我從夢中醒來
眼睛一睜開,發現自己在競技場裡面
即便不曾受過訓練,本能告訴我,活到最後的才能離開
那些來不及反應的可憐蟲已經堆了滿地
我踩在別人的屍體上面,好證明我是那唯一有資格活下去的
最後我發現,原來
將我們關在競技場裡的,是我們自己
那競技場之門未曾闔上
2014年12月8日 星期一
2014年6月12日 星期四
為科學傳統之辯護
科學傳統
基於觀測事實,我們提出理論來詮釋,而此觀測事實的標準在於:用最少的假設來解釋最多的現象
有人會對此科學傳統批判,認為知識體系是由意識形態和權力來支配
這或許是真的,因為科學的詮釋是人為的,沒有人能夠避免受到社會影響
但是真的科學只是如此嗎?
舉個例子,當初希格斯粒子被用於解釋弱交互作用,同時也預言了一種新的粒子-Z玻色子
你可以說當初的競爭理論多如牛毛,如果這樣的一個理論只是純粹掌權者所建構的
那麼尚未被預測,完全沒人想像過的Z玻色子是如何被預測?
同樣的故事在過去不段的發生,可能原本大家公認某理論正確,但是在實驗的驗證下
某不知名理論所預測的完全正確,且符合簡單原則,最終這不為人知的理論成功
我們永遠都要留心意識形態和權力所支配的知識結構,但別忘記在科學我們永遠都有一個相對客觀的仲裁者:實驗,可以被不斷重複驗證的實驗(永無止盡的實驗)
在實驗做出最後的仲裁之前,大家可以爭論不休,可以任意來操作尚未被確認的知識體系
一旦實驗做出仲裁,那麼結果就再明顯也不過,就算你是教宗,是總統,或是愛因斯坦
沒有可以違背此仲裁
這種科學傳統已經使用超過300年了,到目前為止仍然算是成功,至少在整個物質文明的進步上功不可沒,你可以說那不過是另外一種權威體系,沒錯!藉由前人所訂下的規則來行事
但是,你現在有電腦可用,有手機可以用,都是出自這權威體系,我沒有說這科學傳統是真理,也許哪天它會失效,如果你坐在電腦前仍然不相信這樣的科學傳統,那是你的自由,對我們這些科學家而言,到目前為止都很成功,所以在出問題之前我們都繼續使用!
所以學校教育著重在灌輸上我不能認同,因為科學傳統不一定是真理
我個人傾向於,我告訴你它到目前為止多成功,至於你願不願意相信,那是你的自由
2014年5月8日 星期四
2013年10月13日星期日 文章後繼
又一個人被我列入拒絕討論黑名單裡面了
我過去曾經讚賞過的人
在2013年10月13日星期日文章中提到過
信念歸信念
信念不能被證明
所以你跟我說你的信念是什麼就是什麼
這沒什麼
當你企圖把信念放在理性層次討論時
你就必然的要接受批判檢視
這是理性層面的優點,同時也是缺點
當你把所有不利於你論點的可能性都刪除時
你不會得到知識或真理
你只會的到自己永遠是對的結論
當討論雙方無法達成共識,有可能是因為前提或推論過程沒有交集
那討論下去沒有意義,要試著回頭看沒有交集之處
或者直接回到信念層次
當被扣上某種帽子,因為你的智商,你的意識形態時
完全沒有繼續討論下去的意義
因為他們只是要給自己台階下
"因為你是白癡,所以你講的都是錯的,我贏了"
"因為你是某黨的人,所以你說的話都不可信"
你們想要什麼?
想證明自己很聰明永遠不會錯?
想證明自己很有智慧?
你們想要就給你們吧
我不在乎那些東西
我只想知道那些是對的那些是假的
我過去曾經讚賞過的人
在2013年10月13日星期日文章中提到過
信念歸信念
信念不能被證明
所以你跟我說你的信念是什麼就是什麼
這沒什麼
當你企圖把信念放在理性層次討論時
你就必然的要接受批判檢視
這是理性層面的優點,同時也是缺點
當你把所有不利於你論點的可能性都刪除時
你不會得到知識或真理
你只會的到自己永遠是對的結論
當討論雙方無法達成共識,有可能是因為前提或推論過程沒有交集
那討論下去沒有意義,要試著回頭看沒有交集之處
或者直接回到信念層次
當被扣上某種帽子,因為你的智商,你的意識形態時
完全沒有繼續討論下去的意義
因為他們只是要給自己台階下
"因為你是白癡,所以你講的都是錯的,我贏了"
"因為你是某黨的人,所以你說的話都不可信"
你們想要什麼?
