# Quod Erat Demonstrandum

## 2015/09/13

### 一式過

Filed under: Pure Mathematics — johnmayhk @ 5:19 下午
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$-1,1,-1,1,\dots$

$(-1)^n$

## 2015/08/27

### 某 monic 多項式

Filed under: NSS,Pure Mathematics — johnmayhk @ 10:09 上午
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## 2014/02/20

### 平方和

Filed under: Fun — johnmayhk @ 11:44 下午
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$a_1,a_2,a_3,\dots a_n$

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

$(a_1+bi)^m=P+Qi$

## 2012/05/01

### 無聊談通項

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

## 2012/02/26

### 答網友：y=z^2

Filed under: HKALE,Pure Mathematics — johnmayhk @ 6:29 下午
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By considering

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

show that

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

## 2012/01/20

### arctan,pi,complex numbers

Filed under: mathematics,Pure Mathematics — johnmayhk @ 12:02 下午
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$\pi=\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$

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

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

## 2011/11/26

### 利用複數尋寶

Filed under: Fun,Pure Mathematics — johnmayhk @ 4:51 下午
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## 2011/11/18

### polar form

Filed under: Pure Mathematics — johnmayhk @ 11:46 上午
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Convert $z=-\cos \theta -i\sin\theta$ into polar form.

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

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

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

## 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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## 2009/09/29

### 今天雜記

Filed under: Life,mathematics,NSS,School Activities — johnmayhk @ 11:02 下午
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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.

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

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.

– – – – – –

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.

## 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

## 2007/12/11

### [AL][PM][U] 比較複數的大小？

Filed under: HKALE,Pure Mathematics,University Mathematics — johnmayhk @ 6:27 下午
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John sir早前上堂提到 complex number: i 和 0 大小比較的問題

$i > 0$, then $i^2 > 0$,則 $-1 > 0$, contradiction。反之設 $0 > i$, 亦生矛盾。

＂＞＂關係能滿足以下四個性質:
1)對任何兩個不同的數 $a,b$; $a > b$ or $b > a$ 不能同時成立
2)if $a > b$,$b > c$ then $a > c$
3)if $a > b$ then $a + c > b + c$
4)if $a > b$, $c > 0$ then $ac > bc$

by 4) $i^2 > 0$, $-1 > 0$

(因為”>”不一定是實數中規定的含義)←這和阿sir上堂提到的問題有分別。

(這裏並未有將inequality sign倒轉),這裡是否和(*)所提到的有關？
we have $-1 > 0$
by property 3), add 1 into both sides
$-1 + 1 > 0 + 1$,then $0 > 1$
by from above, $0 > 1$ and $1 > 0$,which contradicts property 1)
then we can conclude that $i$ is not $> 0$

1,2,3,…

1 < 2 < 3 < …

1 > 2，2 > 3 等等（參考 order dual）

a + bi < c + di 當且僅當 a < c 或 (a = c 及 b < d)

（當心，在 A-level 千萬不可出現 a + bi < c + di 等等的式子，老師會二話不說給你零蛋。我們現在討論的是一種『序』，隨便你用什麼符號，但用『<』，是讓我們有一種熟悉的感覺而已，它不一定是一般的不等式符號，它只用來表達熟先熟後的標記。所以 Justin 可以放心，在中學，『<』這個符號一定是代表『嚴格小於』，只是到了大學，就要看情況了。）

3 + 100i < 4 + 2i
5 + 6i < 5 + 7i
8 – 2i < 8

（大家領略到好像『查字典』的感覺嗎？）

1. 偏序（partial order）：亦可再細分為弱偏序及嚴格偏序
2. 線序（linear order）或稱全序（total order）
3. 良序（well order）

（自身性 reflexivity）對所有 a $\in$ A, 恒有 a $\leq$ a。
（反對稱性 antisymmetry）對所有 a , b $\in$ A, 若 a $\leq$ b 及 b $\leq$ a，則 a = b。
（傳遞性 transitivity）對所有 a , b , c $\in$ A, 若 a $\leq$ b 及 b $\leq$ c，則 a $\leq$ c。

a = {1}
b = {2}
c = {3}
d = {1,2}
e = {1,3}
f = {2,3}
g = {1,2,3}

x $\leq$ y iff x $\subseteq$ y

Justin 給的序和偏序差不多，即是所謂嚴格偏序（strict order）。

（非自身性 irreflexivity）對所有 a $\in$ A, 無 a $<$ a。
（非對稱性 asymmetry）對所有 a , b $\in$ A, 若 a $<$ b，則無 b $<$ a。
（傳遞性 transitivity）對所有 a , b , c $\in$ A, 若 a $<$ b 及 b $<$ c，則 a $<$ c。

2 $\leq$ 6　［因 2 整除 6］
5 $\leq$ 20　［因 5 整除 20］

（三分律 trichotomy）對所有 a , b $\in$ A, 必有 a $<$ b, b $<$ a 或 a = b。

『線』者，取其把所有『元素』都可如『數線』，按序排列。
『全』者，即謂『全部』元素都可知其先後。

（1）對所有 a , b , c $\in$ F，若 a $<$ b，則 a + c $<$ b + c。
（2）對所有 a , b $\in$ F，若 0 $<$ a 及 0 $<$ b，則 0 $<$ ab。

0 ⊿ i 或 i ⊿ 0 或 i = 0（不合）