Remove ads
ω₁

ω₁

The Long Count
Joined
Messages
334
Reactions
1,027 4
Points
0

Profile posts Latest activity Postings About

  • Baughn
    Hope you're okay.

    (Yes, the moment someone doesn't log on for a few weeks we get worried. You can't *leave* SV.)
    MJ12 Commando
    Are you a secret lawyler on this board? Because your copyright post was definitely very lawyerly.
    ω₁
    ω₁
    I couldn't possibly say yes. If I did, I wouldn't be a secret lawyer anymore, now would I?

    (But no. I just enjoy researching things, and US copyright law has been one of those things on occasion.)
    MJ12 Commando
    MJ12 Commando
    I said lawyler! not lawyer!

    It was probably up to snuff at first glance, which is the important thing. Who's going to spend more effort than that on an interbutt debate?
    ω₁
    ω₁
    a) It was mostly up to snuff at first glance, which is the important thing. ^_^

    b) Well, it's 3:00 in the morning and I should be asleep, so... me, probably.
    Aranfan
    The S thing included more than just translations. Rotations, dialations, and combinations of the three were considered.

    S can't have uncountably many of itself in the euclidean plane unless it's defined as any of a certain shape constructed according to certain rules but with variable parameters. If those parameters vary in the right way, uncountably many S's are possible.
    ω₁
    ω₁
    Oh, just the similarity-translations, then? I never did finish proving it for arbitrary affine transformations, though it seemed to be true.

    I had arbitrarily chosen two 3/4-circular arcs, conjoined only at one pair of endpoints to be C¹. (I did notice that two semicircles joined in the same way can be translated by ε>0 without self-intersection.)
    Aranfan
    Aranfan
    Yeah, that's what I chose too. I can probably dig up the paper and PM it to you if you want.

    The proof is more a sketch than anything formal though.
  • Loading…
  • Loading…
  • Loading…
Remove ads
Back
Top