MathGroup Archive 1997

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Re: Roots of an equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg6644] Re: [mg6582] Roots of an equation
  • From: "Preferred Customer" <sherman.reed at worldnet.att.net>
  • Date: Wed, 9 Apr 1997 00:34:05 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

eq=1/x^4 +1/y^2 +2 - ( 1/(y^2 - x^2) )^(5/3)
resx=Table[i,i,-.1,-.01,.01]
res=Table[0,0,i,-.1,-.01,.01]
For[i=1,i<11,i++,
			sol=NSolve[(eq/.x->resx[[i]])==0,y],
				res[[i,1]]=resx[[i]],
				res[[i,2]]=y/.sol[[1]]
]	

Jaimee,

the above code lets me look at the shape of y(x).  varying resx, the x
unknown
lets one look at the code over any range. res is the solutions.

I use 
ListPlot[res, PlotJoined->True] to see the results.

It appears there is no root.  At x=0, y is indeterminate.

I then set y=0 by

eqy=eq/.y->0.

I then use FindRoot[eqy==0,{y,.1} to see if a solution exists.
there is none.

hope this helps.

sherman reed

----------
> From: Jaimee Tahsiri <jtt at deltanet.com>
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> Subject: [mg6644] [mg6582] Roots of an equation
> Date: Friday, April 04, 1997 1:11 AM
> 
> I appreciate it if someone in your group to write me a short mathematica
> program to solve the real roots of this equation.
> 
> 1/x^4 +1/y^2 +2 = ( 1/(y^2 - x^2) )^(5/3)
> 
> A million thanks to anyone that can help me.


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