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MathGroup Archive 2001

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Re: Who can help me?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26832] Re: [mg26778] Who can help me?
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Thu, 25 Jan 2001 01:13:13 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

Here is one other possible approach to this problem besides the ones
suggested so far. I think it is more expensive in terms of time, but because
it makes no use of high precision arithmetic I thought it worth mentioning.
Bacically, the idea is to use Mathematica's powerful algebra functions to
simplify the expression as much as posible before applying N:

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;

In[2]:=
p = First[RootReduce[FunctionExpand[2 + 2*Cos[(2*Pi)/7]]]][x]

Out[2]=
              2    3
-1 + 6 x - 5 x  + x

In[3]:=
q = PolynomialReduce[poly, p][[2]]

Out[3]=
                                                        2
-883889576185949 + 840711507944414 x - 175083482624375 x

In[4]:=
N[q /. x -> 2 + 2 Cos[2Pi/7] // FunctionExpand // RootReduce]

Out[4]=
0.00108056

-- 
Andrzej Kozlowski
Toyama International University
JAPAN

http://platon.c.u-tokyo.ac.jp/andrzej/
http://sigma.tuins.ac.jp/

on 1/22/01 5:10 PM, Jacqueline Zizi at jazi at club-internet.fr wrote:

> 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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