# Quod Erat Demonstrandum

## 2008/01/24

### Pure Mathematics 補底

Filed under: HKALE,Pure Mathematics,Teaching — johnmayhk @ 3:08 下午

1.
If f(x) is a polynomial of degree 2, and f(1) = f(2) = 0, then it is NOT true to say
f(x) = (x-1)(x-2).
Instead, we should write
f(x) $\equiv$ k(x-1)(x-2) for some non-zero constant k.
[Read : 2004-AL-pure math-paper I-Q.4]

2.
Let f(x) be a polynomial of degree not less than 3.
If f(x) is divided by (x-1)(x-2)(x-3), then it is NOT true to say
f(x) = (x-1)(x-2)(x-3)Q(x) + R
(where R is a constant, the remainder)
Instead, we should write
f(x) $\equiv$ (x-1)(x-2)(x-3)Q(x) + $ax^2 + bx + c$
that is,
we should set the remainder as a polynomial of degree 2
(just less than the degree of the divisor (x-1)(x-2)(x-3) by 1)

3.
Resolve $\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2}$ into partial fraction.
Don’t write the following
$\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} = \frac{A}{x} + \frac{Bx + C}{x^2} + \frac{D}{x - 1} + \frac{Ex + F}{(x - 1)^2}$.
Instead, we should write
$\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} \equiv \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$
[Read : 2000-AL-pure math-paper I-Q.12]

4.
If $x > k$, it is NOT true to draw the following conclusion
$x^2 - x + 1 > k^2 - k + 1$.
But, if we know $x + k - 1 > 0$, then it is true to have $x^2 - x + 1 > k^2 - k + 1$.
Proof: Multiply $x + k - 1 > 0$ and $x - k > 0$, we have $(x^2 - k^2) - (x - k) > 0$ and result follows.
[Read : 1999-AL-pure math-paper I-Q.2]

5.
Let {$x_n$} be a sequence of real numbers.
Knowing that $x_n > n$.
It is NOT true to say
{$x_n$} is increasing.
Just give an example.
$x_1 = 10, x_2 = 9, x_3 = 8, x_4 = 7, x_5 = 6, ...$
Obviously, $x_n > n$ (for n = 1,2,3,4,5), but {$x_n$} is not increasing.
[Read 1999-AL-pure math-paper I-Q.2]

6.
Let {$x_n$} be a sequence of real numbers.
Knowing that $x_n < 1 - \frac{1}{n}$.
We cannot say {$x_n$} is bounded from above by $1 - \frac{1}{n}$.
$1 - \frac{1}{n}$ is NOT an upper bound of the sequence {$x_n$}.
Instead, with a small effort added, we have
$x_n < 1 - \frac{1}{n} < 1$ for all positive integers n.
And now, it is clear that, 1 is an upper bound of {$x_n$}.

7.
(a) Solve $|x| < 3$
Solution: $-3 < x < 3$
(b) Solve $|x| > 3$
Solution: $x > 3$ or $x < -3$

8.
Consider the quadratic equation
$ax^2 + bx + c = 0$
Given that the discriminant $\Delta < 0$.
Then it is NOT true to say that the quadratic equation has no real solution.
Just consider a quadratic equation with roots 0 and i.
That is
$x(x - i) = 0$
$x^2 - ix = 0$
Then $\Delta = (-i)^2 - 4(1)(0) = -1 < 0$.
But there is a real root (x = 0) for the quadratic equation.

9.
“Vector" is out of the syllabus now.
“Linearly independent", “linearly dependent" are out-of-syllabus concepts.

10.
Techniques in evaluating limits
(a) l’hôpital rule
(b) Special limit : $\lim_{x \rightarrow 0}\frac{\sin(x)}{x} = 1$
(c) Special limit : $\lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = e$
(d) Special limit : $\lim_{a \rightarrow 0}(1 + a)^{\frac{1}{a}} = e$
(e) Boundedness : Knowing that {$a_n$} is bounded and $\lim_{n \rightarrow \infty}b_n = 0$, then $\lim_{n \rightarrow \infty}a_nb_n = 0$
(f) Squeezing (Sandwich) principle
(g) Taking limit on the recurrence relation : $a_{n + 1} = 2a_{n}$ and suppose {$a_n$} converges $\Rightarrow \lim_{n \rightarrow \infty}a_{n + 1} = \lim_{n \rightarrow \infty}(2a_{n}) \Rightarrow \lim_{n \rightarrow \infty}a_n = 2\lim_{n \rightarrow \infty}a_n \Rightarrow \lim_{n \rightarrow \infty}a_n$ = 0

