Newsgroups: geometry.research
From: haoyuep@aol.com (Dan Hoey)
Date: 12 Jan 2001 11:52:22 -0500
Subject: Circumference to diameter ratios in Lp metrics

Recall that for 1 <= p <= infinity there is a metric on R^2 called
Lp, for which the distance is given as

  d(p,q) = (|p_x-q_x|^p + |p_y-q_y|^p) ^ (1/p).

In the L2 (Euclidean) metric, the ratio of a disc's circumference to
its diameter is Pi.  But in the L1 (taxicab) metric the ratio is 4.
The Linfinity metric is proportional to L1 rotated 90 degrees, so
of course its ratio is 4 also.

In the Lp metric the ratio is

              Pi/4            p(p-1)          p(p-1)
                       (cos t)       + (sin t)         (1/p)
  R(p) =  4 Integral ( ----------------------------- )        dt
                                p           p  (p+1)
                0      ( (cos t)   + (sin t)  )

according to Mathematica.  The only p for which I've been able to
integrate R(p) to anything recognizable are 1, 2, 1/2, 1/3, -1, and
-1/2, though I haven't gone seriously into the inverse symbolic
calculator.  Not that L_p is a metric for p<1 , but it's still neat
that

  R(1/3) = 4 + 8 Gamma[4/3] Gamma[5/3] ,
  R(1/2) = 4 + Pi ,
  R(-1) = Pi ,
  R(-1/2) = 8 Gamma[4/3] Gamma[5/3].

at least numerically. Can any other integraters do anything with this,
either in general or for other special cases?

The other interesting thing I've found is that, numerically at least,

R(p) = R(p/(p-1)) for p>1,
R(p) = R(p/(p-1)) - 4 for 1>p>0, and
R(p) = R(p/(p-1)) + 4 for p<0.

At least the first one is suggested by the L1 disc rotating to the
rescaled Linfinity disc, and the L2 disc being fixed under rotation,
but in general the Lp norm is not just the L_(p/(p-1)) norm scaled
and rotated.  So what gives?

Dan Hoey
haoyuep@aol.com