[9up數學] 五次方程沒有解?

出文未

中學背晒啲式，sub數計架咋麻，都唔洗知佢點解既，洗唔洗咁串

search root可以用calculus
mean value theorem+Rolle‘s Theorem

唔洗知點解? 似係你本人問題

我好奇佢讀咩

講錯佐
係應該用IVT

我唔係math major不過都鍾意睇吓math

個topic係想搵一個closed form而唔係approximate佢

我頭先只係睇佐post title 下意識留言

咁真係喎，中學冇教過條式點derive出黎，淨係叫你點solve equations,學完都1999咁

推 有心

連登學術post通病，要求每個參與者都好熟悉嗰科，唔容許"stupid question"

3. Roots的洗牌機

證明五次方程無一般的radical solution的其中一個key idea係: 將d roots洗牌

但係我地唔係求其亂咁洗, 有幾樣野希望保留:

(a) 我地淨係想將D roots洗牌, 所以未加料之前的野我地唔希望郁到佢
(b) 我地唔希望破壞加左料之後的field的結構
(c) 既然係洗牌即係我地能夠將洗牌的步驟倒返轉頭

每一種洗牌的方法我地都可以用一個function黎形容, 用嚴謹的寫法將上面的條件重寫一次我地有:

If E is a field extension of F, and f: E -> E be a function shuffling the roots according to the rule above,
(a) f(x) = x for any x in F.
(b) f(xy) = f(x)f(y) and f(x + y) = f(x) + f(y) for any x, y in E.
(c1) for any x, y in E, f(x) = f(y) implies x = y.
(c2) given any y in E, there is x in E such that f(x) = y.

Remark: 如果有學過abstract algebra的話, 如果有個function f: E -> E satisfies (a) & (b), 一係 f 符合 (c1), 一係 f(x) = 0 for all x in E.

而根據呢四條規則就知道,

(i) f(0) = 0
因為 f(0) = f(0 + 0) = f(0) + f(0), 兩邊減去f(0)得出 f(0) = 0
(ii) f(1) = 1
因為 f(1) = f(1^2) = [f(1)]^2, 所以 f(1) = 1 or 0, 但係(c1) rule out 左f(1) = 0呢個可能
(iii) f(-x) = -f(x)
(iv) f(1/x) = 1/f(x) if x ≠ 0

Given一個加左root的field, 我地係可以計到有d咩function(s)符合以上的特性

(a) Q(sqrt{2}) over Q:
Recall Q(sqrt{2}) 係所有 a + b sqrt{2} 咁ge樣的數字 (a, b係rational), 根據 (b)

f(a + b sqrt(2)) = f(a) + f(b sqrt(2)) = f(a) + f(b) f(sqrt(2))

根據(a),

f(a) + f(b) f(sqrt(2)) = a + b f(sqrt(2))

由呢條式我地知道 f 係完全由 f(sqrt(2)) 的value決定

由於我地先前講過我地想 f 將某個polynomial的root洗牌, 簡單計算知道 lowest degree polynomial in Q[x] with leading coefficient 1 which has sqrt{2} as a root係 x^2 - 2, 呢個polynomial的另一個root係 -sqrt{2}, 所以 f(sqrt{2}) 有兩個choices:

(i) f(sqrt{2}) = sqrt{2}, 呢個case就會變左 f(x) = x for all x in Q(sqrt{2}) (呢種map我地叫identity map)
(ii) f(sqrt{2}) = -sqrt{2}, 呢個case就會有 f(a + b sqrt(2)) = a - b f(sqrt(2))

問題6: 試驗證以上兩個 f 都符合 (a), (b), (c1), (c2).

睇完(a)個例子之後, 大家會發覺, 呢D functions係咪完全由加入去的料的value under f黎決定? 答案係係的, 所以之後的例子我地就可以捉住呢D醋同鹽黎玩

(b) Q(sqrt{2}, sqrt{3}) over Q:
經過計算, 包含哂sqrt{2}, sqrt{3}做roots的最低次的polynomial係 (x^2 - 2)(x^2 - 3), 呢個polynomial另外有兩個roots: -sqrt{2}, -sqrt{3}

見到有四個唔同的數字, 如果讀緊(完) DSE core的你, 諗下有幾多種唔同的permutations? 心水清就知係 4P4 = 4! = 24, 但係實際上符上 (a), (b), (c1), (c2) 要求的得4個:

(i) f(sqrt{2}) = sqrt{2}, f(sqrt{3}) = sqrt{3}
(ii) f(sqrt{2}) = -sqrt{2}, f(sqrt{3}) = sqrt{3}
(iii) f(sqrt{2}) = sqrt{2}, f(sqrt{3}) = -sqrt{3}
(iv) f(sqrt{2}) = -sqrt{2}, f(sqrt{3}) = -sqrt{3}

你可能會問, 點解唔可以 f(sqrt{2}) = sqrt{3}?
答案係, square both sides得出

[f(sqrt{2})]^2 = (sqrt{3})^2
f((sqrt{2})^2) = 3
.f(2) = 3
2 = 3

Impossible!
f(sqrt{2}) = -sqrt{3} 同理
==============================================
(待續)

上面有人話太多名詞所以我改緊篇野

Contents:
#1 - Part 1
#9 - Part 2
#115 - Part 3

Errata:
#6: Ignore this

#9: "例如R同Q的algebraic closure係C."
-> "例如R的algebraic closure係C."

咁樣會唔會易睇D

Contents:
#1 - Part 1
#9 - Part 2
#115 - Part 3

Errata:
#6: Ignore this

#9: "例如R同Q的algebraic closure係C."
-> "例如R的algebraic closure係C."

#115: "Remark: 如果有學過abstract algebra的話, 如果有個function f: E -> E satisfies (a) & (b), 一係 f 符合 (c1), 一係 f(x) = 0 for all x in E."
-> "Remark: 如果有學過abstract algebra的話, 如果有個function f: E -> E satisfies (b), 一係 f 符合 (c1), 一係 f(x) = 0 for all x in E."

聽講過
想了解多啲
lm

數學界有好多天才，個人認為當中最天才嘅天才係拉馬努金同黎曼，可惜天妒英才，兩個都活唔過40歲。如果俾佢地活多30年，數學發展至少快100年

好難比 Ramanujan只係掟左一堆野出黎
Galois係直頭創造左現代純數學的五大分支之一

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