Re: request help

*To*: mathgroup at smc.vnet.net*Subject*: [mg116206] Re: request help*From*: Gary Wardall <gwardall at gmail.com>*Date*: Sat, 5 Feb 2011 05:44:38 -0500 (EST)*References*: <iig76m$rl2$1@smc.vnet.net>

On Feb 4, 12:43 am, Berihu Teklu <beri... at gmail.com> wrote: > I need to invert a real function of two real variables Dis[r, n] with > respect to the first variable r, while the second variable n is fixed. > The function is rather difficult, that I couldn't invert it. this is > kindly request you to write me any comments on the attached notebook. > > Many thanks for any help, > > Berihu > > Dis[r_, n_] := > 1/4 (2 (-2 + Sqrt[(1 + 2 n)^2]) Log[-2 + Sqrt[(1 + 2 n)^2]] - > 2 (2 + Sqrt[(1 + 2 n)^2]) Log[ > 2 + Sqrt[(1 + 2 n)^2]] - (-2 + > Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]) Log[-2 + > Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]] + (2 + > Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]) Log[ > 2 + Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]] - (-2 + > Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]) Log[-2= + > Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]] + (2 + > Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]) Log[ > 2 + Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]]) > > Solve[1/4 (2 (-2 + Sqrt[(1 + 2 n)^2]) Log[-2 + Sqrt[(1 + 2 n)^2]] - > 2 (2 + Sqrt[(1 + 2 n)^2]) Log[ > 2 + Sqrt[(1 + 2 n)^2]] - (-2 + > Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]) Log[-2 + > Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]] + (2 + > Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]) Log[ > 2 + Sqrt[(1 + 2 n)^2 Cosh[2 r]^2]] - (-2 + > Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]) Log= [-2 + > Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]] + (= 2 + > Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]) Log= [ > 2 + Sqrt[((1 + 2 n)^2 (1 + 2 n - > 2 Cosh[2 r])^2)/(-2 + (1 + 2 n) Cosh[2 r])^2]]) = == > Dis[r, n], r] Since you can't explicitly solve for r perhaps a numerical method might be of help. For any specific value of n you could try NSolve. Perhaps you could create program to solve a series of equations numerically. Another possibility would be to solve a series of of ODE's. Here's a variation of a problem I worked on earlier today. Given: Sin[x+y+n]+y=x Solve for y. Solution: Let yp be the first derivative of y with respect to x. Then: Cos[x+y+n]*[1+yp]+yp=1 Solving for yp: Cos[x+y+n]+Cos[x+y+n]*yp+yp=1 Cos[x+y+n]+(Cos[x+y+n]+1)*yp=1 (Cos[x+y+n]+1)*yp=1-Cos[x+y+n] yp=(1-Cos[x+y+n])/(Cos[x+y+n]+1) Now find a point on the graph of y, numerically or otherwise: Given particular of x and n find a corresponding value for y. Suppose we have x0,n0, and y0. That is solve for y0 where: Sin[x0+y0+n0]+y=x0 Next solve the a series of ODE's: yp=(1-Cos[x+y+n0])/(Cos[x+y+n0]+1) when y(x0)=y0 and Sin[x0+y0+n0]+y=x0 I hope this will help. Good Luck Gary Wardall