作者pikahacker (死亡筆記 型電腦)
看板IMO_Taiwan
標題Re: [問題] Dumb questions
時間Sat Oct 9 04:16:56 2004
※ 引述《darkseer (公假中)》之銘言：
: ※ 引述《pikahacker (死亡筆記 型電腦)》之銘言：
: : 1. Prove that there exists a UNIQUE function f from set R+ to R+ such that
: : f(f(x))=6x-f(x)
: : 2. For every n in Z+, let Rn be the minimum value of |c-d*3^(1/2)| for all
: : nonnegative integers c and d with c+d=n. Find, with proof, the smallest
: : positive real number g with Rn < or = g for all n in Z+.
: 1.顯然f(x)=2x是一解
: 假如f(x)"偏離"了2x, 若f(a)>2a,設epsilon=E=f(a)-2a(<時做法同).
: 則f(f(a))=6a-f(a)=4a-E=2f(a)-3E.
: 又會有f(f(f(a)))=2f(f(a))+9E, f(f(f(f(a))))=2f(f(f(a)))-27E
: 有f(f(...f(a)...)=2^n * a + E * ( 2^n - (-3)^n ) / 5
: ~n個~
: 後面E的項的成長數率較快,若E>0則必爆(變負的). Q.E.D.
: 2. Let k be 1+sqrt(3). The set { |c-d*sqrt(3)| : c+d=n } is the set { |n|,
: |n-d|, ..., |n-nd| }, whose minimum is d * min( {n/d}, {-n/d} ). Note:{x}=x-[x].
: Since d is irrational, the supremum of the set { min({n/d},{-n/d}) : for all
: positive integers n } is 1/2. Hence g=d/2.
d irrational -> supremum of the set is 1/2.
Seems true enough, but is there a theorem behind this? How do you prove it?
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