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林太史
依家發現其實物理學入面係冇咩真正意義上既infinite series
物理學既fourier 絕大部分都係continuous
Taylor series 通常淨係攞頭一兩個term
QFT 既perturbations 多數都係asymptotic ，開頭好似converge，實際就diverge
物理學既fourier 絕大部分都係continuous
Taylor series 通常淨係攞頭一兩個term
QFT 既perturbations 多數都係asymptotic ，開頭好似converge，實際就diverge
Edelschwarz
咁多位我返嚟喇
喺繼續講theorem之前，我地要介紹吓一種好乖嘅sequence先
佢個名就叫單調數列(monotone sequence)
咁monotone sequence就分兩種嘅
- increasing: 數列入面每一個term都不小於前一個term
- decreasing: 數列入面每一個term都不大於前一個term
如果數列每一個term都大過前一個term，我地叫佢做strictly increasing;
咁大家應該知咩係strictly decreasing
點解話佢地好乖呢？因為佢地只可以向一個方向行，冇得彈入又彈出，所以佢地好容易就converge到喇
有幾容易? Monotone convergence theorem答到你:
大家可以想像吓，啲點只可以向一個方向行，但又唔畀過界，咁會converge都係好自然嘅事
雖然睇落好似好有道理，但係我地都係要證一次先安樂
我地依家假設{an}係increasing, bounded above
咁completeness axiom話畀我地聽sup{an}存在
叫住佢做L先
咁因為L係最細嘅upper bound，所以無論ϵ>0幾細，L-ϵ都唔係一個upper bound, 即係有一個ak>L-ϵ
但係我地知道{an}係increasing，所以ak之後嘅term都夾喺L-ϵ同L之間
所以{an}就converge去L喇
第二part可以用同一個方法證，所以too long didn't write
:^(
喺繼續講theorem之前，我地要介紹吓一種好乖嘅sequence先
:^(
佢個名就叫單調數列(monotone sequence)
咁monotone sequence就分兩種嘅
- increasing: 數列入面每一個term都不小於前一個term
- decreasing: 數列入面每一個term都不大於前一個term
如果數列每一個term都大過前一個term，我地叫佢做strictly increasing;
咁大家應該知咩係strictly decreasing
:^(
點解話佢地好乖呢？因為佢地只可以向一個方向行，冇得彈入又彈出，所以佢地好容易就converge到喇
有幾容易? Monotone convergence theorem答到你:
:^(
大家可以想像吓，啲點只可以向一個方向行，但又唔畀過界，咁會converge都係好自然嘅事
:^(
雖然睇落好似好有道理，但係我地都係要證一次先安樂
:^(
我地依家假設{an}係increasing, bounded above
咁completeness axiom話畀我地聽sup{an}存在
:^(
叫住佢做L先
咁因為L係最細嘅upper bound，所以無論ϵ>0幾細，L-ϵ都唔係一個upper bound, 即係有一個ak>L-ϵ
但係我地知道{an}係increasing，所以ak之後嘅term都夾喺L-ϵ同L之間
所以{an}就converge去L喇
第二part可以用同一個方法證，所以too long didn't write
Edelschwarz
講完monotone sequence，我地就講吓subsequence啦
讀過analysis嘅巴打一定知我想講咩theorem
當我地有一條數列{an}，咁我地就可以從中抽一啲term出嚟，喺唔重複唔打亂次序嘅情況下造出另一條數列
咁呢條新數列就叫做{an}嘅子數列(subsequence)
例如當{an}=n嘅時候
1,3,5,7,9,... 係{an}嘅subsequence
1,4,9,16,... 都係
但 1,1,2,3,5,8,13,...唔係{an}嘅subsequence(因為1重複咗)
1,5,3,7,9,..都唔係(因為次序唔啱)
我地通常用{ank}去表示一條subsequence，而nk就係話畀我地知要攞原本條sequence係幾個term
{ank}個k應該要再細隻過n，但因連登formatting所限做唔到，不過我諗唔會令大家混淆嘅
咁我地就有兩個關於subsequence嘅簡單result:
1. 如果{an} converge去L，咁佢所有subsequence都會converge去L
2. 如果{an}有兩條唔同嘅subsequence converge去兩個唔同嘅數，咁{an}就會係divergent
Proof: Exercise
例子:
{an} = (-1)ⁿ⁺¹ (即係1,-1,1,-1,1,-1,1,-1,1,...)係divergent, 因為我地搵到兩條converge去唔同數嘅subsequence:
{a2n}=1係converge 去1,
但係{a2n+1}=-1係converge 去-1
