On Tue, Jan 5, 2010 at 5:25 PM, Carl Witthoft <c...@witthoft.com> wrote:
> quote:
> > There are certainly formulas for solving polynomials numerically up to
> 4th degree non-iteratively, but you will almost certainly get better results
> using iterative methods.
>
> I must be missing something here. Why not use the analytic formulas for
> polynomials below 5th degree? Once you do so, your answer is as precise as
> the level of precision you enter for the coefficients.
Why do you believe that? Are you assuming you can perform *exact*
arithmetic? Did you read the references I gave?
* George Forsythe, How do you solve a quadratic equation?
* Yves Nievergelt, How (Not) to Solve Quadratic Equations
They show that that isn't even true for quadratic equations without a lot of
care.
Let's try a cubic:
p = 1000000*x^3-998000*x^2-1001999*x+999999
That factors exactly over the integers to:
(x-1001)*(x-1000)*(x-999)
but plugging the floating-point coefficients (which are exactly
representable as floats) into (one version of) the cubic formula (using
Maxima), I get the roots
x = 966.1329834413779+58.65086897690403i
x = 966.1329834413779-58.65086897690403i
x = 1067.734033117244
On the other hand, using an interative approach, I get:
x = 999.0000000278754
x = 999.9999926817675
x = 1001.000007290357
Which looks better to you?
-s
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
