3 ms·

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.

burnt-resistor 7mo ago
That'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 7mo ago
  Right, that's why you need further assumptions on x in order for that simplification to hold.
  - contubernio 7mo ago
    It's not a simplification, it's wrong. Sqrt(square(x)) equals abs(x).
    - MForster 7mo ago
      It also equals x with appropriate assumptions (x > 0).
      - exe34 7mo ago
        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.
        crubier 7mo ago
        sqrt(square(i)) != abs(i) So no, it’s not unconditionally correct either.
      - notarget137 7mo ago
        Well, then sin(x) = x if x is infinitely small
        kccqzy 7mo 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 7mo ago
      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.
- yorwba 7mo ago
  For 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 7mo ago
    for arbitrary complex values sqrt() gives 2 answers with +- signs so sqrt(square(-i)) = +-i, one of which is x
    - syockit 7mo ago
      I've never seen a CAS that gives two answers for sqrt. Mathematica doesn't, sympy doesn't, and IIRC Maxima also doesn't.
    - Armisael16 7mo ago
      The sqrt function returns the principle square root, not both. That’s true for all numbers, positive, negative, and complex alike.
- fph 7mo ago
  It's abs(x) only over the reals, for complex numbers it's more complicated.
- SoftTalker 7mo ago
  That abs(x) (or |x| as we wrote it) used to catch out so many of us in HS trig and algebra.
ogogmad 7mo ago
I 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 7mo ago
  it's literally the prototypical example for `Assuming` https://reference.wolfram.com/language/ref/Assuming.html https://reference.wolfram.com/language/ref/Assuming.html

burnt-resistor 7mo ago

That'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 7mo ago
Right, that's why you need further assumptions on x in order for that simplification to hold.
- contubernio 7mo ago
  It's not a simplification, it's wrong. Sqrt(square(x)) equals abs(x).
  - MForster 7mo ago
    It also equals x with appropriate assumptions (x > 0).
    - exe34 7mo ago
      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.
      - crubier 7mo ago
        sqrt(square(i)) != abs(i) So no, it’s not unconditionally correct either.
    - notarget137 7mo ago
      Well, then sin(x) = x if x is infinitely small
      - kccqzy 7mo 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 7mo ago
    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.
yorwba 7mo ago
For 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 7mo ago
  for arbitrary complex values sqrt() gives 2 answers with +- signs so sqrt(square(-i)) = +-i, one of which is x
  - syockit 7mo ago
    I've never seen a CAS that gives two answers for sqrt. Mathematica doesn't, sympy doesn't, and IIRC Maxima also doesn't.
  - Armisael16 7mo ago
    The sqrt function returns the principle square root, not both. That’s true for all numbers, positive, negative, and complex alike.
fph 7mo ago
It's abs(x) only over the reals, for complex numbers it's more complicated.
SoftTalker 7mo ago
That abs(x) (or |x| as we wrote it) used to catch out so many of us in HS trig and algebra.

ogogmad 7mo ago

I 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 7mo ago
it's literally the prototypical example for `Assuming` https://reference.wolfram.com/language/ref/Assuming.html https://reference.wolfram.com/language/ref/Assuming.html