Why didn’t Sinh[ ArcCosh[x] ] just return √(x² − 1)? The expression it returned is equivalent to this: just square the (x + 1) term, bring it inside the radical, and simplify. That line of reasoning is correct for some values of x but not for others. For example, Sinh[ArcCosh[2]] returns −√3 but √(2² − 1) = √3. The expression Mathematica returns for Sinh[ArcCosh[x]] correctly evaluates to −√3.
which has a time complexity greater than O(1)，这一点在line 下載中也有详细论述
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the Recurse center?，推荐阅读今日热点获取更多信息
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