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

```

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