Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: How to Plot this equation ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg47121] Re: How to Plot this equation ?
  • From: drbob at bigfoot.com (Bobby R. Treat)
  • Date: Fri, 26 Mar 2004 03:56:23 -0500 (EST)
  • References: <c3uekb$9u0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Solving for y doesn't work, but we can solve for x and transpose the
resulting graph. x in terms of y involves ProductLog, which has a
discontinuity at -1/E, so I locate the y values that get us to that
value, then restrict the function to keep it "real". After drawing x
as a function of y, I reverse points in the Line directive to
transpose the graph.

eqn = x*Exp[x*y + 0.8] + Exp[y^2] == 3;
Off[InverseFunction::ifun, Solve::ifun]
inverse[y_] = x /. First[Solve[eqn, x]]
yLimits = (y /. FindRoot[Cases[inverse[y], ProductLog[a_] :> a == -1/
  E], {y, #1}] &) /@ {0, 1}
invPlot = Plot[inverse[y], Evaluate[{y, Sequence @@ yLimits}]];
Show[invPlot /. Line[a_] :> Line[Reverse /@ a]];

(1/y)*ProductLog[
   9.080790118821775*^-9*
    (1.48443789*^8 - 
     4.9481263*^7*E^y^2)*y]
{-0.4657975933929999, 
  1.1455559242943212}

Bobby

m004202002 at yahoo.com (why) wrote in message news:<c3uekb$9u0$1 at smc.vnet.net>...
> Hi,
> 
> How I can Plot tis equation with mathematica?
> 
> x exp(xy+0.8) + exp(y^2)=3 
> 
> Please help me .
> 
> Thank You .


  • Prev by Date: Re: usage / "More..."
  • Next by Date: RE: Animation
  • Previous by thread: RE: How to Plot this equation ?
  • Next by thread: help me