終於都認清自己係咩料

學術
#1 DutalaTortoise
16/12/21 18:25

話說我2005 弱智CE Phy拎B, 2007 AL拎D

以下物理知識我從來無學過,係近日先知:

1. Mirage (refraction of hot air)

2. Black matter is both a good absorber and good (IR) emitter

靠操pastpaper拎番嚟既grade,

呃到人呃唔到神仙

CE靠勤力,AL靠走精面,到頭來得個吉

我懷疑Pure Math 拎A走去讀BBA班上進MK,

佢哋係咪真係明哂個SYLLABUS,

係咪識得計 ln(i) = i·pi/2

#2 DutalaTortoise
16/12/21 18:28

話說我2005 弱智CE Phy拎B, 2007 AL拎D

以下物理知識我從來無學過,係近日先知:

1. Mirage (refraction of hot air)

2. Black matter is both a good absorber and good (IR) emitter

靠操pastpaper拎番嚟既grade,

呃到人呃唔到神仙

CE靠勤力,AL靠走精面,到頭來得個吉

我懷疑Pure Math 拎A走去讀BBA班上進MK,

佢哋係咪真係明哂個SYLLABUS,

係咪識得計 ln(i) = i·pi/2

syllabus一千幾百樣野,

HKEAA從來就只會翻炒或抄坊間Question Bank

問來問去連10%都cover 唔到

也難怪堆MK俾錢去補習社拎TIP可以拎到ABC

中國考試制度已經變哂質

內捲程度超乎我預計

咁嘅環境下,識創新反而難生存

資源錯配愈來愈嚴重

我相信,未來大學TEST-BLIND將會是大趨勢

#3 DutalaTortoise
16/12/21 18:41

玩過玄學都知嚟緊2024香港會轉地運,

由磚頭旺的8運過渡去虛幻旺的9運

而且一轉就會行足2年運

基本上,除非家宅北面向水,否則一定無運行

我成日講,股票樓盤呢啲唔係自己真正的錢,

係好concrete,不過啲錢喺由你以外的人搶劫過來的

文史哲數理化工商知識先係真正屬於自己啲錢,

係好虛幻,但對自己好充實,而且係public goods

當然,玩開炒賣的賤民會當正我傻仔,

那又難怪,我一野否定哂佢哋幾十年既基本信念,

佢哋見到我一定會陷入認知失調暴露自己猙獰畜牲一面

#4 DutalaTortoise
16/12/21 18:42

玩過玄學都知嚟緊2024香港會轉地運,

由磚頭旺的8運過渡去虛幻旺的9運

而且一轉就會行足2年運

基本上,除非家宅北面向水,否則一定無運行

我成日講,股票樓盤呢啲唔係自己真正的錢,

係好concrete,不過啲錢喺由你以外的人搶劫過來的

文史哲數理化工商知識先係真正屬於自己啲錢,

係好虛幻,但對自己好充實,而且係public goods

當然,玩開炒賣的賤民會當正我傻仔,

那又難怪,我一野否定哂佢哋幾十年既基本信念,

佢哋見到我一定會陷入認知失調暴露自己猙獰畜牲一面

而且一轉就會行足20*年運

#5 DutalaTortoise
16/12/21 18:56

喺hierarchy of arguments,

直接指出對方論點錯誤係最高明的手段

但現實上唔係咁

你推翻對方信仰,

對方必然憤怒,

於是會演變成雙方互相攻擊的死局

Dialectics(辯證法)比arguing更高明

但要做到平衡雙方利益肯定難過KO對方

我一路諗,

共產主義要如何才能與資本主義甚至其他宗教教義相容

我目前諗到的,是女人如何主導未來伊斯蘭文明,

然後新共產主義融合伊斯蘭教、佛教等甚至無神論,

以世界文明命運共同體協作collective security

當然,我要面對成千上萬保守的狗公,

因為我的思想正在威脅佢哋既得利益

要在學術上maintain開放的態度不是易事,

但我相信只要有毅力和軟實力終能成成事

#6 DutalaTortoise
16/12/21 19:02

性慾,永遠是女人操縱狗公的手段

所以,要搞好思想,

一定要去除所有有戀愛意識的agents

大家見得多辦公室戀愛(甚至宮廷)如何腐化制度

所以,自由戀愛呢類反經濟的意識形態必然要被推翻

問題係,咁多MK靠戀愛意識做生意

咁樣究竟要消滅佢哋定收編佢哋,

兩樣一齊做,定係兩樣都唔好做,才算是高明手段?

