       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)
• 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
619-5610814 :
alternative email: rlbtftn at netscape.net