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Re: How to get an answer as a Root object?
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 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul http://InternationalMathematicaSymposium.org/IMS2005/