原帖由 Angelfire 於 17/6/2008 21:15 發表
睇完都係唔明.......
中二應該未學呢d
火影忍者有多重影分身術，其實數學界也有這樣的絕技。
# 例一: “同影分身” (1 --> 1’s)
證明 (n+1)^(1/m) < 1 + n/m for all n>=1, m>=2 where m,n (- Z.
解: 由於AM>=GM, 所以
(n+1)^(1/m)=[(n+1)*1*1*….*1]^(1/m) < [n+1+(1+1+…+1)]/[1+m-1]
=(n+1+m-1)/m = (m+n)/m = 1+n/m.
# 例二: “異影分身” (n --> 1’s)
證明 n+1/2+1/3+1/4+ …+1/n > n*[(n+1)/2]^(1/n) for all n>=2 where n (- Z.
解: 由於AM>=GM, 所以
(n+1/2+1/3+1/4+ …+1/n)/n
= [(1+1+1+…+1)+1/2+1/3+1/4+…+1/n]/n
= [1+(1+1/2)+(1+1/3)+…+(1+1/n)]/n
= [1+3/2+4/3+5/4+…+(n+1)/n]/n > [1*3/2*4/3*5/4*…*(n+1)/n]^(1/n)
= [(n+1)/2]^(1/n)
==> n+1/2+1/3+1/4+…+1/n > n*[(n+1)/2]^(1/n).
原帖由 sapphire 於 18/6/2008 05:01 PM 發表
AM = Arithmetic Mean
GM = Geometric Mean
By proving, AM >= GM
Reference : http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
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