Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2001
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2001

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Who can help me? (Instabilty of Evaluation)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26879] Re: [mg26778] Who can help me? (Instabilty of Evaluation)
  • From: Roland Franzius <Roland.Franzius at uos.de>
  • Date: Fri, 26 Jan 2001 01:27:24 -0500 (EST)
  • Organization: Universitaet Osnabrueck
  • References: <001801c084d2$ac9023e0$a7e3e994@dqb2301> <94og7i$ec9@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi again

I was a bit confused when I recognized to be the unique poster with a
different result.
But now, I think I am right. Otherwise I will take a lesson in
approximations. 
Lets do it again:

In:
f[x_] := -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;  

Argument is half angle Cos

In:
y = 2 + 2 Cos[Pi/7];
z = 4 Cos[Pi/14]^2;
y - z // Simplify

Out:	
	0

This looks much more promising

In:
Coeff=CoefficientList[f[4 x], x]

Out:
{0, -71232786711432720, 2223725831049492000, \
-33747509155258917632, 333235768384884453888, \
-2414751330222278023168, 13712555420461799161856, \
-63541731432065355563008, 246714624908086456811520, \
-817187969499268347592704, 2338191870760035165405184, \
-5831646763362143982059520, 12764017106300237787430912, \
-24645123767847232884178944, 42149256443238706956992512, \
-64049533236363795426705408, 86668805002116669993123840, \
-104561005329592884588969984, 112494310494722410497966080, \
-107834631503058167115284480, 91907120293944836803788800, \
-69413102111596198620037120, 46233508633837346171650048, \
-26984081358159822356217856, 13685006274531853720354816, \
-5965042363660181805465600, 2202690401620578973777920, \
-675834011146843183906816, 167697332761460471758848, \
-32339448204222057676800, 4548275335674011320320, \ 
-415051741658464911360, 18446744073709551616}

In:
FactorInteger[Coeff[[-1]], GaussianIntegers -> True]
Out:
{{1 + I, 128}}

In:
ww = TrigFactor[f[ 4 Cos[Pi/14]^2]]

Out:
-3*(-1)^(3/7)*(-I + (-1)^(1/14))^2*(I + (-1)^(1/14))^2*
  (-I + (-1)^(1/7))*(I + (-1)^(1/7))*(1 - (-1)^(1/14) + 
   (-1)^(1/7))*(1 + (-1)^(1/14) + (-1)^(1/7))*
  (-1 + (-1)^(1/7) - (-1)^(4/7))*(-1 + (-1)^(1/7) + (-1)^(4/7))*
  (144028488250734 - 143014322039070*(-1)^(1/7) + 
   140735512734731*(-1)^(2/7) - 138908051041337*(-1)^(3/7) + 
   138908051041337*(-1)^(4/7) - 140735512734731*(-1)^(5/7) + 
   143014322039070*(-1)^(6/7))

In:
N[ww]

Out:
156012.  - 0.360791 I

In:
N[ww, 100]

Out:
156010.480152622649895682270126021905222214727164022359842011452550020304833769506087\
975757897182836878984012952`100 -0*^-137 I


So this will be a simple Fourier sum

In:
qq1 = (f[4 Cos[Pi/14]^2] // TrigToExp) // ExpToTrig 

Out:
-4210946951736 + 189057241621814*Cos[Pi/7] - 
  128092274251632*Cos[(2*Pi)/7] + 42260253935594*Cos[(3*Pi)/7] + 
  49848112491644*Cos[(4*Pi)/7] - 129966324569880*Cos[(5*Pi)/7] + 
  183806279699252*Cos[(6*Pi)/7]

In:
N[qq1, 100]

Out:
156010.48015262264989568227012602190522221472716402235984201145255002030483376\
95060879757578971828369

Now Cos[n Pi/7]= -Cos[ (7-n)Pi/7] making two cancellations.

