For the period T ( α ) of a simple pendulum with the length L and the amplitude (the initial elongation) α ∈ (0, π ), a strictly increasing sequence T n ( α ) is constructed such that the relations T1(α )=2Lgπ − 2+1ϵ ln1+ϵ 1− ϵ +π 4− 23ϵ 2,Tn+1(α )=Tn(α )+2Lgπ wn+12− 22n+3ϵ 2n+2,$$\begin{array}{c} \displaystyle T_1(\alpha)=2\sqrt{\frac{L}{g}}\left[\pi-2+\frac{1}{\epsilon} \ln\left(\frac{1+\epsilon}{1-\epsilon}\right)+\left(\frac{\pi}{4}-\frac{2}{3}\right)\epsilon^2\right],\\ \displaystyle T_{n+1}(\alpha)=T_n(\alpha)+2\sqrt{\frac{L}{g}}\left(\pi w_{n+1}^2 - \frac{2}{2n+3}\right)\epsilon^{2n+2}, \end{array}$$ and 0< T(α )− Tn(α )T(α )< 2ϵ 2n+2π (2n+1),$$\begin{array}{} \displaystyle 0 \lt \frac{T(\alpha)-T_n(\alpha)}{T(\alpha)} \lt \frac{2\epsilon^{2n+2}}{\pi(2n+1)}\,, \end{array}$$ holds true, for α ∈ (0, π ), n ∈ ℕ, wn:=∏ k=1n2k− 12k$\begin{array}{} \displaystyle w_n:=\prod_{k=1}^n\frac{2k-1}{2k} \end{array}$ (the n th Wallis’ ratio) and ϵ = sin( α /2).

中文翻译：

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简单摆的周期如何随着振幅的增加而增长？

对于长度为L且振幅为（初始伸长率）α∈（0，π）的简单摆的周期T（α），构造了严格增加的序列T n（α），使得关系T1（α） =2Lgπ− 2 + 1ϵ ln1 + ϵ 1− ϵ +π4−23ϵ 2，Tn + 1（α）= Tn（α）+2Lgπwn + 12− 22n + 3ϵ 2n + 2，$$ \ begin {array} {c} \ displaystyle T_1（\ alpha）= 2 \ sqrt {\ frac {L} {g}} \ left [\ pi-2 + \ frac {1} {\ epsilon} \ ln \ left（\ frac {1 + \ epsilon} {1- \ epsilon} \ right）+ \ left（\ frac {\ pi} {4}-\ frac {2} {3} \ right）\ epsilon ^ 2 \ right]，\\ \ displaystyle T_ {n + 1}（\ alpha）= T_n（\ alpha）+2 \ sqrt {\ frac {L} {g}} \ left（\ pi w_ {n + 1} ^ 2-\ frac {2} { 2n + 3} \ right）\ epsilon ^ {2n + 2}，\ end {array} $$和0 <T（α）− Tn（α）T（α）<2ϵ 2n +2π（2n + 1）， $$ \ begin {array} {} \ displaystyle 0 \ lt \ frac {T（\ alpha）-T_n（\ alpha）} {T（\ alpha）} \ lt \ frac {2 \ epsilon ^ {2n + 2} } {\ pi（2n + 1）} \ ,, \ end {array} $$成立，对于α∈（0，π），n∈ℕ，wn：= ∏ k = 1n2k− 12k $ \ begin {array} {} \ displaystyle w_n：= \ prod_ {k = 1} ^ n \ frac {2k-1} {2k} \ end {array} $（第n个Wallis ratio = sin（α/ 2）。