[Date Index]
[Thread Index]
[Author Index]
How to get an answer as a Root object?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg57137] How to get an answer as a Root object?
*From*: "David W. Cantrell" <DWCantrell at sigmaxi.org>
*Date*: Tue, 17 May 2005 01:20:00 -0400 (EDT)
*Sender*: owner-wri-mathgroup at wolfram.com
This question must have an obvious answer!
A certain expression, given below my signature, involves nothing worse than
some nested square roots. It equals 0 for a value of the variable s near
2.707, easily approximated using FindRoot. However, I would like to get
that value expressed precisely as a Root object. Trying to use Solve or
Reduce to do that, Mathematica seems to take "forever".
Of course, I could do this by hand; in the days before CASs -- before I got
spoiled by them! -- I wouldn't have hesitated to do so. But surely there
must be an easy way to get Mathematica to give that value as a Root object.
How?
David Cantrell
(-1 + Sqrt[2] - Sqrt[(1 + 2*Sqrt[2] - s)*(-1 + s)] + s)*(-1 - Sqrt[2] +
Sqrt[2]*s - Sqrt[-1 - 2*Sqrt[2] + 2*(2 + Sqrt[2] - s)*s]) -
(-1 - Sqrt[2] + Sqrt[2]*s + Sqrt[-1 - 2*Sqrt[2] + 2*(2 + Sqrt[2] - s)*s])*
(-2 - 2*Sqrt[2] + 2*s + 2*Sqrt[1 - (1/4)*(-1 - Sqrt[2] + Sqrt[(1 +
2*Sqrt[2] - s)*(-1 + s)] + s)^2])
Prev by Date:
**Re: UpSetDelayed and N**
Next by Date:
**Re: Show two vector fields with different arrow colors**
Previous by thread:
** Re: Modifying displayed form of an expression?**
Next by thread:
**Re: How to get an answer as a Root object?**
| |