2023 DSE Maths 終於過左喇,如果大家想睇下答案都可以睇下以下既 suggested solutions 架!
當然唔少得 M1 同埋 M2 啦!
港大數學系 | 專補 DSE 數學(Core / M1 / M2)| 2018 DSE 超過 40% 學生考取 5* 或以上,超過 65% 學生考取 5 或以上(只計算出席超過四堂常規課程的學生)
2023 DSE Maths 終於過左喇,如果大家想睇下答案都可以睇下以下既 suggested solutions 架!
由於小卒好似暫時上唔到,所以我 post 埋之前打過嘅 DSE Suggested Solutions 黎呢邊等同學溫書啦~
Maths Core
2021 DSE Paper 1 2021 DSE Paper 2
2020 DSE Paper 1 2020 DSE Paper 2
2019 DSE Paper 1 2019 DSE Paper 2
2018 DSE Paper 1 2018 DSE Paper 2
M1
M2
不如我地試下做少少 experiment 先,求其寫兩個 positive integers,睇下佢地嘅 greatest common divisor(gcd,或者香港學生見開既 H.C.F.)係 1 嘅 probability,比如抽到 $21$ 同 $34$ 就係屬 favourable outcome,抽到 $21$ 同 $35$ 嘅話就唔係一個 favourable outcome。
由於電腦出 random integers 未能做到冇 upper bound 嘅情況,所以我退而求其次,去睇下唔同 upper bound $M$ 嘅結果會係點啦!
$M$ |
$100$ |
$10000$ |
$1000000$ |
The total number of pairs ($N$) |
$100000$ |
$100000$ |
$100000$ |
The number of pairs whose gcd is $1$ ($n$) |
$60949$ |
$60894$ |
$60647$ |
The probability of getting a coprime pair
($P=\frac{n}{N}$) |
$0.60949$ |
$0.60894$ |
$0.60647$ |
$\sqrt{\frac{6}{P}}$ |
$3.137562119$ |
$3.138978736$ |
$3.145364377$ |
不如我地將呢個 process 各重覆 $30$ 次,睇下會點~eh... 咁熟口面嘅,好似同我地熟悉既 π 好似喎!
$M$ |
$100$ |
$10000$ |
$1000000$ |
Sample mean of $\sqrt{\frac{6}{P}}$ |
$3.139835545$ |
$3.141452666$ |
$3.142048756$ |
Sample standard deviation of $\sqrt{\frac{6}{P}}$ |
$0.004702442$ |
$0.003096638$ |
$0.003385599$ |
哇,真係好似喎!因為宜家每組有 $30$ 個 data,我地可以 assume 佢地係 normal,咁樣只有 $M=100$ 嘅 case 計到嘅 sample mean 係 significant smaller than $\pi$。
究竟點解呢個數字會咁接近 $\pi$ 呢?
我地可以睇下呢個 probability 可以點樣計出黎先~
首先,我地搵兩個數嘅 gcd 首先會搵左呢兩個數嘅 prime factorization(即小學教嘅質因數連乘式),點樣做到兩個數嘅 gcd 係 1?就係佢地唔會同時係 2 嘅倍數、唔會同時係 3 嘅倍數、唔會同時係 5 嘅倍數、唔會同時係 7 嘅倍數…… 即佢地唔會同時係同一個質數嘅倍數;而因為質數 $p$ 嘅倍數係每 $p$ 就會有一個,所以一個數字係質數 $p$ 嘅倍數嘅 probability 就會係 $\frac{1}{p}$,因此兩個數都唔係質數 $p$ 嘅倍數嘅 probability 就會係 $\frac{1}{p^{2}}$;考慮哂個咁多個 prime numbers,就會得到 $$P=\prod_{p\textbf{ prime}}\left(1-\frac{1}{p^{2}}\right)= \left(1-\frac{1}{2^{2}}\right)\times\left(1-\frac{1}{3^{2}}\right)\times\left(1-\frac{1}{5^{2}}\right)\times\cdots。$$
跟住我地就可以嘗試爆開上面,因為 $0<\frac{1}{p^{2}}<1$,根據 GS sum to infinity,我地知道 $$\frac{1}{1-\frac{1}{p^{2}}} = \sum_{k=0}^{\infty}\frac{1}{p^{2k}}=1+\frac{1}{p^{2}}+\frac{1}{p^{4}}+\frac{1}{p^{6}}+\cdots,$$ 所以我地可以將上面個 product 寫成 $$\frac{1}{P}=\prod_{p\textbf{ prime}}\left(\sum_{k=0}^{\infty}\frac{1}{p^{2k}}\right) = \left(1+\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}\right) \times \left(1+\frac{1}{3^{2}}+\frac{1}{3^{4}}+\frac{1}{3^{6}}\right) \times \left(1+\frac{1}{5^{2}}+\frac{1}{5^{4}}+\frac{1}{5^{6}}\right) \times \cdots$$
因為我地小學已經學左嘅 Fundamental Theorem of Arithmetic,即每個整數都可以寫成一個 unique 嘅 prime factorization,我地就會見到其實爆開出黎嘅 terms 都係 $\frac{1}{m^{2}}$,而 $m$ 係任何 integers 嘅樣,而且唔會重覆。比如因為 $12=2^{2}\times 3$,所以 $\frac{1}{12^{2}}$ 就可以由第一個 bracket 揀 $\frac{1}{2^{4}}$、第二個 bracket 揀 $\frac{1}{3^{2}}$ 同埋剩餘嘅 brackets 都揀 $1$ 得出,所以 $$\frac{1}{P}=\sum_{k=1}^{\infty} \frac{1}{k^{2}}=1^{2}+2^{2}+3^{2}+\cdots。$$
1735 年,數學家 Euler 發現左 $$\sum_{k=1}^{\infty} \frac{1}{k^{2}}=1^{2}+2^{2}+3^{2}+\cdots=\frac{\pi^{2}}{6},$$ 而呢條式就可以幫我地計到 $$P=\frac{6}{\pi^{2}},$$ 亦因此我地就知道 $$ \pi=\sqrt{\frac{6}{P}} $$ 喇!
如果想知上面條式點黎,我可以下次打番架!
當然同學都可以諗下點解當 $M$ 係細數嘅時候,點解 estimate 出黎嘅 $\pi$ 通常會細過 actual value?
人數
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11月15日或之前 Early Bird 八折優惠
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正價
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不購買 2019
DSE Mock Exam Set
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購買 2019
DSE Mock Exam Set
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不購買2019
DSE Mock Exam Set
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購買 2019
DSE Mock Exam Set
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1
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$232
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$320
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$290
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$400
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2
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$216
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$304
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$270
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$380
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3
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$200
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$280
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$250
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$350
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4
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$184
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$256
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$230
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$320
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5
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6 – 10
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> 20
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