我會利用餘生繼續思索

傳宗接代唔係我的選項,因為教育工作是更好的選項

#7 DutalaTortoise
16/12/21 19:07

望下成堆動漫迷

佢哋有靈魂嗎?

不,牠們只會管自己嘅利益

如果當權者全部都好似動漫迷咁嘅生態

Publix goods會無哂

Social Benefits 全部唔見哂

咁政府呢個角色自然會崩潰

這也許是無良新自由主義資本家樂於見到的

當人民99%財富被呢啲1%畜牲搶哂,

我相信人間將會連地獄都不如,人人變成行屍走肉

#8 DutalaTortoise
16/12/21 19:08

望下成堆動漫迷

佢哋有靈魂嗎?

不,牠們只會管自己嘅利益

如果當權者全部都好似動漫迷咁嘅生態

Publix goods會無哂

Social Benefits 全部唔見哂

咁政府呢個角色自然會崩潰

這也許是無良新自由主義資本家樂於見到的

當人民99%財富被呢啲1%畜牲搶哂,

我相信人間將會連地獄都不如,人人變成行屍走肉

如果當權者全部都好似動漫迷咁嘅心*態

Public* goods會無哂

#9 窮人唔好生仔
16/12/21 23:41

You is pure autism. People in 郊den will not have intrest on you.

#10 窮人唔好生仔
16/12/21 23:43

What you talk is self speak self language without concerning other's felling. That's why people in linden fuck you.

#11 DutalaTortoise
19/12/21 18:29

What you talk is self speak self language without concerning other's felling. That's why people in linden fuck you.

Thanks for your earnest reminder.

#12 DutalaTortoise
19/12/21 18:30

You is pure autism. People in 郊den will not have intrest on you.

No. MK are the most authentic autists.

#13 夢追人
25/12/21 02:08

烏sir好耐無見

見有人問起你特登過嚟呢邊開個account

無咗你少咗好多樂趣

#14 窮人唔好生仔
25/12/21 08:06

烏sir好耐無見

見有人問起你特登過嚟呢邊開個account

無咗你少咗好多樂趣

歡迎蒞臨郊登,我係烏生朋友窮人唔好生仔,多啲黎啦,叫埋連登啲人黎。

#15 重治
25/12/21 11:45

烏SIR又發瘟

見你上面講埋D PURE野

不如整D數你做下 等你知道咩叫PURE啦

1. Find a general expression of Sqrt[i]

2.

(c) Hence, prove that the series {1/n} diverges.

#16 重治
25/12/21 11:46

烏SIR又發瘟

見你上面講埋D PURE野

不如整D數你做下 等你知道咩叫PURE啦

1. Find a general expression of Sqrt[i]

2.

(c) Hence, prove that the series {1/n} diverges.

你要搵梳士一定會搵到

就睇下你有冇學數學果種自我挑戰同追求啦

#17 DutalaTortoise
25/12/21 12:42

烏sir好耐無見

見有人問起你特登過嚟呢邊開個account

無咗你少咗好多樂趣

受寵若驚

其實我CE到AL一條PASTPAPER都無操過

做CE AMATH題目都係靠背題解

所以班主任一早話我無料讀大學

BTW, 連登太多細路,尤其是御理嗰堆

所以我先叫連狗CLOSE ACCOUNT

#18 DutalaTortoise
25/12/21 12:46

z=r(cosx + isinx)=re^ix

=e^(lnr + ix)=e^lnr·e^ix

i=e^[i(pi/2)]=e^[i(2npi+ pi/2)] where n is integer

sqrt(i) = i^0.5 = e^[i(2npi+ pi/2)/2]

#19 DutalaTortoise
25/12/21 12:49

integration 無與趣做

見到substitution 同 by parts 就覺得好撚煩

同玩Gaussian Elimination一樣咁煩

#20 DutalaTortoise
25/12/21 12:54

For the last part,

ln n is continuous for n>0.

