
(1 + \omega^3)^{n} = C_{0}^{n} + C_{1}^{n}\omega^3 + C_{2}^{n}\omega^6 + \dots + C_{n}^{n}\omega^{3n}

Sum up the above 4 equations and use the trivial results just mentioned, we have

4(C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007})
= 2^{2007} + (1 + \omega)^{2007} + (1 + \omega^2)^{2007} + (1 + \omega^3)^{2007}

Hey, we can simplify it further because we may take \omega = i, \omega^2 = -1, \omega^3 = -i, hence

C_{0}^{2007} + C_{4}^{2007} + C_{8}^{2007} + \dots + C_{2004}^{2007}
= \frac{1}{4}(2^{2007} + (1 + i)^{2007} + (1 - 1)^{2007} + (1 - i)^{2007})
= \frac{1}{4}(2^{2007} + 2^{\frac{2007}{2}}(2\cos\frac{2007\pi}{4}))
= \frac{1}{4}(2^{2007} + 2^{\frac{2007}{2}}(2\cos\frac{\pi}{4}))
= 2^{1002}(2^{1003} + 1)

Solution to Q.3

The complex numbers corresponding to the vertices of a regular n-gon inscribed in a unit circle are roots of z^n = 1. Hence, we may let

1, \omega, \omega^2, \dots, \omega^{2006} be the complex numbers corresponding to the vertices of the regular 2007-gon; where \omega = \cos\frac{2\pi}{2007} + i\sin\frac{2\pi}{2007}.

It is easy to have \omega, \omega^2, \dots, \omega^{2006} are roots of the equation

x^{2006} + x^{2005} + \dots + x + 1 = 0 – – – (1)

Suppose we translate the regular 2006-gon to the left by 1 unit, then the vertices will become 0, \omega - 1, \omega^2 - 1, \dots, \omega^{2006} - 1, and hence

P_1P_2 = |\omega - 1| = |z_1| (say)
P_1P_3 = |\omega^2 - 1| = |z_2|
P_1P_4 = |\omega^3 - 1| = |z_3|

P_1P_{2007} = |\omega^{2006} - 1| = |z_{2006}|

Now, if we can find out an equation whose roots are z_1, z_2, \dots , z_{2006}, then P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2006} is simply the modulus of the product of roots.

By (1), just set z = x - 1 then an equation with roots z_1, z_2, \dots , z_{2006} is

(z + 1)^{2006} + (z + 1)^{2005} + \dots + (z + 1) + 1 = 0
z^{2006} + \dots + 2007 = 0

Since the product of roots of the above equation = 2007,
P_1P_2 \times P_1P_3 \times \dots \times P_1P_{2006} = 2007

Not enough? If you want to know how powerful in using complex numbers for solving elementary mathematics problems, I suggest an old Chinese popular mathematics for your own leisure reading: 神奇的複數─如何利用複數解中學數學難題

I do admire Mr. Siu’s concept : “I don’t teach: I share". Read his blog to learn better English!
http://hk.myblog.yahoo.com/siu82english

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