跟住我地就有一條難少少嘅Lemma
(Lemma通常係指前往我地想證嘅野嘅踏腳石，但唔一定冇乜用)
就係每條sequence都有一條monotone subsequence
要證明呢個lemma，我地首先要屈啲野出嚟先(所以有少少難)
喺一條sequence {an}入面，如果ak不小於之後嘅term，我地就叫ak做dominant
咁一條sequence就只有兩個可能性
一係有無限咁多個dominant term，一係得有限多個dominant term
射十二碼一係入一係唔入
如果我地有無限咁多個dominant term, 咁呢啲dominant term就砌成一條decreasing subsequence
相反，如果我地得有限多個dominant term，咁我地就會有一個最後嘅dominant term as
咁an1=as+1就唔係dominant, 即係有一個term an2係大過an1
而an2都唔係dominant，即係有一個term an3係大過an2
如此類推，{ank}就會係一條increasing subsequence
搞掂
用呢條Lemma，我地就可以證到一條有名嘅定理(雖然我個人覺得佢冇乜用)
站在巨人的肩膊上，呢條theorem就變得好易證
我地假設{an}係一條bounded sequence
咁上面個Lemma就話我地有一條monotone subsequence (呢樣唔關bounded事)
而monotone convergence theorem就我地知呢條subsequence係converge嘅
搞掂
用呢句代替QED好似唔錯
下集我地開始講Cauchy呢條仆街仔
讀過analysis嘅巴打一定知我想講咩theorem
當我地有一條數列{an}，咁我地就可以從中抽一啲term出嚟，喺唔重複唔打亂次序嘅情況下造出另一條數列
咁呢條新數列就叫做{an}嘅子數列(subsequence)
例如當{an}=n嘅時候
1,3,5,7,9,... 係{an}嘅subsequence
1,4,9,16,... 都係
但 1,1,2,3,5,8,13,...唔係{an}嘅subsequence(因為1重複咗)
1,5,3,7,9,..都唔係(因為次序唔啱)
我地通常用{ank}去表示一條subsequence，而nk就係話畀我地知要攞原本條sequence係幾個term
{ank}個k應該要再細隻過n，但因連登formatting所限做唔到，不過我諗唔會令大家混淆嘅
咁我地就有兩個關於subsequence嘅簡單result:
1. 如果{an} converge去L，咁佢所有subsequence都會converge去L
2. 如果{an}有兩條唔同嘅subsequence converge去兩個唔同嘅數，咁{an}就會係divergent
Proof: Exercise
:^(
例子:
{an} = (-1)ⁿ⁺¹ (即係1,-1,1,-1,1,-1,1,-1,1,...)係divergent, 因為我地搵到兩條converge去唔同數嘅subsequence:
{a2n}=1係converge 去1,
但係{a2n+1}=-1係converge 去-1
跟住我地就有一條難少少嘅Lemma
(Lemma通常係指前往我地想證嘅野嘅踏腳石，但唔一定冇乜用)
就係每條sequence都有一條monotone subsequence
:^(
要證明呢個lemma，我地首先要屈啲野出嚟先(所以有少少難)
喺一條sequence {an}入面，如果ak不小於之後嘅term，我地就叫ak做dominant
:^(
咁一條sequence就只有兩個可能性
一係有無限咁多個dominant term，一係得有限多個dominant term
射十二碼一係入一係唔入
如果我地有無限咁多個dominant term, 咁呢啲dominant term就砌成一條decreasing subsequence
相反，如果我地得有限多個dominant term，咁我地就會有一個最後嘅dominant term as
咁an1=as+1就唔係dominant, 即係有一個term an2係大過an1
而an2都唔係dominant，即係有一個term an3係大過an2
如此類推，{ank}就會係一條increasing subsequence
搞掂
:^(
用呢條Lemma，我地就可以證到一條有名嘅定理(雖然我個人覺得佢冇乜用)
:^(
站在巨人的肩膊上，呢條theorem就變得好易證
我地假設{an}係一條bounded sequence
咁上面個Lemma就話我地有一條monotone subsequence (呢樣唔關bounded事)
而monotone convergence theorem就我地知呢條subsequence係converge嘅
搞掂
:^(
用呢句代替QED好似唔錯
下集我地開始講Cauchy呢條仆街仔
:^(
Elias
證一個space係complete有時都要靠佢
btw BW-theorem 係重要, 因爲佢明確確立左Cauchy sequence => converges (i.e. R is complete)
而之前所謂的sup/inf property仲未有咁明顯的確立
而證一個space係complete好多時都係抽subsequetial limit
subsequential limit係一樣好有用嘅嘢...