想證明自己很聰明永遠不會錯?
想證明自己很有智慧?
你們想要就給你們吧
我不在乎那些東西
我只想知道那些是對的那些是假的
給科學家的一封信
什麼才是成為優秀科學家的條件?
數學能力?紀律?智商?
在著名的科學家中
有些人數學及強,有些人極差
有些人一絲不苟,有些人生活糜爛
但是都有一個特點
如果他們都覺得某個領域太過無聊
那麼他們根本就不會投注時間在那領域,更何況有重大發現
所以,答案很明顯了
就是因為我們笨,我們好奇,所以我們是科學家
數學能力?紀律?智商?
在著名的科學家中
有些人數學及強,有些人極差
有些人一絲不苟,有些人生活糜爛
但是都有一個特點
如果他們都覺得某個領域太過無聊
那麼他們根本就不會投注時間在那領域,更何況有重大發現
所以,答案很明顯了
就是因為我們笨,我們好奇,所以我們是科學家
2014年2月14日 星期五
今天是希爾伯特的忌日
特此紀念
Hilbert's twenty-three problems are:
| Problem | Brief explanation | Status | Year Solved |
|---|---|---|---|
| 1st | The continuum hypothesis (that is, there is no set whose cardinality is strictly between that of the integers and that of the real numbers) | Resolved. Proven to be impossible to prove or disprove within the Zermelo–Fraenkel set theory with or without the Axiom of Choice (provided the Zermelo–Fraenkel set theory with or without the Axiom of Choice is consistent, i.e., contains no two theorems such that one is a negation of the other). There is general consensus that this solves the problem, although there have been proposals which would give a definitive truth value (see Ω-logic). | 1963 |
| 2nd | Prove that the axioms of arithmetic are consistent. | There is no consensus on whether results of Gödel and Gentzen give a solution to the problem as stated by Hilbert. Gödel'ssecond incompleteness theorem, proved in 1931, shows that no proof of its consistency can be carried out within arithmetic itself. Gentzen proved in 1936 that the consistency of arithmetic follows from the well-foundedness of the ordinal ε₀. | 1936? |
| 3rd | Given any two polyhedra of equal volume, is it always possible to cut the first into finitely many polyhedral pieces which can be reassembled to yield the second? | Resolved. Result: no, proved using Dehn invariants. | 1900 |
| 4th | Construct all metrics where lines are geodesics. | Too vague to be stated resolved or not.[n 1] | – |
| 5th | Are continuous groups automatically differential groups? | Resolved by Andrew Gleason, depending on how the original statement is interpreted. If, however, it is understood as an equivalent of the Hilbert–Smith conjecture, it is still unsolved. | 1953? |
| 6th | Mathematical treatment of the axioms of physics | Partially resolved depending on how the original statement is interpreted.[13] In particular, in a further explanation Hilbert proposed two specific problems: (i) axiomatic treatment of probability with limit theorems for foundation of statistical physics and (ii) the rigorous theory of limiting processes "which lead from the atomistic view to the laws of motion of continua". Kolmogorov’s axiomatics (1933) is now accepted as standard. There is some success on the way from the "atomistic view to the laws of motion of continua".[14] | 1933-2002? |
| 7th | Is a b transcendental, for algebraic a ≠ 0,1 and irrational algebraic b ? | Resolved. Result: yes, illustrated by Gelfond's theorem or the Gelfond–Schneider theorem. | 1935 |
| 8th | The Riemann hypothesis ("the real part of any non-trivial zero of the Riemann zeta function is ½") and other prime number problems, among them Goldbach's conjecture and the twin prime conjecture | Unresolved. | – |
| 9th | Find the most general law of the reciprocity theorem in any algebraic number field. | Partially resolved.[n 2] | – |
| 10th | Find an algorithm to determine whether a given polynomial Diophantine equation with integer coefficients has an integer solution. | Resolved. Result: impossible, Matiyasevich's theorem implies that there is no such algorithm. | 1970 |
| 11th | Solving quadratic forms with algebraic numerical coefficients. | Partially resolved.[citation needed] | – |
| 12th | Extend the Kronecker–Weber theorem on abelian extensions of the rational numbers to any base number field. | Unresolved. | – |
| 13th | Solve 7-th degree equation using continuous functions of two parameters. | The problem was partially solved by Vladimir Arnold based on work by Andrei Kolmogorov. [n 4] | 1957 |