The series {1/n} exceeds the upper bound for each ln(n), so the series does not converge.

#21 DutalaTortoise
25/12/21 13:10

integration 無與趣做

見到substitution 同 by parts 就覺得好撚煩

同玩Gaussian Elimination一樣咁煩

睇錯

ln(r/r-1)= lnr - ln (r-1)

ln(1/r-1)= ln 1 - ln2 + ln 2 - ln3 + ... - ln(r-1)

之後懶得再prove

我識嘅gimmick 太少

#22 DutalaTortoise
25/12/21 13:15

1/(r-1)= -[(r - 1) - r]/(r-1)

= r/(1-r) - 1

之後懶得再plug

#23 DutalaTortoise
25/12/21 16:54

烏SIR又發瘟

見你上面講埋D PURE野

不如整D數你做下 等你知道咩叫PURE啦

1. Find a general expression of Sqrt[i]

2.

(c) Hence, prove that the series {1/n} diverges.

你要搵梳士一定會搵到

就睇下你有冇學數學果種自我挑戰同追求啦

e^x > x + 1

equality holds iff x=0

it intersect at (0, 1)

so take the inverse function,

lnx and x-1 intersect at (1,0)

Thought inspired by Integration Bee.

#24 DutalaTortoise
25/12/21 16:55

其實我想睇full solution

由中一至中七做親數都俾人pp-1

#25 DutalaTortoise
25/12/21 17:08

烏SIR又發瘟

見你上面講埋D PURE野

不如整D數你做下 等你知道咩叫PURE啦

1. Find a general expression of Sqrt[i]

2.

(c) Hence, prove that the series {1/n} diverges.

你要搵梳士一定會搵到

就睇下你有冇學數學果種自我挑戰同追求啦

e^x > x + 1

equality holds iff x=0

it intersect at (0, 1)

so take the inverse function,

lnx and x-1 intersect at (1,0)

Thought inspired by Integration Bee.

BTW, e^x 一用Taylor series 爆就知大過 1+x

同理, cos x > 1 - x^2/2

#26 重治
25/12/21 17:09

烏SIR又發瘟

見你上面講埋D PURE野

不如整D數你做下 等你知道咩叫PURE啦

1. Find a general expression of Sqrt[i]

2.

(c) Hence, prove that the series {1/n} diverges.

你要搵梳士一定會搵到

就睇下你有冇學數學果種自我挑戰同追求啦

e^x > x + 1

equality holds iff x=0

it intersect at (0, 1)

so take the inverse function,

lnx and x-1 intersect at (1,0)

Thought inspired by Integration Bee.

OK seems like you make use of Macurin series of e^x, I may accept that

then you use inverse function of each function...

you kind of make use of graphs to interpret your results, which is not sound at this moment, remember this is just pure math syllabus.

and this is where you lose pp marks

If you wanna know the full steps in terms of algebra or pure math approach, try to use differentiation on f(x)=lnx-x+1

#27 重治
25/12/21 17:11

For the last part,

ln n is continuous for n>0.

The series {1/n} exceeds the upper bound for each ln(n), so the series does not converge.

the word continous is wrong here

n is a natural number, which is pointwise

you need to use increasing instead

to be convergent you need to find an upper bound which is independent of n

#28 重治
25/12/21 17:14

z=r(cosx + isinx)=re^ix

=e^(lnr + ix)=e^lnr·e^ix

i=e^[i(pi/2)]=e^[i(2npi+ pi/2)] where n is integer

sqrt(i) = i^0.5 = e^[i(2npi+ pi/2)/2]

#29 夢追人
25/12/21 21:20

z=r(cosx + isinx)=re^ix

=e^(lnr + ix)=e^lnr·e^ix

i=e^[i(pi/2)]=e^[i(2npi+ pi/2)] where n is integer

sqrt(i) = i^0.5 = e^[i(2npi+ pi/2)/2]

