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 .