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RE Graphical strange pairs

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  • Jose Ramón Brox
    ... From: Jose Ramón Brox A last annotation: If we assume x_m, y _m, then the biggest D for a strange pair of (m,n) is: D(m ,n ) =
    Message 1 of 2 , Dec 31, 2005
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      ----- Original Message -----
      From: "Jose Ramón Brox" <ambroxius@...>

      A last annotation: If we assume x_m, y'_m, then the biggest D for a strange pair of (m,n)
      is:

      D(m',n') = D(m+x_m,n-y'_m) = D(G,G) = 2·G^2 + 2·G = 2[m·n+Sqrt(m·n)] and then:

      D2 - D1 = 2[m·n+Sqrt(m·n)] - (2·m·n + m + n) = 2·Sqrt(m·n) - (m+n)

      is the biggest difference in norms expected for a pair (m,n) and it's biggest strange
      pair.

      ==============================================

      Brox reporting from 2006!

      What you see above is a mistake: if you think about the strange pairs the way I do,
      graphically, then we have y in the y-axis and x in the x-axis, with a smooth curve that
      starts at (0,0) an ends at (n-m,0). The values (x,y) that generate psp's are those lattice
      points BELOW the curve (but with positive y). Therefore, the optimal point of the curve
      does not generate the biggest psp, but is the point that has more psp's below itself.

      If you check what I wrote in the other email, you'll see that D(G,G) - D1 is the geometric
      mean of (m,n) minus its arithmetic mean, and this result must be always negative or 0!
      meaning that there would never be psp's in my mistaken interpretation. This comes from
      thinking that (G,G) is the biggest possible psp for (m,n): but it's not.

      What we have to do is to compare n - G with G - m (that is, m-n). If there's more than 1
      of difference, for big numbers then there will be psp for (m,n). For example:

      m = 10^4, n = 4·10^4
      G = Sqrt(10^4·4·10^4) = 2·10^4
      x_m = G - m = 10^4
      y'_m = n - G = 2·10^4

      Since y'_m - x_m >> 1, there are a lot of psp's, we can take any y' such that x_m < y' <
      y'_m, with x_m:

      y' = 15·10^3

      m' = m+x_m = 2·10^4
      n' = n - y' = 25·10^3

      We'll verify that (m',n') is a psp for (m,n):

      1) m' + n' = 2·10^4 + 25·10^3 = 4.5·10^4 > 5·10^4 = 10^4 + 4·10^4 = m + n. OK
      2) D(m',n') = 2·2·2.5·10^8 + (2+2.5)·10^4 = 10·10^8 + 4.5·10^4
      D(m,n) = 2·1·4·10^8 + (1+4)·10^4 = 8·10^8 + 5·10^4 < D(m',n')
      (m',n') > (m,n). OK

      There wil probably be a lot more of psps with other x's different from x_m.

      We knew that y'_m > x_m +1 is a neccesary condition for (m,n) to have psp's, because it's
      the maximum of the curve below which we have to find solutions. But we see that it
      trivially is also a sufficient condition, because if there's one below, then there are
      solutions.

      One implication of this is that the pairs of the form (m-1,m), (m,m) and (m,m+1) don't
      have positive strange pairs: if we want bigger pairs, we must do one of these two things:
      a) Allow the first component to reduce to m', but increasing the second component to n' in
      such a way that m'+n' > m+n and assuring that (m,n) is not a psp for (m',n').
      b) Keep the first component and increase the second, or increase both (remember the
      symmetry).

      But what about nsp's (negative strange pairs)? Well, for these two types it's easy:
      (m-a,m+b) = (m+b,m-a), since (m+b,m-a) is not a psp for (m,m), (m-a,m+b) is not a nsp;
      (m-a,m+b+1) = (m+b+1,m-a) and (m+b+1,m-a) is not a psp for (m,m+1) because it has not
      psp's. A similar argument goes for (m-1,m).

      Let's see what happens around the interval formed by these two pairs, (m,m) and (m,m+1):

      Consider the pairs (m-i,m+j):

      If j-i <=0, then (m-i,m+j) < (m,m), because m-i+m+j < 2m and (m,m) doesn't have sp's.
      If j-i =1, then (m-i,m+j) < (m,m+1) because the mean is the same and the deviance is lower
      for (m,m+1)

      Inside the interval can be other pairs as well, those that are bigger than (m,m) ( (m,m)
      is not a nsp for them) but smaller than (m,m+1) ( (m,m+1) is a psp for them). We may study
      them other day if we find a way to do so.

      I know this is tiring, I'm tired my self :P

      Jose
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