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Why Mathematica does not simplify sinh(arccosh(x))
- hnarayanan 8mo agoThis 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.
- burnt-resistor 8mo agoThat's not what it simplifies to using a real or complex number domains for x, it's abs(x). CAS need type inference assumptions and/or type qualifiers to be more powerful. Edit: Fixed stuff.
- jstanley 8mo agoRight, that's why you need further assumptions on x in order for that simplification to hold.
- contubernio 8mo agoIt's not a simplification, it's wrong. Sqrt(square(x)) equals abs(x).
- MForster 8mo agoIt also equals x with appropriate assumptions (x > 0).
- exe34 8mo agoso 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.
- crubier 8mo agosqrt(square(i)) != abs(i) So no, it’s not unconditionally correct either.
- notarget137 8mo agoWell, then sin(x) = x if x is infinitely small
- kccqzy 8mo ago> 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.
- oh_my_goodness 8mo agoNot 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.
- yorwba 8mo agoFor x = -i, square(x) = -1, sqrt(square(x)) = i. Meanwhile, abs(x) = 1. You're right that it simplifies to abs(x) for real x, but that no longer holds for arbitrary complex values.
- NooneAtAll3 8mo agofor arbitrary complex values sqrt() gives 2 answers with +- signs so sqrt(square(-i)) = +-i, one of which is x
- syockit 8mo agoI've never seen a CAS that gives two answers for sqrt. Mathematica doesn't, sympy doesn't, and IIRC Maxima also doesn't.
- Armisael16 8mo agoThe sqrt function returns the principle square root, not both. That’s true for all numbers, positive, negative, and complex alike.
- fph 8mo agoIt's abs(x) only over the reals, for complex numbers it's more complicated.
- SoftTalker 8mo agoThat abs(x) (or |x| as we wrote it) used to catch out so many of us in HS trig and algebra.
- ogogmad 8mo agoI think you would get sqrt(x^2) = x, if x belonged to the natural domain of sqrt, which is a Riemann surface, that may also be defined using the language of "sheaves". I don't know how to connect this to the article or Mathematica.
- mathisfun123 8mo agoit's literally the prototypical example for `Assuming` https://reference.wolfram.com/language/ref/Assuming.html https://reference.wolfram.com/language/ref/Assuming.html
- noosphr 8mo agoMore 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.
- jmyeet 8mo agoI think "simplify" is pretty clear here. For trigonometric functions you would expect a trig function and an inverse trig function to be simplified. We all know what we'd expect if we saw sin(arcsin(x)) (ie x). If we saw cos(arcsin(x)) I'll spoil it for you: it simplifies to sqrt(1-x^2). Hyperbolic functions aren't used as much but the same principle applies. Here the core identity is cosh^2(x) = sinh^2(x) = 1 so: sinh(arccosh(x)) = sqrt(1 + cosh^2(arccosh(x)) = sqrt(1 + x^2) You should absolutely expect that from "simplify".
- noosphr 8mo agoHow is going from two functions with one variable to three functions with a variable and a constant a simplification?
- jmyeet 8mo agoIf you can't recognize how much simpler the simplified version is, I'm not sure exactly what to tell you. But let's think about it in terms of assembly steps: 1. Multiply the input by itself 2. Add 1 3. Take the square root. There is often a fast square root function available. The above is a fairly simply sequence of SIMD instructions. You can even do it without SIMD if you want. Compare this to sinh being (e^x - e^-x) / 2 (you can reduce this to one exponentiation in terms of e^2x but I digress) and arccosh being ln(x + sqrt(s^2 - 1)) and you have an exponentiation, subtraction, division, logarithm, addition, square root and a subtraction. Computers generally implement e^2 and logarithm using numerical method approximations (eg of a Taylor's series expansion).
- noosphr 8mo ago
- bryango 8mo agoI 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.
