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MathGroup Archive 2001

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Re: Who can help me?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26783] Re: Who can help me?
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Wed, 24 Jan 2001 04:18:29 -0500 (EST)
  • References: <94gslj$r0h@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Jacqueline,
Machine precision is not good enough

But, since this is an exact expression, we can ask N to try and return the
answer with precision, say 30, which must be greater then machine precision.
Thus

poly = -17808196677858180 x + 138982864440593250 x^2 -
      527304830550920588 x^3 + 1301702220253454898 x^4 -
      2358155595920193382 x^5 + 3347791850698681436 x^6 -
      3878279506351645237 x^7 + 3764566420106299695 x^8 -
      3117324712750504866 x^9 + 2229873533973727384 x^10 -
      1390372935143028255 x^11 + 760794705528035032 x^12 -
      367240961907017721 x^13 + 157018216115380477 x^14 -
      59650776196609992 x^15 + 20179153653354540 x^16 -
      6086251542996201 x^17 + 1637007669992780 x^18 - 392300104078670 x^19 +
      83589038962550 x^20 - 15782712151030 x^21 + 2628070696678 x^22 -
      383466859804 x^23 + 48618908986 x^24 - 5298021900 x^25 +
      489095520 x^26 - 37516324 x^27 + 2327268 x^28 - 112200 x^29 +
      3945 x^30 - 90 x^31 + x^32;


N[poly /. x -> 2 + 2 Cos[2Pi/7], 30]

0.00108056072343889043621555323644

Please note that if we ask for precision <= machine precision then machine
arithmetic is used and the result is not guaranteed.

--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Jacqueline Zizi" <jazi at club-internet.fr> wrote in message
news:94gslj$r0h at smc.vnet.net...
> I'm working on this polynomial linked to the truncated icosahedron:
>
>         -17808196677858180 x +
>         138982864440593250 x^2 - 527304830550920588 x^3 +
>         1301702220253454898 x^4 - 2358155595920193382 x^5 +
>         3347791850698681436 x^6 - 3878279506351645237 x^7 +
>         3764566420106299695 x^8 - 3117324712750504866 x^9 +
>         2229873533973727384 x^10 - 1390372935143028255 x^11 +
>         760794705528035032 x^12 - 367240961907017721 x^13 +
>         157018216115380477 x^14 - 59650776196609992 x^15 +
>         20179153653354540 x^16 - 6086251542996201 x^17 +
>         1637007669992780 x^18 - 392300104078670 x^19 +
>         83589038962550 x^20 - 15782712151030 x^21 +
>         2628070696678 x^22 - 383466859804 x^23 + 48618908986 x^24 -
>         5298021900 x^25 + 489095520 x^26 - 37516324 x^27 +
>         2327268 x^28 - 112200 x^29 + 3945 x^30 - 90 x^31 + x^32;
>
> I'm interested at its value for x-> 2 + 2 Cos [2 [Pi] / 7].
> Taking N [] gives  3.2628184 10^7
>
> But if I simplify  first and then take N[] it gives -0.0390625 +
> 0.0195313 [ImaginaryI]
>
> As it is a polynomial with integer coefficients, and 2 + 2 Cos [2 pi /
> 7] is real too, the result should be real.  So I prefer the 1st
> solution,  but for another reason, I'm not so sure of this result.
>
> A Plot between 3 and 3.5, does not help me  neither to check if the
> value 3.2628184  is good and If I do : polynomial /. x -> 3.2628184
> 10^7, it gives 2.7225238332205106`^240
>
> How could I check the result 3.2628184 10^7 ?
>
> Thanks
>
> Jacqueline
>
>
>




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