| 14th | Is the ring of invariants of an algebraic group acting on a polynomial ring always finitely generated? | Resolved. Result: no, counterexample was constructed by Masayoshi Nagata. | 1959 |
| 15th | Rigorous foundation of Schubert's enumerative calculus. | Partially resolved.[citation needed] | – |
| 16th | Describe relative positions of ovals originating from a real algebraic curve and as limit cycles of a polynomial vector field on the plane. | Unresolved. | – |
| 17th | Express a nonnegative rational function as quotient of sums of squares. | Resolved. Result: yes, due to Emil Artin. Moreover, an upper limit was established for the number of square terms necessary.[citation needed] | 1927 |
| 18th | (a) Is there a polyhedron which admits only an anisohedral tiling in three dimensions? (b) What is the densest sphere packing? | (a) Resolved. Result: yes (by Karl Reinhardt). (b) Widely believed to be resolved, by computer-assisted proof (by Thomas Callister Hales). Result: Highest density achieved by close packings, each with density approximately 74%, such as cubic close packing and hexagonal close packing.[n 5][citation needed] | (a) 1928 (b) 1998 |
| 19th | Are the solutions of regular problems in the calculus of variations always necessarily analytic? | Resolved. Result: yes, proven by Ennio de Giorgi and, independently and using different methods, by John Forbes Nash. | 1957 |
| 20th | Do all variational problems with certain boundary conditions have solutions? | Resolved. A significant topic of research throughout the 20th century, culminating in solutions[citation needed] for the non-linear case. | ? |
| 21st | Proof of the existence of linear differential equations having a prescribed monodromic group | Resolved. Result: Yes or no, depending on more exact formulations of the problem.[citation needed] | ? |
| 22nd | Uniformization of analytic relations by means of automorphic functions | Resolved.[citation needed] | ? |
| 23rd | Further development of the calculus of variations | Unresolved. | – |
2014年2月1日 星期六
2014年1月13日 星期一
Paul Mccartney
Maybe I'm Amazed
Baby I'm amazed at the way you love me all the time
Maybe I'm afraid of the way I love you
Baby I'm amazed at the the way you pulled me out of time
Hung me on a line
Maybe I'm amazed at the way I really need you
Baby I'm a man and maybe I'm a lonely man
Who's in the middle of something
That he dosen't really understand
Babe I'm a man and maybe you're the only woman
Who could ever help me
Baby won't you help to me understand
Baby I'm a man and maybe I'm a lonely man
Who's in the middle of something
That he dosen't really understand
Babe I'm a man and maybe you're the only woman
Who could ever help me
Baby won't you help me understand
Baby I'm amazed at the way you're with me all the time
Maybe I'm afraid of the way I leave you
Baby I'm amazed at the way you help me sing my song
You right me when I'm wrong
Maybe I'm amazed at the way I really need you
Baby I'm amazed at the way you love me all the time
Maybe I'm afraid of the way I love you
Baby I'm amazed at the the way you pulled me out of time
Hung me on a line
Maybe I'm amazed at the way I really need you
Baby I'm a man and maybe I'm a lonely man
Who's in the middle of something
That he dosen't really understand
Babe I'm a man and maybe you're the only woman
Who could ever help me
Baby won't you help to me understand
Baby I'm a man and maybe I'm a lonely man
Who's in the middle of something
That he dosen't really understand
Babe I'm a man and maybe you're the only woman
Who could ever help me
Baby won't you help me understand
Baby I'm amazed at the way you're with me all the time
Maybe I'm afraid of the way I leave you
Baby I'm amazed at the way you help me sing my song
You right me when I'm wrong
Maybe I'm amazed at the way I really need you
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