烏air仲拎住我唔知14定15年喺uwants比你個答案

#30 夢追人
25/12/21 21:24

烏sir好耐無見

見有人問起你特登過嚟呢邊開個account

無咗你少咗好多樂趣

歡迎蒞臨郊登,我係烏生朋友窮人唔好生仔,多啲黎啦,叫埋連登啲人黎。

其實我唔係跟得好貼

膠登係點解變咗做郊登

定係唔同嘢嚟 但膠登又好似無死到 有啲亂

但呢邊又唔係用膠登account

#31 DutalaTortoise
25/12/21 21:55

z=r(cosx + isinx)=re^ix

=e^(lnr + ix)=e^lnr·e^ix

i=e^[i(pi/2)]=e^[i(2npi+ pi/2)] where n is integer

sqrt(i) = i^0.5 = e^[i(2npi+ pi/2)/2]

烏air仲拎住我唔知14定15年喺uwants比你個答案

無印象

個uwants accounts俾人ban撚埋

#32 DutalaTortoise
25/12/21 21:59

烏sir好耐無見

見有人問起你特登過嚟呢邊開個account

無咗你少咗好多樂趣

歡迎蒞臨郊登,我係烏生朋友窮人唔好生仔,多啲黎啦,叫埋連登啲人黎。

其實我唔係跟得好貼

膠登係點解變咗做郊登

定係唔同嘢嚟 但膠登又好似無死到 有啲亂

但呢邊又唔係用膠登account

香港郊登 hkGalden,又名郊登討論區,簡稱「郊登」,是一個從膠登討論區(下稱「膠登」)分裂出來,目前為獨立於並繼承後者的討論區。郊登的系統與其他登不相似,例如以50回覆為一頁而非25回覆,沒有1001上限,頻道內包含多個標籤等。由於目前膠登充斥著影印post和舊post,再加上鞭屍情況猖獗,被戲稱為「墳場」。而大部份膠登會員已經轉場至郊登,因此郊登亦被稱為「墳場後花園」。

2018年9月1日,前膠登admin hydrogen早前曾表示會建立方舟取代無人管治的膠登討論區,該日討論區正式完工啟用。膠登用戶可直接使用膠登帳號登入郊登。

無人更新

#33 DutalaTortoise
25/12/21 22:02

For the last part,

ln n is continuous for n>0.

The series {1/n} exceeds the upper bound for each ln(n), so the series does not converge.

the word continous is wrong here

n is a natural number, which is pointwise

you need to use increasing instead

to be convergent you need to find an upper bound which is independent of n

唔識🙈

#34 重治
26/12/21 09:16

For the last part,

ln n is continuous for n>0.

The series {1/n} exceeds the upper bound for each ln(n), so the series does not converge.

the word continous is wrong here

n is a natural number, which is pointwise

you need to use increasing instead

to be convergent you need to find an upper bound which is independent of n

唔識🙈

continuous is for function / curve, with an interval containing all values of x

series is only point by point, n=1, 2, 3, ..., you will not say n=1.2

so we call this as pointwise, not continuous

#35 DutalaTortoise
26/12/21 11:18

the word continous is wrong here

n is a natural number, which is pointwise

you need to use increasing instead

to be convergent you need to find an upper bound which is independent of n

唔識🙈

continuous is for function / curve, with an interval containing all values of x

series is only point by point, n=1, 2, 3, ..., you will not say n=1.2

so we call this as pointwise, not continuous

我係問點hence prove that the series {1/n} diverges, AL 唔俾用 ratio test🙈

#36 DutalaTortoise
26/12/21 11:30

唔識🙈

continuous is for function / curve, with an interval containing all values of x

series is only point by point, n=1, 2, 3, ..., you will not say n=1.2

so we call this as pointwise, not continuous

我係問點hence prove that the series {1/n} diverges, AL 唔俾用 ratio test🙈

係咪就咁話lnx 無 upper bound and strictly increasing

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