作者znmkhxrw (QQ)
看板trans_math
標題Re: [積分] 幾題積分
時間Sun Oct 9 22:49:39 2011
看了一開始的不等式後
發現這根本是beta函數耶XD
(借用一下式子)
: Proof of Wallis' formula:
: 當0＜x＜π/2 ,
: 2n+1 2n 2n-1
: 0＜sin (x)＜sin (x)＜sin (x)
so Β(n,1/2) ＜ Β( n+(1/2) , 1/2) ＜ Β(n+1 , 1/2)
Γ(n)Γ(1/2) Γ(n+(1/2))Γ(1/2) Γ(n+1)Γ(1/2)
so ─────── ＜ ───────── ＜ ───────
Γ(n+(1/2)) Γ(n+1) Γ(n+(3/2))
1.Γ(1/2)都砍掉
2.Γ(n) = (Γ(n+1)) /n (套用在左式)
3.Γ(n+(3/2)) = (n+(1/2)) * Γ(n+(1/2)) (套用在右式)
4.左右的Γ(n+1)都除過來中式
5.左右的Γ(n+(1/2))都乘過來中式
6.左式的n乘到中式與右式
7.三方開根號
we have
Γ(n+(1/2)) n
1 ＜ ────── * n^(1/2) ＜ (────)^(1/2)
Γ(n+1) n+(1/2)
denoted by
A ＜ B ＜ C
where
Γ(n+(1/2)) (n-(1/2))*(n-(3/2))*....*(1/2)*Γ(1/2)
────── = ────────────────────
Γ(n+1) n!
(2n-1)*(2n-3)*...*1*Γ(1/2)
= ────────────── (同乘2^n)
2^n * n!
(2n-1)!!*Γ(1/2)
= ──────── (by def of double fractorial)
2^n * n!
(2n)!*Γ(1/2)
= ───────── (同乘(2n)!!)
2^n * n! * (2n)!!
(2n)!*Γ(1/2)
= ───────── (因為(2n)!! = 2^n * n!)
2^(2n) * (n!)^2
by Squeeze principle , B→1 as n→inf
(2n)!*Γ(1/2)
so lim B = lim ──────── * n^(1/2) = 1
n→inf n→inf 2^(2n) * (n!)^2
之後Γ(1/2)=π^(1/2) 代進去後取倒數即可
(原題目是要求lim 1/B = π^(1/2) )
n→inf
如果要嚴謹證明的話
Let f(x) = 1/x is continuous at {π^(1/2)}
so for B→π^(1/2) , f( lim B ) = lim f(B) = lim 1/B = π^(1/2)
n→inf n→inf n→inf
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打的有點詳細...如果真的考試有出 計算過程不用寫出來啦~
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