※ 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. --

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