# 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).
f(x) $\equiv$ k(x-1)(x-2) for some non-zero constant k.

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)
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
$\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} = \frac{A}{x} + \frac{Bx + C}{x^2} + \frac{D}{x - 1} + \frac{Ex + F}{(x - 1)^2}$.
$\frac{x^3 - x^2 -3x +2}{x^2(x - 1)^2} \equiv \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1} + \frac{D}{(x - 1)^2}$

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.

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.

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 {$x_n$}.
$x_n < 1 - \frac{1}{n} < 1$ 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.
$ax^2 + bx + c = 0$
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)

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)

16.
(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 $x < a$

$f'(x) = f_1'(x)$ 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, $\lim_{n \rightarrow \infty}\int_{0}^{1}\frac{x^{n}}{1 + x^2}dx = L = 0$

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

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 | 回應

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 | 回應

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
3learning.net
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。

迴響 由 zen — 2008/02/20 @ 2:41 上午 | 回應

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 | 回應

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 | 回應

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 | 回應