       Re: problem getting the area under a parmetric curve

• To: mathgroup at smc.vnet.net
• Subject: [mg52860] Re: problem getting the area under a parmetric curve
• From: Roger Bagula <tftn at earthlink.net>
• Date: Wed, 15 Dec 2004 04:26:20 -0500 (EST)
• References: <cpauof\$ipf\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Another of these distribution like pairs is:
{(1-t^7)/(1+t^7),2^1/6*t*(7+35*t^14+21*t^28+t^42)^1/7/(1+t^7)}
When fitted in the previous area programs,
my computer doesn't compute the area of the normalization factor.
n=6, 8 are again a squared circles.
It appears that these can be done for general n case
and the odd ones give the distributioin like parametric.
Roger Bagula wrote:

>I worked on this late last night.
>I had trouble even having the curve well defined
>finding the area under it.
>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.
>
>(* cubic pair and rotated distribution function*)
>Clear[x0,y0,ang,x,y,z]
>x0=(1-t^3)/(1+t^3)
>y0=2^(1/3)*t*(3+t^6)^(1/3)/(1+t^3)
>ParametricPlot[{x0,y0},{t,-2*Pi,2*Pi}]
>Simplify[x0^3+y0^3]
>ang=4
>x=Cos[Pi/ang]*x0-Sin[Pi/ang]*y0
>y=Cos[Pi/ang]*y0+Sin[Pi/ang]*x0
>NSolve[x==0,t]
>N[y/.t->0.486313]
>N[x/.t->0.486313]
>ParametricPlot[{x,y},{t,-2*Pi,2*Pi}]
>f[t_]=x
>norm=Integrate[-y*f'[t],{t,0.486313,Infinity}]
>a0=N[2*norm]
>g1=ParametricPlot[{x,y}/(a0),{t,-2*Pi,2*Pi}]
>g2=Plot[Exp[-t^2/2]/Sqrt[2*Pi],{t,-Pi,Pi}]
>Show[{g1,g2}]
>Respectfully, Roger L. Bagula
>
>tftn at earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
>alternative email: rlbtftn at netscape.net
>
>
>

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