11.
Symbols about left and right hand limits
$x \rightarrow 3^+$" does NOT mean “$x \rightarrow +3$"
$x \rightarrow 3^-$" does NOT mean “$x \rightarrow -3$"
See the difference? Try to obtain the following by sight.
(a) $\lim_{x \rightarrow 3^+}\sqrt{x - 3} = 0$
(b) $\lim_{x \rightarrow 3^-}\sqrt{x - 3}$ does not exist.
(c) $\lim_{x \rightarrow +3}\sqrt{x - 3}$ does not exist (since the left and right lmits are not the same)
(d) $\lim_{x \rightarrow -3}\sqrt[3]{x + 3} = 0$

12.
To evaluate $\lim_{x \rightarrow \infty}[f(2004 + x) - f(x)]$, please do NOT split it as
$\lim_{x \rightarrow \infty}f(2004 + x) - \lim_{x \rightarrow \infty}f(x)$
No! Both limits may NOT exist!
You may try “rationalization", “sum to product", “MVT" etc.

13.
How to evaluate $\lim_{x \rightarrow \infty}[f(\sqrt{2004 + x}) - f(\sqrt{x})]$ = ?
Try the Mean Value Theorem (of course, $f$ should be differentiable at certain interval), that is
$\lim_{x \rightarrow \infty}[f(\sqrt{2004 + x}) - f(\sqrt{x})]$
= $\lim_{x \rightarrow \infty}f'(c)(\sqrt{2004 + x} - \sqrt{x})$ for some $c \in [\sqrt{x} , \sqrt{2004 + x}]$
= $\lim_{x \rightarrow \infty}f'(c)[\frac{2004 + x - x}{\sqrt{2004 + x} + \sqrt{x}}]$ for some $c \in [\sqrt{x} , \sqrt{2004 + x}]$
= $\lim_{x \rightarrow \infty}f'(c)[\frac{2004}{\sqrt{2004 + x} + \sqrt{x}}]$ for some $c \in [\sqrt{x} , \sqrt{2004 + x}]$
(if $\lim_{x \rightarrow \infty}f'(c)$ is bounded, then the limit is zero, read 10.(e))

14.
If $\lim_{x \rightarrow 0}\frac{g(x)}{x}$ exists, then $\lim_{x \rightarrow 0}\frac{g(x)} = 0$.
e.g. Given $\lim_{x \rightarrow 0}\frac{f(x) - 1}{x} = 1$, evaluate $\lim_{x \rightarrow 0}f(x)$. (ANS: 1)
[Read : 2002-AL-pure math-paper II-Q.3]

15.
Be careful, the following may NOT be true!
$\lim_{x \rightarrow a}(f(x) + g(x)) = \lim_{x \rightarrow a}(f(x) + \lim_{x \rightarrow a}g(x))$
e.g. Given $\lim_{x \rightarrow 0}\frac{f(x) - 1}{x} = 1$, evaluate $\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}$
It is WRONG to write
$\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}$
= $\lim_{x \rightarrow 0}\frac{f(x) - 1}{x \times 1}$ (for $\lim_{x \rightarrow 0}e^x = 1$)
= 1
You should give somthing like
$\lim_{x \rightarrow 0}\frac{f(x) - e^x}{xe^x}$
= $\lim_{x \rightarrow 0}\frac{f(x) - 1 + 1 - e^x}{xe^x}$
= $\lim_{x \rightarrow 0}(\frac{f(x) - 1}{xe^x} + \frac{1 - e^x}{xe^x})$
= $\lim_{x \rightarrow 0}(\frac{f(x) - 1}{xe^x} + \frac{-e^x}{xe^x + e^x})$ (l’hôpital rule)
= 0 (see, NOT 1)
[Read : 2002-AL-pure math-paper II-Q.3]

16.
Please please please DON’T make the following silly mistakes!
(a) $\int\tan^n(x)dx \neq \frac{tan^{n + 1}}{n + 1} + C$
(b) $\int e^{x^2}dx \neq e^{x^2} + C$
For (a), try the reduction formula.
For (b), there is no closed form, just stop there.

