[R] Solving 100th order equation
Rolf Turner
r.turner at auckland.ac.nz
Mon May 26 01:48:24 CEST 2008
On 26/05/2008, at 10:38 AM, Hans W. Borchers wrote:
> Rolf Turner <r.turner <at> auckland.ac.nz> writes:
>
>>
>> [...]
>>
>> However uniroot(p,c(1.995,2.005)) gives
>>
>> $root
>> [1] 1.999993
>>
>> $f.root
>> [1] -4.570875e+24
Whoops! I read that e+24 as e-24, so scrub all that I said.
(You'd think, that having seen Bill Venables make a similar
error --- which he corrected in the follow-up posting to which
I was replying --- I would've been more careful. Well, you'd
think wrong. Actually it *was* e-24 when I posted; then the
Gremlins got in and changed everything. :-) )
>>
>> $iter
>> [1] 4
>>
>> $estim.prec
>> [1] 6.103516e-05
>>
>> What a difference 7.214144e-06 makes! When you're dealing with
>> polynomials of degree 100.
>>
>
> I don't think R is the right tool to solve this kind of questions
> that belong to
> the realm of Computer Algebra Systems. Yacas is much to weak for
> such high-order
> polybomials, but we can apply more powerful CASystems, for instance
> the free
> Maxima system. Applying
>
> (%i) nroots(x^100 - 2*x^99 + 10*x^50 + 6*x - 4000, minf, inf)
>
> immediately shows that there are only *two* real solutions, and then
>
> (%i) a: realroots(x^100 - 2*x^99 + 10*x^50 + 6*x - 4000, 1e-15);
> (%i) float(a)
>
> (%o) [x=-1.074126672042147,x=1.999999999999982]
>
> will provide real solutions with 15 decimals (does not change when
> more decimals
> are used). So the difference that counts is actually much smaller.
>
> The complex roots -- if needed -- will require special treatment.
>
> I believe it would be fair to point such questions to Computer
> Algebra mailing
> lists and not try to give the appearance they could be solved
> satisfyingly with
> R. The same as a complicated statistics question in a Matlab or
> Mathematica
> mailing list should be pointed to R-help.
Be that as it were, it doesn't look like Maxima is doing so wonderfully
either. Consider:
> library(PolynomF)
> x <- polynom()
> p <- x^100-2*x^99+10*x^50+6*x-4000
> a <- 1.999999999999982
> p(a)
[1] -1.407375e+14
And notice the + sign in the exponent. :-)
I tried doing
> uniroot(p,c(1.9995,2.0005),tol=.Machine$double.eps)
and got
$root
[1] 2
$f.root
[1] 912
$iter
[1] 5
$estim.prec
[1] 8.881784e-16
Well, 912 is a lot closer to 0 than -4.570875e+24, or even -1.407375e
+14.
But it still ain't 0. Also that ``2'' of course isn't really 2.
I set r <- uniroot(p,c(1.9995,2.0005),tol=.Machine$double.eps)$root
I get p(r) = 912.
But print(r,digits=22) gives 1.999999999999982.
And print(a,digits=22) gives the identical result.
Although p(r) and p(a) are wildly different.
I guess it's possible that p(a) --- where ``a'' is as it appears,
written out to 14 decimal places --- really is quite close to zero,
and that -1.407375e+14 is due to round-off error in the calculation
process. But it's also possible that p(a) really is pretty close to
-1.407375e+14, i.e. Maxima isn't really finding the correct root
either.
Or something in between.
I guess that to get a really ``correct'' answer, and check it properly,
you need a system that will do ``arbitrary precision arithmetic''.
Which
isn't R. I don't know if Maxima was using arbitrary precision
arithmetic
in the calculations shown.
Anyhow, I think the real point is that solving 100-th degree
polynomials
is a pretty silly thing to do, whatever package or system or
language you use.
That is, R may be the wrong tool for the job, but it's the wrong job.
And if you *must* do it, then you should do some careful checking
and investigation
of the results --- again irrespective of whatever package or system
or language
you use.
cheers,
Rolf Turner
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