Re: [分析] 一題高微

[math1209 (將心比心)]

※ 引述《pobm (Rafale)》之銘言：
: Rudin p.141 18 的一部分
:                        1
: Let γ(t)=e^(2πit sin---) define on [0,2π]
:                        t             ^^^
:                                       ?
: Show that γ(t) and f(t)=e^(it) have the same range
: 我想問的是任給定θ在[0,2π]間
:                              1
: 如何找到一個t使得θ=2πt sin---
:                              t
: 這是不是高中數學阿XD
: 謝謝^^

 看起來不是你所說的這回事，關於 t = 0 的地方，是需要被定義為γ(t) = 0 .

 [以下是我 N 年前寫過的東西，可能有錯，你斟酌一下]

 18. Let γ_1,γ_2,γ_3 be curves in the complex plane, defined on [0,2π] by

   γ_1(t) ＝ e^(it), γ_2 ＝ e^(2it), γ_3 ＝ e^{(2πit)．sin(1/t)}.

    Show that these three curves have the same range, that γ_1 and γ_2 are
    rectifiable, that the length of γ_1 is 2π, that the length of γ_2 is
    4π, and that γ_3 is not rectifiable.

Proof. First, it is clear γ_1, and γ_2 have the same range. In addition,
       it is also clear that by Theorem 6.27, the length of γ_1 is 2π,
       and the length of γ_2 is 4π.

       So, let us study γ_3 as folows. In order to show three curves have
       the same range, it suffices to show that the range of

                            f(t) :＝ t sin (1/t)

       defined on (0,2π] contains a closed interval I with |I| ＝ 1.
       Note that sinx ＞ x － x^3/(3!) if x in (0,π/2), then we consider

       f(2π) － f(2/(3π)) ＝ {2π sin (1/(2π))} － {2/(3π) sin (3π/2)}

                            ＝ {2π sin (1/(2π))} ＋ 2/(3π)

                                        1      1      1
                            ＞ 2π．(----- － ---．-------) ＋ 2/(3π)
                                       2π     6    8π^3

                            ＝ 1 ＋ 5/(8 π^2)

                            ＞ 1.

       So, the range of f(t) contains a closed interval I with |I| ＝ 1.

       From above sayings, we have proved that these three curves have the
       same range.

       In order to show that γ_3 is not rectifiable, we consider

               γ_3 ＝ (cos {2πf(t)}, sin {2πf(t)}).

       If γ_3  is rectifiable, then so are cos {2πf(t)}, and sin {2πf(t)}.

       Consider g(t)：＝ sin {2πf(t)} defined on (0,1], and define a new
       sequence {a_n ＝ f( 2/(nπ) )}, then

       a_1 ＝ 2/π ; a_2 ＝ 0, a_3 ＝ －2/(3π), a_4 ＝ 0, a_5 ＝ 2/(5π),

                                  ．．．．．．

       It is clear that for k＝0,1,2,3,...

       a_(4k＋1) ＝ 2/{(4k＋1)π},   ＝＞ g(a_(4k＋1))＝sin {4/(4k＋1)}.
       a_(4k＋2) ＝ 0,               ＝＞ g(a_(4k＋2))＝0.
       a_(4k＋3) ＝ －2/{(4k＋3)π}, ＝＞ g(a_(4k＋3))＝We dont care.
       a_(4k＋4) ＝ 0,               ＝＞ g(a_(4k＋4))＝0.

       Let P ＝ {a_1,a_2,.....,a_n,..}, then consider


         Σ_[j＝1,∞] |Δg_j|

      ＝ Σ_[j＝1,∞] |g(a_j) － g(a_(j＋1))|

      ≧ Σ_[k＝1,∞] sin {4/(4k＋1)} ＝ ＋∞,

       by limit comparison test and lim_{x→∞} (sin x)/x ＝ 1.

       So, γ_3 is not rectifiable.

NOTE. In order to make preceding proof more clear, it is better to draw the
      graph of γ_3, f and g, roughly.

--
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推 pobm:請問為什麼可以把需要包函(0,1]轉換成只要有個|I|=1就好?
→ math1209:r_1 與 r_2 的 range 是單位圓的圓周 對吧?
→ math1209:別忘了 γ_3 ＝ e^{(2πit)．sin(1/t)} = e^( 2πif(t) )
→ pobm:其實我大概知道m大的意思 負的可以補足正的不到1的部分
→ pobm:可是該怎麼把這裡寫清楚呢>"< 謝謝
→ math1209:Sorry, 我不懂你的意思 = =
推 pobm:我的意思是說如果是一般的函數g(3t)要onto g(t)的range
→ pobm:所需的區間只要g(t)的1/3 如e^t要佈滿圓要0~2pi e^(2t)則只要
→ pobm:0~pi (我剛剛說錯是e^(it)和e^(2it) )
→ pobm:所以e^(2πit) 所需的是0~1 對吧
→ pobm:但是你給的f(t)恆小於1所以不在0~1 但是卻可以跑到-的而且實
→ pobm:際上的range比1 還大 這是因為在圓週上跑到負的角度剛好可以
→ pobm:補足逆時針方向沒跑完對吧
→ pobm:我的意思是我不知道怎麼把上面這一大段寫出來(還是有錯?)
→ math1209:我有寫: f(2π) － f(2/(3π)) ＞ 1.
推 pobm:抱歉我表達能力不太好 我的意思是"it suffices to show that"
→ pobm:該如何說明 只要包含一個大小大於1的區間即可
→ pobm:感謝m大我OK了^^