Hi Frank,
thanks for you Answer!
I am not sure if FindRoot is able to do what I want to do. Or I am doing something wrong still, since I am getting cryptic error messages - it doesnt like the starting point term. I will add the code of the concrete function I am given, so it is easier to follow: I have given the relation r = rfunc(R) with
rfunc[R_] := (2 R + M + Sqrt[4 R^2 + 4 M R + 3 M^2])/
4*(((4 + 3 Sqrt[2])*(2 R - 3 M))/(
8 R + 6 M + 3*Sqrt[8 R^2 + 8 M R + 6 M^2]))^(1/Sqrt[2]);
M := 1;
Plot[rfunc[R], {R, 0, 6}]
, and where M = 1.
As one can see from the plot, R goes to 3/2 as r goes to 0, and R goes to infinity as r goes to infinity. Inbetween the function is one-to-one, so it should be possible to invert the function numerically on this domain. So I want an expression R = somefunction.
What I tried before is,
NSolve[rfunc[R]==r,R]
but this does not work.
Can FindRoot do this, and if yes, how? Or can NSolve do this, with specifying additional assumptions on the domain and/or range? If so, how would I do this?
The Mathematica Help to the assumption functions is not really helpfull here for me. : /