- MinimalAction 8mo agoThat blackbox being their entire moat, I would assume they'd never want to open-source any function. Mathematica as a front-end has innumerable frustrating bugs, but its CAS is top-notch. Especially combined with something like Rubi for integration, for me nothing comes close to Mathematica for algebraic computations.
- eulerboiler 8mo agoMany functions source ate viewable. Use https://resources.wolframcloud.com/FunctionRepository/resources/PrintDefinitions https://resources.wolframcloud.com/FunctionRepository/resour...
- gdelfino01 8mo agoMany built-in functions are open source too. Use the "PrintDefinitions" ResourceFunction to see the code of functions that are implemented in Wolfram Language itself.
- mkl 8mo agoSource available? The license is still proprietary, right?
- LegionMammal978 8mo agoYes, it is all proprietary, but there are still ways to inspect most of the WL-implemented functions since the system does not go to extreme pains to keep them hidden from introspection. It is not unlike Maple in that sense.
- SillyUsername 8mo agoI'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.
- kzrdude 8mo agoSimple rule to keep in mind that even math savvy people seem to forget about is that: sqrt(x²) = |x| with bars for absolute value. For a programmer, it's clear that we have lost the sign information but not the magnitude. Simple. Makes most sign and solution reasoning explicit instead of implicit when solving quadratics or otherwise working with square roots.
- tzs 8mo ago> Simple rule to keep in mind that even math savvy people seem to forget about is that: sqrt(x²) = |x| with bars for absolute value. i would disagree with that (pun intended).
- derf_ 8mo agoThis 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.
- johndcook 8mo agoThanks. That was a mess. Don't know what happened, but I fixed it this morning.
- ogogmad 8mo agoDoes 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.
- densh 8mo agoFor me Mathematica is much more akin to numpy+sympy+matplotlib+... with absolutely crazy amount of batteries included in a single coherent package with IDE and fantastic documentation. In a way numpy ecosystem already "won" industry users over, yet Wolfram stack is still appealing to me personally for small experiments. Coq/Lean target very different use cases.
- kccqzy 8mo agoDo you want your LLM to generate one hundred lines of code to do things using open source libraries, or five lines in Mathematica? This is actually subjective. For the vibe coding folks, they don’t care if the code is long winded and verbose. For others, the conciseness is part of the point; see APL and Notation as a Tool of Thought.
- Almondsetat 8mo agoAnd yet it incorrectly simplifies f(x) = x/x with f(x) = 1
- rfc3092 8mo agoI believe this is correct: x/x = 1 everywhere except 0, where it has a removable singularity. So you can extend x/x holomorphically to full C. This is completely different from the phenomenon described in the article: arccosh discontinuity can’t be dealt the same way. In fact complex analysis prefers to deal with it my making functions path-dependent (multi-valued).
- newobj 8mo agoPLEASE explain "So you can extend x/x holomorphically to full C" to someone with only a BSc in math/cs; something about this thread is giving me an existential crisis right now.
- rfc3092 8mo ago- function extension is defining a function where it is not defined - <Adj> function extension is an extension that keeps (or gives) Adj property - extended function is usually treated as originals if extension is good enough. Real analysis starts with defining real numbers and extending familiar functions onto them - in this particular case we do not need C - even continuous extension on R works and agrees with x/x = 1 at 0 - holomorphic (analytic) extension makes function infinitely differentiable at every point of C - because of the nature of discontinuity you can’t extend the simple arccosh in any reasonable way on C without introducing multivalued or path-dependent functions - this continuity makes x/x=1 a reasonable simplification for CAS imo but not for complex functions as in the OP - many things with point singularities in R have more structure in C, but x/x is not one of them. Even 1/x is of a different nature. “You do not divide by zero” that forces you to carry x != 0 is more of a high-school construct than a real thing. Physicists ignore even more important stuff, and in the end their formulas work “just fine”.
- 8mo ago
- useftmly 8mo ago[dead]
- rfc3092 8mo agoThis 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.
- meatmanek 8mo ago> 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]]?