Re: request help
- To: mathgroup at smc.vnet.net
- Subject: [mg116207] Re: request help
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 5 Feb 2011 05:44:49 -0500 (EST)
You'd have call Solve this way:
Solve[Dis[r, n] == z, r]
but Mathematica (of course) cannot solve it. I doubt that god could.
You might try an approximation such as
series = Normal@Series[Dis[r, n], {r, 0, 5}];
Solve[series == z, r]
but that may not be useful to you.
Otherwise, you'll have to give n and z numeric values and use NSolve or
FindRoot.
Bobby
On Fri, 04 Feb 2011 00:41:22 -0600, Berihu Teklu <berihut 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]
>
--
DrMajorBob at yahoo.com