17.
The following definitions MUST be memorized!
(a) “$f(x)$ is well-defined at $x = a$" means “$f(a)$ can be found".
(b) “$f(x)$ is continuous at $x = a$" means “$\lim_{x \rightarrow a}f(x) = f(a)$“.
(c) “$f(x)$ is differentiable at $x = a$" means “$\lim_{h \rightarrow 0}\frac{f(x + h) - f(x)}{h}$ exists".

18.
Q: At a look at the definitions above, why there is no mention of the left and right hand limits? When should we consider the left and right hand limits?
A: When $f(x)$ has DIFFERENT expressions for $x \ge a$ and $x < a$ (say)
e.g.
(a) Let $f(x) = x + 1$ for $x \ge 3$ and $f(x) = 2(x - 1)$ for $x < 3$. Then $f(x)$ is continuous at $x = 3$. Proof: $\lim_{x \rightarrow 3^+}f(x) = \lim_{x \rightarrow 3^+}(x + 1) = 4$ while $\lim_{x \rightarrow 3^-}f(x) = \lim_{x \rightarrow 3^-}2(x - 1) = 4 = f(3)$ Hence $f(x)$ is continuous at $x = 3$.
(b) Let $f(x) = x + 1$ for $x > 3$ and $f(x) = 2(x - 1)$ for $x < 3$ with f(3) = 5 Then $f(x)$ is NOT continuous at $x = 3$. (why?)
(c) Let $f(x) = x|x|$. To prove $f(x)$ is continuous at $x = 0$, it is NOT necessary to “divide cases", since both left and right hand limits are zero. i.e. $\lim_{x \rightarrow 0^+}x = \lim_{x \rightarrow 0^-}x = 0$, hence $\lim_{x \rightarrow 0}x = 0$ and therefore, $\lim_{x \rightarrow 0}x|x| = 0 = f(0)$. Done.

19.
Given

$f(x) = f_1(x)$ for $x \ge a$
$f(x) = f_2(x)$ for $x < a$

(both $f_1(x)$ and $f_2(x)$ are ‘nice’ functions)
Then, to find $f'(x)$, the problem only appears at $x = a$, hence DON’T write

$f'(x) = f_1'(x)$ for $x \ge a$
$f'(x) = f_2'(x)$ for $x < a$

Instead, you should write

$f'(x) = f_1'(x)$ for $x > a$
$f'(x) = f_2'(x)$ for $x < a$

And try to check the value of $f'(x)$ at $x = a$ by finding

$f_{+}'(a) = \lim_{h \rightarrow 0^+}\frac{f_1(a + h) - f(a)}{h}$
$f_{-}'(a) = \lim_{h \rightarrow 0^-}\frac{f_2(a + h) - f(a)}{h}$

and if $f_{+}'(a) = f_{-}'(a)$, then $f'(a)$ is the common limit.

20.
Techniques in integrations: substitution
(a) Involves $\sqrt{a^2 - x^2}$ , substitute $x = a\sin(\theta)$ or $x = a\cos(\theta)$
(b) Involves $\sqrt{x^2 - a^2}$ , substitute $x = a\sec(\theta)$
(c) Involves $x^2 + a^2$ , substitute $x = a\tan(\theta)$

21.
Techniques in integrations: partial fractions
(a) Involves $x^2 - a^2$ , partial fractions may help
(b) Involves $x^4 + a^4$ , don’t forget
$x^4 + a^4$
= $x^4 + 2x^2a^2 + a^4 - 2x^2a^2$
= $(x^2 + a^2)^2 - (\sqrt{2}ax)^2$
= $(x^2 + \sqrt{2}ax + a^2)(x^2 - \sqrt{2}ax + a^2)$
and then partial fraction resolving

22.
Techniques in integrations: others
(a) Integration by parts
(b) Taking limits
e.g. $\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx$ = ?
Let $I_n = \int_{0}^{1}\frac{x^{n}}{1 + x^2}dx$, then
$0 \le I_n = \int_{0}^{1}\frac{x^{n}}{1 + x^2}dx \ge \int_{0}^{1}\frac{x^{n + 1}}{1 + x^2}dx = I_{n + 1}$
That is {$I_n$} is decreasing and bounded from below by zero.
Hence $\lim_{n \rightarrow \infty}I_n$ exists, says L.
For sufficiently large $n$, write
$\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx$ = $\int_{0}^{1}\frac{x^{n - 2}(x^2 + 2x + 1 - 2x - 1)}{1 + x^2}dx$ = $\int_{0}^{1}(x^{n - 2} -\frac{2x^{n - 1}}{1 + x^2} - \frac{x^n}{1 + x^2})dx$ = $\frac{1}{n - 1} -2\int_{0}^{1}\frac{x^{n - 1}}{1 + x^2}dx - \int_{0}^{1}\frac{x^n}{1 + x^2}dx$
Taking $n \rightarrow \infty$ from both sides,
$L = 0 - 2L - L$
Hence, $\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx = L = 0$
[Read : 1990-AL-pure math-paper II-Q.8(a)]

