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MathGroup Archive 2001

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Re: Solve InterpolatingFunction problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg31908] Re: Solve InterpolatingFunction problem
  • From: "Jasem Mutlaq" <mutlaqja at ku.edu>
  • Date: Tue, 11 Dec 2001 01:33:44 -0500 (EST)
  • References: <9uss9d$cl6$1@smc.vnet.net> <9uvh0a$i18$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Thanks Brian! I already used FindRoot, and since there exists two roots that
I wanted to obtain, I just plotted the curve and made the starting guess
somewhere near the value of interest and was able to get pretty precise
solutions!

Regards,

Jasem Mutlaq.

"Brian Higgins" <bghiggins at ucdavis.edu> wrote in message
news:9uvh0a$i18$1 at smc.vnet.net...
> Jasem"  use FindRoot instead of Solve .  For example
>
> mySin = Interpolation[Table[{x, Sin[x]}, {x, 0, 5, .2}]];
>
> In[4]:=FindRoot[mySin[x] == .5, {x, 1}]
>
> Out[4]={x -> 0.523618}
>
> Cheers,
>
> Brian
>
> "Jasem Mutlaq" <mutlaqja at ku.edu> wrote in message
news:<9uss9d$cl6$1 at smc.vnet.net>...
> > Hello. I obtained an interpolated function from a set of data
representing a
> > velocity curve (increasing & decreasing). I used the following
> >
> > vr =Interpolation[data];
> > Solve[vr[x]==0.6,x]
> >
> > InverseFunction::ifun: Inverse functions are being used. Values may be
lost
> > for multivalued inverses.
> > Out[221]={{x->1.
> > InverseFunction[InterpolatingFunction[{{0.,0.95}},<>],1,1][0.6]}}
> > I'm expecting something like x = 3 or x = 9 (two solutions).
> >
> > Any suggestions are highly appreciated!
> >
> > Jasem Mutlaq.
>






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