※ [本文轉錄自 Dirichlet 信箱]
作者: pretend.bbs@lalala.twbbs.org ("Weak and only weak")
標題: π為無理數
時間: Sat Jul 19 21:44:06 2014
作者: pretend (Big Lion) 站內: pretend
標題: π為無理數
時間: Fri Nov 5 09:24:43 2004
(2k) (2k)
1 n (-1)^k * [f (0) + f (1)]
Claim 1 : π∫f(x)sin(πx)dx = Σ --------------------------- = I
0 k=0 π^2k
其中 f(x) = x^n(1-x)^n/n! 由部分積分 2n 次可立即得證
(k) (k) (k) (k)
Claim 2 : f(x)=f(1-x) ==> f (x) = (-1)^k * f (1-x) ==> f (0) = (-1)^k * f (1)
for any positive integer k
(k) k k (i) (k-i)
Claim 3 : f (x) = 1/n!Σ {C [x^n] * [(1-x)^n] }
i=0 i
0 , 0≦i<n (k)
x^n 在 x=0 時的 i 階導數為 = n! , i = n ==> f (0) 是整數
0 , i > n
p^2n
Claim 4 : lim -------- = 0 , 其中 p 為正整數
n->oo n!
(p^2n)/n! -> 0 , as n -> oo <==> k^n/n! -> 0 , n -> oo (p^2 = k)
k^n/n! = (k^k/k!)(k/k+1)*...*(k/n!) < (k^k/k!)[k/(k+1)]^n-k -> 0
as n -> oo
pf: If π is a rational number ==> π = p/q , gcd(p,q) = 1 , p,q belong
to N , ==> I*P^2n is a integer (By Claim 1,2,3)
Moreover , since 0<x<1 , So 0≦[x^n][(1-x)^n]sin(πx)≦1 ==>
0 ≦ I*P^2n ≦ (π*P^2n)/n! , By Claim 4 we have (π*P^2n)/n! -> 0
as n -> oo , it follows that 0 ≦ I*P^2n ≦ 1 , if n is sufficient
large , but this violates that I*P^2n is a integer (=><=)
Hence , π is a irrational numbber.
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＜ From : 220-142-8-99.dynamic.hinet.net ＞
→ pretend 修改本文於 Fri Nov 5 09:31:46 2004
※ 發信站: 批踢踢實業坊(ptt.cc)
※ 轉錄者: Dirichlet (111.255.113.174), 07/19/2014 21:46:56