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Why Mathematica does not simplify sinh(arccosh(x))

143 points by ibobev

by derf_

1 subcomments

This sentence confused me: "For example, Sinh[ArcCosh[-2 + 0.001 I]] returns 11.214 + 2.89845 I but Sinh[ArcCosh[-2 + 0.001 I]] returns 11.214 - 2.89845 I," not the least of which because the two input expressions are the same, but also because we started out by saying Sinh[ArcCosh[-2]] = -Sqrt[3], which is not at all near 11.214 +/- 2.89845 I.
I think the author meant to say, "ArcCosh[-2 + 0.001 I] returns 1.31696 + 3.14102 I but ArcCosh[-2 - 0.001 I] returns 1.31696 - 3.14102 I," because we are talking about defining ArcCosh[] on the branch cut discontinuity, so there is no need to bring Sinh[] into it (and if we do, we find the limits are the same: the imaginary component goes to zero and Sinh[ArcCosh[-2 +/- t*I]] approaches -Sqrt[3] as t goes to zero from above or below). I am not sure what went wrong to get what they wrote.

by hnarayanan

2 subcomments

This is a general pattern in CAS. For a more basic case, it’s not obvious sqrt(square(x)) will simplify to x without any further assumptions on x.

by bryango

2 subcomments

I really wish Mathematica would open-source the heuristics behind these core functions (including common mathematical functions, Simplify, Integrate, etc.). The documentation is good, but it still lags behind the actual implementation. It would be much easier if we could peek inside the black box.

by noosphr

4 subcomments

More generally it's not at all clear what 'simplify' means.
Is x*x simpler than x^2? Probably? Is sqrt(5)^3 simpler than 5^(3/2)? I don't know.
It entirely depends on what you're going to be doing with the expression later.

by meatmanek

0 subcomment

> For example, Sinh[ArcCosh[2]] returns −√3 but √(2² − 1) = √3. The expression Mathematica returns for Sinh[ArcCosh[x]] correctly evaluates to −√3
but the expression given is sqrt((x-1)/(x+1))(x+1), which for x=2 would be sqrt(1/3)*3 = sqrt(3)
did you mean Sinh[ArcCosh[-2]]?

by SillyUsername

1 subcomments

I've only an A-Level in Further Maths from 1997, but understand complex numbers and have come across complex inverse trig functions before.
My takeaway for other people like me from this is "computer is correct" because the proof shows that we can't define arccosh using a single proof across the entire complex plane (specifically imaginary, including infinity).
The representation of this means we have both complex functions that are defined as having coverage of infinity, and arccosh, that a proof exists in only one direction at a time during evaluation.
This distinction is a quirk in mathematics but means that the equation won't be simplified because although it looks like it can, the underlying proof is "one sided" (-ve or +ve) which means the variables are fundamentally not the same at evaluation time unless 2 approaches to the range definition are combined.
The QED is that this distinction won't be shown in the result's representation, leading to the confusion that it should have been simplified.

by rfc3092

0 subcomment

This is what differentiates (pun intended) between Complex Algebra and Complex Analysis: complex functions in analysis are multivalued (or path dependent in some schools). Even a simple concept of value of F at complex point x becomes a topic of several lectures.
I’m algebraist at heart and training, but I remember beautiful many-layered surfaces of ordinary complex functions in books and on blackboards.

by Almondsetat

1 subcomments

by useftmly

0 subcomment

by ogogmad

2 subcomments

Does anyone else think that the latest LLMs - some of which can be used locally for free - combined with proof-verifying software like Coq or Lean for mistake-detection, might make many uses of Computer Algebra Systems like Mathematica obsolete?
Certainly, people don't need Wolfram Alpha as much.
On another point, it sucks to know what this means for Algebraic Geometry (the computational variant), which you could partly motivate, until now, for its use in constructing CASes.