證一個space係complete有時都要靠佢
btw BW-theorem 係重要, 因爲佢明確確立左Cauchy sequence => converges (i.e. R is complete)
而之前所謂的sup/inf property仲未有咁明顯的確立
而證一個space係complete好多時都係抽subsequetial limit
:^(
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Elias
對比上面幅圖簡直小菜一碟
同埋first course in analysis只係嚴謹少少的微積分姐
睇到呢個位已經覺得maths唔係人類的語言，係外星文 :^(
用心去感受一下數學（尤其係analysis）嘅偉大
你會發現其實唔難
對比上面幅圖簡直小菜一碟
:^(
同埋first course in analysis只係嚴謹少少的微積分姐
算子代數
Elementary real analysis其實已經算淺白sosad
睇到呢個位已經覺得maths唔係人類的語言，係外星文 :^(
用心去感受一下數學（尤其係analysis）嘅偉大
你會發現其實唔難
對比上面幅圖簡直小菜一碟
同埋first course in analysis只係嚴謹少少的微積分姐
Elementary real analysis其實已經算淺白sosad
Edelschwarz
啱
不過證完呢個之後
你再見到佢都只會係Euclidean space嘅加強版
subsequential limit係一樣好有用嘅嘢...
證一個space係complete有時都要靠佢
btw BW-theorem 係重要, 因爲佢明確確立左Cauchy sequence => converges (i.e. R is complete)
而之前所謂的sup/inf property仲未有咁明顯的確立
而證一個space係complete好多時都係抽subsequetial limit
啱
:^(
不過證完呢個之後
你再見到佢都只會係Euclidean space嘅加強版
:^(
Edelschwarz
如果你喺大學讀數，你會成日見到呢條契弟:
無錯，呢條友正是Augustin Louis Cauchy
其實唔使入大學，各位A-level嘅同學仔都可能見過佢
Cauchy-Schwarz Inequality
咁佢同我地要講嘅野有咩關係呢?
事緣有一日，佢覺得要證一條sequence係converge要首先搵到佢嘅limit係一件好麻煩嘅事，所以佢就想搵吓有冇啲唔使知道條sequence嘅limit嘅方法
於是佢就諗到下面呢個定義:
翻譯蒟蒻: A sequence {an} is a Cauchy sequence, if for all ϵ>0, there exists a natural number N, such that |am -an|<ϵ for all m,n≥N.
呢個定義即係話第N個之後嘅term都同其他(第N個之後嘅)term好近
咁有冇例子呢?
其實，是但一條convergent sequence都係例子，因為我地有下面嘅Theorem:
Theorem: A convergent sequence is a Cauchy sequence.
點解?
如果{an}係一條convergent sequence, 咁佢就會converge去一個數，譬如話L
咁對於是但一個ϵ，我地就會搵到一個N，令到第N個之後嘅term都喺L-ϵ/2同L+ϵ/2之間 (呢度其實我地擺咗ϵ/2落個定義度)
咁我地喺第N個之後嘅term求其揀兩個am,an出嚟
咁因為佢地都喺L-ϵ/2同L+ϵ/2之間
所以佢地嘅差|am -an|就細過(L+ϵ/2)-(L-ϵ/2) = ϵ喇
搞掂
咁呢個定義有咩用呢?
我地想搵一個唔需要事先知道limit嘅方法去證一條sequence係convergent，所以如果將條theorem調轉(我地叫converse)都啱嘅話就好喇
好好彩，喺實數R上面，呢條theorem的確可以調轉:
Theorem: A Cauchy sequence {an} is convergent.
喺證明佢之前，我地需要一條Lemma:
Lemma: A Cauchy sequence is bounded.
Proof: Exercise
提示: 諗吓我地點證明convergent sequence係bounded
跟住Bolzano-Weierstrass theorem(可能係最後一次喺度見到佢)就話{an}有一條subsequence converge去某個L
咁對於是但一個ϵ，我地就會搵到一個K，令到{ank}第K個之後嘅term都係L-ϵ/2同L+ϵ/2之間
而因為{an}係Cauchy sequence，所以我地就會搵到一個N，令到第N個之後嘅term之後任意term嘅差都細過ϵ/2
我地依家老屈個數 s = max{K, N} 出嚟
咁我地就有ns≥N
所以當n≥N嗰陣，我地就有呢條不等式:
|an-L| = |an-ans+ans-L|≤|an-ans|+|ans-L|<ϵ/2+ϵ/2=ϵ
所以an係convergent嘅
搞掂
所以我地就有以下嘅結論:
佢有用嘅地方就係我地唔再需要知道sequence嘅limit去證佢convergent
考慮呢條sequence: a1=1, an+1 = an + (-1)ⁿ/(n+1)
咁佢好明顯係Cauchy sequence (因為第n個之後嘅term都係an-1同an之間)
所以佢係convergent，即使其實我地無辦法一眼睇得出佢個limit
(其實佢converge去 ln 2)
但係點解我地要強調我地喺R入面呢?
