14/08/28 21:28:52.55
√√…√(1/(n+m)) (√はm+1個)
=(1/(n+m))^(1/2)^(1/2)^…^(1/2)
=(1/(n+m))^((1/2)*(1/2)*…*(1/2))
=(1/(n+m))^((1/2)^(m+1))
=e^(log1/(n+m))^((1/2)^(m+1))
=e^((log1/(n+m))*((1/2)^(m+1)))
=e^(-log(n+m)/2^(m+1))
ここまで丁寧に書けば分かってもらえるか？
14/08/28 21:28:52.55
√√…√(1/(n+m)) (√はm+1個)
=(1/(n+m))^(1/2)^(1/2)^…^(1/2)
=(1/(n+m))^((1/2)*(1/2)*…*(1/2))
=(1/(n+m))^((1/2)^(m+1))
=e^(log1/(n+m))^((1/2)^(m+1))
=e^((log1/(n+m))*((1/2)^(m+1)))
=e^(-log(n+m)/2^(m+1))
ここまで丁寧に書けば分かってもらえるか？