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MathGroup Archive 2004

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Re: problem getting the area under a parmetric curve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52837] Re: problem getting the area under a parmetric curve
  • From: Roger Bagula <tftn at earthlink.net>
  • Date: Tue, 14 Dec 2004 05:59:38 -0500 (EST)
  • Reply-to: tftn at earthlink.net
  • Sender: owner-wri-mathgroup at wolfram.com

Dear Narasimham,
I have been able to get answers to these nth pair problems
and so far two of the odd pairs have given distribution type
of answers ( n=4 gives a squared like ellipse/ circle
{(1-t^4)/(1+t^4,2^3/4*t*(1+t^8)^1/4/(1+t^4)})
This result appears to be new geometry.  Most elliptic based solutions
also seem to make measure assumptions?

While doing projective pairs research on the
Fermat type measures, I went to the definition of curvature at math world
and found it was made by Gauss/ Riemann and it assumed a quadratic type
measure:
m(x,y)=Sqrt[x2+y2]
or a Riemanian metric of the type:
ds2=a*dx2+b*dx*dy+c*dy2
It is my impression from reading  that generalized geometry
will allow such measues as:
M(,x,y,n)=(x^n+y^n)^(1/n)
on metrics that have a form with ( n+1) terms:
ds^n=Sum[a[i]*dx^i*dy^(n-i),{i,0,n}]
and n Gaussian curvatures analogs of the form:
K(n)=d^nf[x]/dx^n/(1+(df[x]/dx)^n)^((n+1)/n)
I haven't checked this equation completely,
but it has the right form for what the answer should be.
K(n)=1/a^(n-1)
(so that K(2)=1/a)
would be the sphere analogs in the higher measure spaces.
I don't think that using a K(2) type curvature will work well with my
n based pair functions?
Narasimham wrote:

 >Roger Bagula wrote:
 >
 >
 >
 >>I used the symmetry to integrate the side
 >>that was easiest. There is only one real zero for the x parametric
 >>in t  which helps. An Infinite integral does appear to exist for
 >>the distribution.
 >>
 >>
 >...
 >
 >Dear Roger,
 >
 >May be problem with upper limit -> Infinity,hypergeometric function
 >singularity as an improper integral for norm. Did you also try using
 >NDSolve?
 >
 >If you have already not corrected it( :)next morning),the following may
 >work partially if stopped ahead of Infinity at a large enough number.
 >
 >norm = Integrate[-y*f'[t], {t, 0.486313, 100}]
 >a0 = N[Abs[2*norm]]
 >g1 = ParametricPlot[{x, y}/(a0), {t, 0.486313, 100}]
 >g2 = Plot[Exp[-t^2/2]/Sqrt[2*Pi], {t, -Pi, Pi}]
 >Show[{g1, g2}]
 >
 >Regards,  Narasimham
 >
 >
 >

-- 
Respectfully, Roger L. Bagula
tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 
619-5610814 :
alternative email: rlbtftn at netscape.net
URL :  http://home.earthlink.net/~tftn







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