開始忙了

今日好充實，朝早上 (訓) 完 MATH151 再去再聽 eating style (HL001)，然後食午餐再備課備到點半上 MATH204。嚴民 d 堂講得好快，又易明，可惜一個星期得兩堂，如果一星期有四堂我都會肯去。落左堂之後同 204 同學做功課同備課去到 5:45，六點去 MATH204 tutorial lesson。

但奇怪嚴民教授明明 o係 course webpage 寫住

(5/2/2010) Tutorial times set at Tuesday, 18:00 - 18:50, in room 3215.

好在當時 save 左 ust 搵房網址入電腦 (出發時先醒起 3215 房間)，點我，慳左好多功夫，但原來 3215 係一個錯 ge 地址 (3215 係 TA 房)，有 4 至 5 個都走錯左地方。搵房果陣有個同班的人搵我講野...，o係果時先知道原來成日同班內地生傾計果個人都識講廣東話... (堔圳學生)，俾佢呃左好耐 = =，搞到我一直都唔敢搵佢講野。最後大家都要返返去 3584 上堂。

今次 tutorial 堂好奇怪 (奇怪在變返正常)，真係有個 PG 黎做 tutor...，平時都係教授一手包辦埋 tutorial 堂。老實，有少少失望。

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I have come arcoss a question in uwants, prove that $\tan 1^{\circ}$ is irrational, and here is the discussion.
I suppose that $\tan 1^{\circ}$ is rational, then by the identity
$\displaystyle \tan (30 \theta)=\frac{\sum_{k=1}^{15}(-1)^{k+1}\binom{30}{2k-1}\tan^{2k-1}\theta}{\sum_{k=0}^{15}(-1)^k\binom{30}{2k}\tan^{2k}\theta}$

We have by putting $\theta = 1^{\circ}$, $\displaystyle \frac{1}{\sqrt{3}} =\tan (30^{\circ})=\frac{\sum_{k=1}^{15}(-1)^{k+1}\binom{30}{2k-1}\tan^{2k-1}1^{\circ}}{\sum_{k=0}^{15}(-1)^k\binom{30}{2k}\tan^{2k}1^{\circ}}$, this shows that $\displaystyle\frac{1}{\sqrt{3}}$ is rational, a contradiction. I hope this is one of the way that apook (koopa) thinks of.