This gives unexpectedly 


In:
(f[2 + 2Cos[Pi/7]] // TrigToExp) // ExpToTrig//FullSimplify


Out:
Root[-15098642952733854197440281 + 94480339550807903553 #1 + 
        14738314331076 #1^2 + #1^3 &, 3]

oops?

In:
N[Solve[-15098642952733854197440281+ 94480339550807903553 x + 
          14738314331076 x^2 + x^3 == 0, x], 50])

Out:

{{x -> 156010.480152622649895682270126021905222214727164022359842\
01145162732`50 + 2.08821`0*^-100*I}, 
  {x -> -1.473830792054766224316657196353304219511536013659192473\
68303829106`50*^13 + 8.33719`0*^-47*I}, 
  {x -> -6.566538817909456077932149227930906545085622802427191825\
309514007`50*^6 + -6.214281`0*^-90*I}}


If you want to plot the function f on a logarithmic scale, take the
following expression one in order to convert numbers to rationals before
Evaluation in all places 

Plot[
ArcSinh@
    f[N[Rationalize[x, 10^-50] + 2 + 
          2 Rationalize[Cos[Rationalize[Pi, 10^-50]/7], 10^-50], 50]], 
{x, -1,  1}   ]

However, you have to trust the rational approximations Mathematica does
internally while expanding the Cos and Pi


Regards 
Roland


Jacqueline Zizi wrote:
> 
> Thanks to everybody that answers me. Tomas Garza points out very deep important
> facts (from my intuitive point of view).
> 
> The Tomas Garza  property (see below) looks indeed very important and might
> help to solve an old well known open problem in this field.
> 
> Here is now some explanations about this polynomial. Coloring a graph G with
> n-vertices means coloring the vertices such as deux vertices linked can't take
> the same color. The function x colors -> number of ways of coloring G with x
> colors, is shown to be a polynomial and it is called the chromatic polynomial
> of G. Notation p(x). For example for a complete graph p(x)= x(x-1)... (x-n+1).
> 
> The 4th- color problem is: the chromatic polynomial of every planar graph has
> an integer value X less or equal 4, such that p(X)>0.  For non planar graphs,
> this first value such that p(X) >0  can be higher. For example for a complete
> graph this number is n; Notice that indeed in this special case, for 1 to n-1
> it is 0.
> 
> There is 2 well known ways for prooving of the 4th-color problem. The one done
> and the other one by chromatic polynomials. People gave up facing large
> polynomials and the NP-Hard difficulty of computing the polynomials.
> Nevertheless, properties of these invariants and some more general invariants
> generalizations of it are promising in different areas.
> 
> Now one interesting graph configuration for lots of confluent reasons from
> different areas (group's theory, chemistry for example), is the truncated
> icosahedron. Hall and his students took more than 15 years to find the
> corresponding chromatic polynomial. This is our beast.
> 
> If still reading this, notice that there is a more important fact :  this Tomas
> Garza property is also true, in more general situations. Indeed I'm just
> looking at this and  if we believe Mathematica,  this is glorious!
> 
> .................................................
> 
> Nevertheless my main questions remains:
> 
> 1) How could I check the results given by Mathematica?
> 
> 2) If people trust the proof of the 4th color problem, why should not they
> trust this too?
> 
> 3) Is there a mean to proove that in a mathematical way?
> 
> Jacqueline Zizi
> 
> Tomas Garza wrote:
> 
> > Further comments to my previous message. Other than the numerical problem, I
> > found an interesting (or so it seems to me) fact. I haven't the faintest
> > idea what your original problem is about, but the polynomial has the
> > property that if you take the first j terms, then its numerical value for
> > x-> 2 + 2 Cos [2 Pi / 7] is the negative of the numerical value of the last
> > 32 - j terms for that same x, for j = 1 to 31.


-- 
Roland Franzius

  +++ exactly <<n>> lines of this message have value <<FALSE>> +++


  • Prev by Date: Re: Who can help me?
  • Next by Date: Re: Q: lefthand/right hand side of equation
  • Previous by thread: Re:
  • Next by thread: Any better way for finding frequencies of list entries?