因為喺其他地方，呢個結論未必啱:
1, 1.4, 1.41, 1.414, 1.4142, ... 呢條sequence喺Q入面都係Cauchy，但喺Q入面佢係唔converge(因為佢converge去sqrt(2), which is 唔喺Q入面)
下回預告: Harmonic series
:^(
無錯，呢條友正是Augustin Louis Cauchy
其實唔使入大學，各位A-level嘅同學仔都可能見過佢
Cauchy-Schwarz Inequality
咁佢同我地要講嘅野有咩關係呢?
事緣有一日，佢覺得要證一條sequence係converge要首先搵到佢嘅limit係一件好麻煩嘅事，所以佢就想搵吓有冇啲唔使知道條sequence嘅limit嘅方法
於是佢就諗到下面呢個定義:
:^(
翻譯蒟蒻: A sequence {an} is a Cauchy sequence, if for all ϵ>0, there exists a natural number N, such that |am -an|<ϵ for all m,n≥N.
呢個定義即係話第N個之後嘅term都同其他(第N個之後嘅)term好近
咁有冇例子呢?
:^(
其實，是但一條convergent sequence都係例子，因為我地有下面嘅Theorem:
Theorem: A convergent sequence is a Cauchy sequence.
點解?
如果{an}係一條convergent sequence, 咁佢就會converge去一個數，譬如話L
咁對於是但一個ϵ，我地就會搵到一個N，令到第N個之後嘅term都喺L-ϵ/2同L+ϵ/2之間 (呢度其實我地擺咗ϵ/2落個定義度)
咁我地喺第N個之後嘅term求其揀兩個am,an出嚟
咁因為佢地都喺L-ϵ/2同L+ϵ/2之間
所以佢地嘅差|am -an|就細過(L+ϵ/2)-(L-ϵ/2) = ϵ喇
搞掂
:^(
咁呢個定義有咩用呢?
我地想搵一個唔需要事先知道limit嘅方法去證一條sequence係convergent，所以如果將條theorem調轉(我地叫converse)都啱嘅話就好喇
:^(
好好彩，喺實數R上面，呢條theorem的確可以調轉:
Theorem: A Cauchy sequence {an} is convergent.
喺證明佢之前，我地需要一條Lemma:
Lemma: A Cauchy sequence is bounded.
Proof: Exercise
:^(
提示: 諗吓我地點證明convergent sequence係bounded
跟住Bolzano-Weierstrass theorem(可能係最後一次喺度見到佢)就話{an}有一條subsequence converge去某個L
咁對於是但一個ϵ，我地就會搵到一個K，令到{ank}第K個之後嘅term都係L-ϵ/2同L+ϵ/2之間
而因為{an}係Cauchy sequence，所以我地就會搵到一個N，令到第N個之後嘅term之後任意term嘅差都細過ϵ/2
我地依家老屈個數 s = max{K, N} 出嚟
咁我地就有ns≥N
所以當n≥N嗰陣，我地就有呢條不等式:
|an-L| = |an-ans+ans-L|≤|an-ans|+|ans-L|<ϵ/2+ϵ/2=ϵ
所以an係convergent嘅
搞掂
:^(
所以我地就有以下嘅結論:
:^(
佢有用嘅地方就係我地唔再需要知道sequence嘅limit去證佢convergent
考慮呢條sequence: a1=1, an+1 = an + (-1)ⁿ/(n+1)
咁佢好明顯係Cauchy sequence (因為第n個之後嘅term都係an-1同an之間)
所以佢係convergent，即使其實我地無辦法一眼睇得出佢個limit
(其實佢converge去 ln 2)
但係點解我地要強調我地喺R入面呢?
因為喺其他地方，呢個結論未必啱:
1, 1.4, 1.41, 1.414, 1.4142, ... 呢條sequence喺Q入面都係Cauchy，但喺Q入面佢係唔converge(因為佢converge去sqrt(2), which is 唔喺Q入面)
下回預告: Harmonic series
:^(
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讀fourier 應該開頭會聽過 cesaro mean, 即係堆partial sum ge mean
Sigma_n = sum s_i / n
Cesaro summable 即係 lim Sigma_n exists and finite
咁樣有好多divergent series 都被賦予意義