Author 
Comment/Response 
David

03/18/08 3:41pm
Hi all,
I am trying to fit parameters in an equation that has no closed form solution. I am trying to deal with this by using a FindRoot function rather than an explicit equation in the equation part of the FindFit command. It appears as if FindFit is not correctly passing the guess values of the parameters back to FindRoot; at least that is my interpretation of what is going on. Any advice for how to deal with this problem would be appreciated. More details are below, and the program is attached. Thanks
The governing equation is:
Rho == 1  Exp[((\[Rho]^2 + Pred)/Tred)  \[Rho] (1  1/r + vH)]
I have data for Rho that I wish to fit this equation to by adjusting parameters Pred, Tred,, vH, and r.
In order to do this, I am trying to solve the equation with FindRoot inside of the FindFit function, as follows.
FindFit[data,
dens /. FindRoot[\[Rho] == EOS, {dens, 1000}], {{\[Rho]star,
853}, {Pstar, 475}, {Tstar, 518}, {So, 16.6}}, {T}].
where EOS is the right hand side of the equation given above. However, I am getting the following error:
FindRoot::nlnum: "The function value {1.+2.71828^(((1. Tstar \
(101350. Power[<<2>>]+1.*10^6 Power[<<2>>]))/T)(1000. \
(1.<<1>>+<<1>>))/\[Rho]star)+1000./\[Rho]star} is not a list of \
numbers with dimensions {1} at {dens} = {1000.`}."
ReplaceAll::reps: "{FindRoot[\[Rho]==EOS,{dens,1000}]} is neither a \
list of replacement rules nor a valid dispatch table, and so cannot \
be used for replacing."
Is it not possible to combine these two commands this way? If so, why not, and is there a way for me to alter the FindFit function to make this work?
Thanks
Attachment: water fitting.nb, URL: , 
