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Re: Is it feasible? ver 7.

  • To: mathgroup at smc.vnet.net
  • Subject: [mg105407] Re: [mg105367] Is it feasible? ver 7.
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 1 Dec 2009 04:15:40 -0500 (EST)
  • References: <200911301108.GAA13110@smc.vnet.net>

On 30 Nov 2009, at 20:08, ynb wrote:

> Find the minimal polynomial p[z] of
>
> s=Sqrt[2] + 3^(1/3) + 5^(1/4) + 7^(1/5) +
> 11^(1/6) + 13^(1/7) + 17^(1/8) + 19^(1/9) over Q.
>
> RootReduce[s]=_________(<----Is it feasible?
>
>                       mathematica ver 7)
>
> p[s]=________.(=0 <----Is it feasible?
>                  mathematica ver 7)
>
> =
--------------------------------------------------------------------------
-------
>
> Find the minimal polynomial p[z]
> p[z]=_________________________________________
>
>
>
>
>
>

I think none of these questions can be answered with any confidence. 
Getting exact answers to these questions is obviously going  to be very 
hard - but the answer also depends on such things like do you have a 
super-computer at your disposal etc.

You can certainly get approximate answers, good to a very high degree of 
accuracy, which may be OK for certain questions and if you are lucky 
that might even be the exact answers (but this is unlikely in this 
case). For example:

s = Sqrt[2] + 3^(1/3) + 5^(1/4) + 7^(1/5) + 11^(1/6) + 13^(1/7) +
   17^(1/8) + 19^(1/9);
v = RootApproximant[N[s, 500]];

Now

f = MinimalPolynomial[v, x]

will give you a polynomial of degree 40 with large integer coefficients. 
This is an irreducible polynomial and s is very close to being a root of 
it:

 N[f /. x -> s, 500]

During evaluation of In[15]:= N::meprec: Internal precision limit 
$MaxExtraPrecision = 50. reached while evaluating 
-465083107727-77132702464 (

 1.065716524368700929644172333737165712466626785244845*10^-445

so you have got an extremly small residue but still this is unlikely to 
be the maximal polynomial. You can see what sort of precision would be 
needed by looking at a simpler example. Take

s = Sqrt[2] + 3^(1/3) + 5^(1/4);

In this case RootReduce works fast:
 f = RootReduce[s]

 Root[#1^24-24 #1^22-24 #1^21+234 #1^20+288 #1^19-1388 #1^18-1944 
#1^17+9927 #1^16-57960 #1^15-38928 #1^14+445824 #1^13+2162 #1^12-1070568 
#1^11-2510400 #1^10-1856328 #1^9+7229025 #1^8+8106696 #1^7+14935732 
#1^6-10731960 #1^5-31538628 #1^4-12201072 #1^3-2666976 #1^2+54232416 
#1-30291344&,4]

Using the approximate method with 100 digits of precision will not give 
the right answer:

 f1 = RootApproximant[N[s, 100]]

Root[#1^40-4 #1^39-#1^38-#1^37-5 #1^36-4 #1^35+3 #1^34+4 #1^33+2 #1^32-3 
#1^31+28 #1^30-2 #1^29-23 #1^28-55 #1^27+28 #1^26-25 #1^25+#1^24-86 
#1^23+54 #1^22-39 #1^21+69 #1^20+97 #1^19+16 #1^18+52 #1^17-44 #1^16-91 
#1^15+82 #1^14+40 #1^13+68 #1^12+22 #1^11-91 #1^10-18 #1^9+114 #1^8-42 
#1^7+6 #1^6+41 #1^5+53 #1^4+38 #1^3-128 #1^2-18 #1+86&,2]

500 digits is enough

f2 = RootApproximant[N[s, 500]]

Out[24]= Root[#1^24-24 #1^22-24 #1^21+234 #1^20+288 #1^19-1388 
#1^18-1944 #1^17+9927 #1^16-57960 #1^15-38928 #1^14+445824 #1^13+2162 
#1^12-1070568 #1^11-2510400 #1^10-1856328 #1^9+7229025 #1^8+8106696 
#1^7+14935732 #1^6-10731960 #1^5-31538628 #1^4-12201072 #1^3-2666976 
#1^2+54232416 #1-30291344&,4]

 f2 == f
True

Trying only 400 also works

 f3 = RootApproximant[N[s, 400]]

Root[#1^24-24 #1^22-24 #1^21+234 #1^20+288 #1^19-1388 #1^18-1944 
#1^17+9927 #1^16-57960 #1^15-38928 #1^14+445824 #1^13+2162 #1^12-1070568 
#1^11-2510400 #1^10-1856328 #1^9+7229025 #1^8+8106696 #1^7+14935732 
#1^6-10731960 #1^5-31538628 #1^4-12201072 #1^3-2666976 #1^2+54232416 
#1-30291344&,4]

f3 == f

True

You can see that once you get the right answer it will not change as you 
increase the precision (at least I hope so). So in this way you could be 
(almost) sure when that you got the exact answer. But in your example 
the required number of digits is likely to be huge so using this 
approximate method may be no better than using the exact one.

Andrzej Kozlowski=


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