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如果我设置x=0,然后计算: ((x+1)/(x-1))^(1/3); 我得到 NaN
((x+1)/(x-1))^(1/3)
但如果我这样做,我不会遇到问题:
-1^(1/3)
以下两种方式也会导致错误:
((0+1)/(0-1))^(1/3) ((1)/(-1))^(1/3)
而且,当然, 的输出 ((x+1)/(x-1))是 -1
((x+1)/(x-1))
有谁知道R是怎么回事?