23.
Many questions involve the Mean Value Theorem in Cauchy form, that is
both $f(x)$ and $g(x)$ are continuous on [a , b] and differentiable on (a , b) with $g(a) \neq g(b)$, then
then $\frac{f(a) - f(b)}{g(a) - g(b)} = \frac{f'(c)}{g'(c)}$ for some $c \in$ (a , b).
DON’T write the following as a so-called proof

$f(a) - f(b) = f'(c)(a - b)$
$g(a) - g(b) = g'(c)(a - b)$
$\Rightarrow \frac{f(a) - f(b)}{g(a) - g(b)} = \frac{f'(c)}{g'(c)}$ for some $c \in$ (a , b)

NO! We CANNOT ensure both $c$ in $f'(c)$ and $g'(c)$ are the same!!! That is, all we have may be the following

$f(a) - f(b) = f'(c_1)(a - b)$ for some $c_1 \in$ (a , b)
$g(a) - g(b) = g'(c_2)(a - b)$ for some $c_2 \in$ (a , b)

and then we can do nothing at all…How to prove? Try the read
[Read 2006-AL-pure math-paper II-Q.11(a)]

24.
Get tired to give details, just give topics in major techniques:
(a) Integration by parts
(b) Fundamental theorem of integral calculus
(c) Inequalities
(d) Riemann sum

25.
Partial fractions : there is a short-cut for distinct linear factors, just in case you may forget…

$\frac{6x^2 - 18x + 6}{x(x - 2)(x - 3)}$
= $\frac{6(0)^2 - 18(0) + 6}{x(0 - 2)(0 - 3)} + \frac{6(2)^2 - 18(2) + 6}{(2)(x - 2)(2 - 3)} + \frac{6(3)^2 - 18(3) + 6}{3(3 - 2)(x - 3)}$
= $\frac{1}{x} + \frac{3}{x - 2} + \frac{2}{x - 3}$

## 8 則迴響 »

1. John Sir未必太謙了，我認為上述的notes不只是for補底人士。
推推推~

問: Paper 2有何靈方指點？ Long Question經常有"坐多過做"的感覺。

迴響 由 Wong Hon — 2008/01/24 @ 11:26 下午 | 回覆

2. Notes 其實是同學在測驗中犯的錯誤，所以我是靠他們給我靈感，教學相長也。

Paper II 技巧不多，但變化頗大。長題結構多以 (a) 及 (b) 推 (c)。難一概論之，有時間再談。

迴響 由 johnmayhk — 2008/01/25 @ 11:53 上午 | 回覆

3. 3learning.net誠意邀請您成為我們的教學成員
您好！！
我是來自 3learning.net的管理員。
在看過你精彩的教學後，認為您有著教學熱誠。
我們正搜尋像您這樣的教學人材，並請求您加入
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一個免費的教學論壇，成為我們的教學成員。
在本論壇中，你可以輕鬆的上傳並分享你的教學。

About 3learning.net
http://forum.3learning.net/viewthread.php?tid=7&extra=page%3D1

迴響 由 zen — 2008/02/20 @ 2:41 上午 | 回覆

4. Thank you for the invitation. However, it may be too much for me to maintain my blog and forum, not to mention to be a person-in-charge in one of the sections in your forum. Instead, I ‘d registered as one of your members and I’ll answer questions or just share something in your forum ocassionally.

迴響 由 johnmayhk — 2008/02/20 @ 10:47 上午 | 回覆

5. thank for your response
we have upgraded your usergroup in our forum to
teacher to use more functions in my forum

we are also creating mathematic notes

if you have math typing problems, we are willing to help you to type your hand-write notes to be math-typing files.
certainly,this materials’copyright is owned by yourself.

anyway,thank you again~ =)

迴響 由 zen — 2008/02/20 @ 4:21 下午 | 回覆

6. 真係唔明….點解我以前無呢D 野……..

迴響 由 WKK — 2009/04/30 @ 12:47 上午 | 回覆