戒菸

2013年2月1號 中午12點
人生第二還第三次戒菸
第一次在一年多前被小英國壓著走到戒菸診所
莫名其妙拿了藥
很有用 但失敗
這一次心裡想著
也差不多了吧勻勻
到現在一個多月了 中間抽了幾根但沒影響戒菸的決心

當初開始抽煙想著的就是 戒菸的毅力我應該有吧？
過了四年
這毅力似乎來了
但做研究的毅力似乎比這更難
難怪成功的幾位教授 聽說以前論文的一個算式都要幾百包菸才能成就

在當兵之後
每天拉肚子 輕重都有
體力變差
呼吸變很不順
很容易累
怎麼運動 體重還是不斷上升
唱歌的主題曲從外套變睡覺
毛病不斷
都以為是老了 或 當兵讓身體變差

在戒菸一個月後發現
馬的
這些狀況都沒了 或 慢慢恢復成大學時候的情況

看了有機會減肥成功囉！

Proof that e^i(pi) = -1

Proof that e^i(pi) = -1 Date: 06/02/99 at 00:48:10 From: Nathan White Subject: e^[i(pi)] = -1 How can it be proven that e^[i(pi)] = -1? And why does it matter? Date: 06/02/99 at 06:28:07 From: Doctor Mitteldorf Subject: Re: e^[i(pi)] = -1 Dear Nathan, When you do algebraic manipulations, you want to be sure that everything that you do legally "makes sense," which means that equations should have solutions, so you're not just casting around in vain. You can easily write real equations that don't have real solutions. But if you expand to complex numbers, then all algebraic equations become solvable. How then do you make sense out of complex numbers in the exponent? The Rosetta stone is the Euler equation: e^(ix) = cos(x) + i sin(x) You can verify that this equation makes a sensible definition by expanding the two sides as Taylor series in x. You can also differentiate both sides and see that the answer is self-consistent. Thirdly, you can use the formula for cos(2x) and sin(2x) to show that the right side has the property you expect from an exponential, so that e^[i(2x)] = [e^(ix)]^2. Once you believe the Euler equation, you have a pretty easy time of it. If you let x = pi/2, then the real part is zero e^[i(pi)] = cos(pi) + i sin(pi) = -1 + 0*i If you are happy to accept integration of complex functions, a quick and reasonably rigorous proof follows (from Dr. Anthony's answer in the Dr. Math archives): Euler Equation http://mathforum.org/dr.math/problems/dale9.13.97.html Let z = cos(x) + i.sin(x) dz/dx = -sin(x) + i.cos(x) = i[cos(x) + i.sin(x)] dz/dx = i.z so dz/z = i.dx INT[dz/z] = INT[i.dx] ln(z) = i.x + const z = e^(i.x + const) Now from z = cos(x) + i.sin(x) when x=0, z=1 so the constant = 0 Thus z = e^(i.x) That is cos(x) + i.sin(x) = e^(i.x) When x = pi -1 = e^(i.pi) 0 = e^(i.pi) + 1 or e^(i.pi) + 1 = 0 This equation combines the 'famous five' most important numbers in mathematics, 0, 1, e, i, pi. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/

pi

The Monty Hall Problem

https://www.youtube.com/watch?feature=player_embedded&v=mhlc7peGlGg#!

so tired recently

Staying in the library all day, I am so tired, tiered inside. 

Right now, I am just back in my dorm and listening to Radiohead's house of card, and I have a sense of fullness in my soul. Probably, that is because I just talked to my sister on the phone. 

Anyway, life is still on its own way, you can choose to stay on the same spot or walk through that. 

GRE

Finally, I finished C.....

機率問題

Q: 有一試劑對某疾病有九成五的正確性，此疾病每千人有一人感染。
     你測得的結果為陽性，你得病的機率為多少？

A: 2%（若測一千人，有五十人會是陽性，而只有一人得此病。）

為什麼測一千人會有五十人是陽性？