Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by contubernio on Sunday March 15, 2026 @05:14AM
(#47385612)
It's not a simplification, it's wrong. Sqrt(square(x)) equals abs(x).
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Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by MForster on Sunday March 15, 2026 @06:09AM
(#47385896)
It also equals x with appropriate assumptions (x > 0).
Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by notarget137 on Sunday March 15, 2026 @07:40AM
(#47386431)
Well, then sin(x) = x if x is infinitely small
Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by kccqzy on Sunday March 15, 2026 @09:43PM
(#47394146)
> Assuming[x == 0, Simplify[Sin[x] == x]]
Mathematica returns True. And any middle schooler will also tell you it's true.
The only reasonable interpretation of "infinitely small" is that it's zero.
Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by exe34 on Sunday March 15, 2026 @07:08AM
(#47386254)
so there's an unconditionally correct answer (it's also equal to abs(x) for x>0), and then there is an answer that is only correct for half the domain, which requires an additional assumption.
Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by crubier on Sunday March 15, 2026 @10:02AM
(#47387483)
sqrt(square(i)) != abs(i)
So no, itβs not unconditionally correct either.
Re:Why Mathematica does not simplify Sinh[ArcCosh[x]]
by oh_my_goodness on Sunday March 15, 2026 @01:29PM
(#47389611)
Not in general. As people have pointed out elsewhere, it's true if x is real. That isn't always a helpful assumption. (When x is real you can plug that assumption into Mathematica. Then Mathematica should agree with you.)
But consider sqrt(i) = sqrt(exp(i\pi/2)). That's exp(i\pi/4). Your rule would give 1 as the answer. It's not helpful for a serious math system to give that answer to this problem.
When I square 1 I don't get i.