18 Replies to “PoSWW 25: Awe Inspiring”

  1. Is the programming considering complex domains for some reason?

    Even though I’ve checked and this still happens when you set the device to Real.

    My other theory is that the calculator checks to see if f(a) exists and if it does, assumes that the limit is equal to f(a) for some reason.

    1. I didn’t take the screenshot, although I’ve had a number of people check it. I don’t think anyone has succeeded in bashing sense into the machine.

      1. The CAS simplifies the principle square root of a negative number to i times the square root of its absolute value:

            \[\sqrt{-x} = i\sqrt{x}\,, \; x>0\]

        This extends the domain of sqrt to all real numbers. Given this, the limit becomes

            \[\begin{aligned} \lim_{x\to 0^-} \sqrt{x} &= \lim_{y\to 0^+} \sqrt{-y} \,, \quad y=-x>0 \\ &= i \lim_{y\to 0^+} \sqrt{y} = i \times 0 = 0 \end{aligned}\]

        Wolfram Alpha gives a similar result – saying that the limit of sqrt(x) as x->0 is 0 for real x.

        As long as we are only taking limits for real x then is there actually a problem? I always get a little nervous when branch cuts are too close – and am getting flash-backs of Richard Fateman’s classic paper: Why computer algebra systems sometimes can’t solve simple equations

        1. Are you stating that is what the machine is doing or are you guessing that is what it is doing? In any case, yes, it is obviously a problem.

          1. I’m *guessing* that is what it’s doing – I haven’t decompiled/looked at the source code for the limit evaluation. But it is reasonable that symbolic simplification is made using assumptions that can be drawn from the limit first. That assumption and simplification is what I demonstrated.

            You say it is “obviously” a problem, in the way some graduate texts say something is obvious. It depends on the audience – my complex analysis training was not so solid – it’s on my ‘to do’ list.

            My view is that if you are taking the limit from the left, then the variable must be real. If you are using the common convention that sqrt(-x) = i sqrt(x) for x>0 [which then means you have to be careful with many log and exponential identities] then the square root function is now continuous on R but not differentiable at 0, so there is no problem… Why am I wrong?

            I vaguely remember you prefer a multivalued definition rather than choosing a principle root. Is that the root (pun intended) of the problem? Or is it just that 0 is the branch point – which I’m not sure is a problem if we constraint x to be real.

            1. Thanks, Simon. Two quick comments.

              (1) As you note “it depends upon the audience”. The audience here is senior school students (and their teachers), many of whom have not, and will not ever, see complex numbers. So, even if the CAS is in some sense working reasonably in a complex world, the effect for this audience is simply crazy.

              (2) The possibility that the CAS is doing complex number stuff also occurred to me, but I have a feeling that may not be the case. Personally, I never approach CAS without a hammer in hand. But I think the people I discussed this with all tried and failed to get the idiot machine to give a DNE real number answer. Also, for anybody who cares about such things, I was told by one person that “the old black TI” reutrns 0, but “the new blue TI” gives DNE.

              CAS poisons everything.

              1. Ah, I see now. And you’re right there is still a problem. If the Nspire CAS (newest version) is set to “real” mode, then the sqrt(-x)|x>0 raises an “Non-real result” error but the limit of sqrt(x) as x->0- is still outputting zero. As Red Five already noted.

  2. The white CAS (Casio Classpad II) gives it as Undefined. A rare case of white CAS actually being better (comparatively) than black and blue CASes perhaps.

  3. Ah, this must be why all my teachers and lecturers have stated that the limit of a square root exists at all points on its domain, including the endpoint, even though you’d naturally think it wouldn’t as the limit from the left side shouldn’t exist. But all hail our lord and saviour the CAS, who has become a part of VCE for our sins, who has shown us the truth: that the limit from the left side does in fact exist! This explains everything.

    1. Possibly, although it could be for a separate, sane reason.

      It is common to declare, for example, that f(x) = √x is continuous on the domain [0,1]. The point is, it is natural to only consider limits and continuity within the domain of the function. But this bending/altering of the rules for such functions doesn’t provide any legitimacy for the CAS idiocy above.

      1. The CAS engine probably follows a set of rules which it uses to evaluate expressions. If you program sqrt(x) to be continuous at x=0 the rules will pretty much always specify that the limit of sqrt(x) at x=0 from the left side is 0, unless you include a bunch of other edge-cases in the programming. I don’t know what the convention is but it seems a lot cleaner just to have sqrt(x) be discontinuous at x=0.

        In my opinion it’s these kind of ambiguities that lead to confusion both for students and teachers. If you say that the limit of a function exists only if the limit exists from the left and the right side and they are equal, and then state that a function is continuous only if the function evaluated at that point equals its limit at that point, then it follows that if f(x) = sqrt(x) is continuous at x=0, then the limit of f(x) from the left side at x=0 is 0. However, maybe my teachers were negligent in mentioning edge cases with definitions.

        Edit: Okay, now that I’ve had a think about it kind of makes sense to have sqrt(x) be continuous at x=0, if I think about it in terms of the neighbourhoods of 0 within the set [0, infinity). But I have no idea if my line of thought is correct, or if I even understand what a neighbourhood is, as the majority of my “knowledge” of mathematics comes from reading wikipedia. It would be best if all these ambiguities were cleared up when we are actually taught complex topics.

  4. hi,

    presumably the calculator’s settings have been corrupted or the Mathematica is “up the Kyber” …

    Wonder what it gives for limit of 1/x for as x –> 0- and x–>0+

    steve r

    1. It gives positive and negative infinity as expected
      lim(((1)/(x)),x,0,−1) → −∞
      lim(((1)/(x)),x,0,1) → ∞

  5. I think this will be an issue for all CAS machines when they are ‘calculating’ one-sided limits towards a branch point.

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