Math Help, non-euclidean geometry.

  • Ritournelle

    Posts: 134

    Oct 26, 2010 2:12 AM GMT
    I know there's a few high level physics and math people around here so I though I'd at least give it a shot, I'm stumped and the assignment is late...

    So, we have a partition Xbar of the plane X=R^2 consisting of all subsets A (subset of X) of the form

    A= I. {(x,y),(-x,y) x=/=0}
    OR II. {(0,y)}

    Let dbar euclidean (euclidean distance on the partition) be the quotient semi-metric d euclidean ofX=R^2. We want to analyze (XBAR, Dbar euc).

    That is the set up.
    d. For P, Q ∈ X+, consider a discrete walk w from P ̄ to Q ̄ consisting of the pointsP =P1,Q1 ∼P2,Q2 ∼P3,...,Qn−1 ∼Pn,Qn =QofX. Showthat there is another discrete walk w′ from P ̄ to Q ̄ consisting of points P = P1′, Q′1 =P2′,Q′2 =P3′,...,Q′n−1 =Pn′,Q′n =Q,such that ldeuc(w')<=ldeuc(w) and such that all points Pi′ and Q′i are in X+. Hint: Part c.
    e. For the discrete walk w′ of Part d, show that ldeuc(w')=>deuc(P, Q).
    I have D by induction, but I'm not totally sure how to get the case in E.
  • Posted by a hidden member.
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    Oct 26, 2010 3:47 AM GMT
    What a joke. I can't believe you can't figure that out.
  • Ritournelle

    Posts: 134

    Oct 26, 2010 5:02 AM GMT
    I'm about to cry myself to sleep over it
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    Oct 26, 2010 5:09 AM GMT
    Math is awesome lol
  • danisnotstr8

    Posts: 2579

    Oct 26, 2010 5:21 AM GMT
    "Blowing in the wind."
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    Oct 26, 2010 5:23 AM GMT
    why not use the Euclidean space structure to define a metric... this has always been the easiest for me.

    a8372135e7d87a4eed18847555a374b3.png
  • Ritournelle

    Posts: 134

    Oct 26, 2010 5:44 AM GMT
    The metric is already defined... It's euclidean, and it's 2-dimensional, but that's irrelevant anyways. I already used transitivity and induction to demonstrate part D (also before that, I know that because we have a space of a folded euclidean plane, (X+), the reverse triangle inequality including proof by the Taylor-Schwarz inequality definitely proves that dbar is deuc(P, Q), then I believe it must imply a strict equality (or at least an iff statement) for d euc (P,Q)= d euc (P',Q')
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    Oct 26, 2010 6:39 AM GMT
    LOVALOT said *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| • *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ *<|=^D ▁▂▃▄ l|̲̅̅●̲̅̅|̲̅̅=̲̅̅|̅●̲̅̅| ▄▃▂▁ १|˚-˚|५ !!!!!!!˚∆˚!!!!!!! ☀‿☀♫ ˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙˙❤‿❤˙ ◠◡◠|〇◡〇|◠◡◠ ✌ { ◕ ‿ ◕ } ✌ ༗_ゝ༗ [◠し_◠ฺ ] ☛`ิิ‿ゝ´ิ ☛ ªʘ̀‿‿ʘ̀ª |'L'| •❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ❐_❑• ✌▓▒░ ❤‿‿❤░▒▓✌ ✌⊂(✰‿✰)つ✌ ☜Ҩ.¬_¬.Ҩ☞ ◠


    I disagree with your Mel Gibson statements.
  • ursa_minor

    Posts: 566

    Oct 26, 2010 7:13 AM GMT
    what planet are you from?


    my brain hurtsicon_rolleyes.gif
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    Oct 26, 2010 7:21 AM GMT
    What was part c? Hint might be useful.

    edit: nvm, just saw that you already got d
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    Oct 26, 2010 7:34 AM GMT
    Wow this thread seriously turns me on.

    Next semester I will be covering these subjects in an advanced modern physics class--I've seen problems like this before but do not yet know how to solve them.

    Sorry but I have to go jack off now! lol
  • cromi

    Posts: 489

    Oct 26, 2010 1:14 PM GMT
    i consider myself as an ok pure math student and judging by the question... i think ur... screwed lol.

  • Ritournelle

    Posts: 134

    Oct 26, 2010 3:54 PM GMT
    Ugh, I didn't finish the assignment, but I got a little bit further!
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    Oct 27, 2010 7:28 AM GMT
    (You didn't define X+. Is that the half-plane consisting of points (x, y) with x >= 0?)

    Doesn't part (e) follow from the triangle inequality? If your notation is correct, then ldeuc(w') by definition is the sum of the travel distances Sum {deuc(Pi', Qi')}, and we're done. On second thought, since we're talking about Euclidean distance, the inequality ldeuc(w') >= deuc(P, Q) holds for any discrete walk between points P and Q, no matter where the intermediate points (or P and Q themselves) live ...maybe I'm missing something. If part (e) is supposed to be nontrivial then there's an error in notation somewhere.

    EDIT: Duhhh, the inequality ldeuc(w') >= deuc(P, Q) is not synonymous with the triangle inequality, and does not hold for every discrete walk. Here's how to prove (e): Write deuc(P,Q) as deuc(P1', Qn'). The triangle inequality asserts deuc(P1', Qn') <= deuc(P1', Q1') + deuc(Q1', P2') + deuc(P2', Q2') + ... + deuc(Qn-1',Pn') + deuc(Pn', Qn'). For discrete walk w' all the terms of the form deuc(Qi-1', Pi') are zero because Qi-1' = Pi' by construction, and we're left with the sum deuc(Pi', Qi'), which is by definition ldeuc(w').