Mike Schilling

Mike has been a software engineer far longer than he would like to admit. He has strong opinions on baseball, software, science fiction, comedy, contract bridge, and European history, any of which he's willing to share with almost no prompting whatsoever.

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16 Responses

  1. Avatar Jaybird says:

    I thought I mostly understood the Order! post and flattered myself by coming up with a proof for ” if a > b and 0 > c, then b * c > a * c” in my head that worked enough for me without writing it down but when we start getting into e territory, my brain locks up.Report

    • Avatar Mike Schilling in reply to Jaybird says:

      Limits are hard. I didn’t want to double the size of the post by explaining them in detail, but they might be a good topic for next week.Report

      • Avatar Ken in reply to Mike Schilling says:

        I was tutoring my niece in calculus. She’s going into engineering, so I explained limits as something that mathematicians invented so they can justify dividing by zero or adding up an infinite number of terms, but that engineers never have to use. Once we pass Calculus I and II, that is.Report

  2. Avatar Boegiboe says:

    The field of quaternions is also complete, then. Quaternions with a norm of unity are used to represent three-dimensional rotations. Another method of representing rotations is as a series of three Euler angles. Unlike with quaternions, for whichever type of Euler angle representation you choose (e.g. 3-1-3), there are singularities in the field of Euler angle rotation sets; some rotations can be represented by an infinite number of sets of Euler angles. Does this mean an Euler angle representation is not complete? Or is it not even a field? I think maybe it is a field but is not complete. What do you think?Report

    • Avatar Chris in reply to Boegiboe says:

      I hate both you and Mike, in the same way that I generally hate people who are smarter than I.Report

    • Avatar Mike Schilling in reply to Boegiboe says:

      Quaternions are certainly complete, by the same argument as complex numbers. (Since C1′ uses only subtraction and absolute value, for those purposes complex numbers are effectively R2 and quaternions effectively R4.)

      I’m no expert on Euler angles, but they look to me like a group with a single operation (composition), not a field or a metric space, so I don’t see how completeness would apply.Report

      • What I find curious here is that any set of Euler angles can be converted into a quaternion by a generic operation–generic for a given order of rotations (e.g. 3-1-3). So, whatever operations can be one with quaternions, one could say the Euler-angle equivalent can be done by converting to a quaternion, doing the operation, and then converting back. But here’s where the problem comes in, because converting back is not unambiguous. So, my question is really about this ambiguity.Report

    • Avatar lukas in reply to Boegiboe says:

      Like Mike said, rotations are not a field, but a group, because there is only one single associative operation.

      That said, there is a lot of structure to this group. Locally, it looks very much like a R3, since you can describe it using three parameters (e.g. Euler angles), which makes it an example of what is more generally known as a Lie group.

      This group (generally called SO(3)) looks like R3 wherever you look at it, so it is complete, just like R3 itself.Report

  3. Avatar Ken S says:

    C1′ does not imply C1. There are ordered fields in which the only convergent sequences are eventually constant. Such fields are not isomorphic to the real numbers, and so cannot be complete in the sense of C1.Report

    • Avatar Ken S in reply to Ken S says:

      Oops. What I meant to say is that there are ordered fields in which all Cauchy sequences are eventually constant, and so they all converge. These fields all satisfy C1′, but not C1.Report

      • Avatar Mike Schilling in reply to Ken S says:

        I can’t picture what those fields would look like (if all Cauchy sequences are eventually constant, it’s pretty discrete, which makes it seem like the greatest member of a set would also be its least upper bound) but I’ll take your word for it. I do think that I have a correct proof that C1 implies C1′.Report

      • Avatar Ken S in reply to Ken S says:

        The construction goes like this: Given any ordered field F, an extension of F in which F is bounded is the field F(x) of rational functions with coefficients in F, where a(x) > b(x) if a(r) > b(r) for all sufficiently large r in F. Start from any ordered field F_0 you like, and and iterate this extension process aleph_1 times, to obtain the ordered field G. Since every sequence in aleph_1 is bounded, every increasing sequence in G is bounded. By taking reciprocals, every decreasing sequence of positive members of G has a positive lower bound, and so every Cauchy sequence must be eventually constant.Report