Re: How to get an answer as a Root object?
- To: mathgroup at smc.vnet.net
- Subject: [mg57150] Re: How to get an answer as a Root object?
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Thu, 19 May 2005 03:08:20 -0400 (EDT)
- Organization: The University of Western Australia
- References: <d6bv8r$pcu$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <d6bv8r$pcu$1 at smc.vnet.net>,
"David W. Cantrell" <DWCantrell at sigmaxi.org> wrote:
> This question must have an obvious answer!
Possibly ...
> 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".
I expect that this is because the required intermediate simplifications
are not being done.
> 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?
You can use FindRoot to high precision together with
NumberTheory`Recognize` to show that
Root[Function[t, 256 t^12 - 3072 t^11 + 15872 t^10 - 46080 t^9 +
81425 t^8 - 86152 t^7 + 43568 t^6 + 10832 t^5 - 32136 t^4 +
21024 t^3 - 6112 t^2 + 512 t + 64], 8]
is the value that you are after.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
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