设xn = i tantn
则xn+1 = 2i tantn /(1-(tantn)^2)
=i tan2tn
若x0 = i tant
则xn = i tan 2^n t
附：sin ix = i(e^x - e^-x)/2 ,cos ix = (e^x + e^-x)/2
具体查复变函数
【数学竞赛】由递推关系式求通项公式
设xn = i tantn
则xn+1 = 2i tantn /(1-(tantn)^2)
=i tan2tn
若x0 = i tant
则xn = i tan 2^n t
附：sin ix = i(e^x - e^-x)/2 ,cos ix = (e^x + e^-x)/2
具体查复变函数