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MathGroup Archive 2001

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Re: Re: Who can help me?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg26804] Re: Re: [mg26778] Who can help me?
  • From: Tomas Garza <tgarza01 at prodigy.net.mx>
  • Date: Wed, 24 Jan 2001 04:18:43 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

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. That is, simply evaluate

In[1]:=
Table[{Take[a1, j - 32].Take[a2, j - 32] /. x -> 2 + 2 Cos[2.0*Pi/7],
    Take[a1, j].Take[a2, j] /. x -> 2 + 2 Cos[2.0*Pi/7]}, {j, 1, 31}]

Out[2]=
\!\({{5.782285160715823`*^16, \(-5.782285139199467`*^16\)}, \
{\(-1.407456330327663`*^18\),
      1.4074563305476733`*^18}, {1.664350897746699`*^19, \
\(-1.6643508977122853`*^19\)}, {\(-1.2804361627651423`*^20\),
      1.2804361627673385`*^20}, {7.230361392464404`*^20, \
\(-7.230361392461333`*^20\)}, {\(-3.200121970208588`*^21\),
      3.200121970208726`*^21}, {1.1556808440293227`*^22, \
\(-1.1556808440292984`*^22\)}, {\(-3.495373870315845`*^22\),
      3.4953738703158645`*^22}, {9.01004052138053`*^22, \
\(-9.010040521380498`*^22\)}, {\(-2.0035257611223482`*^23\),
      2.00352576112235`*^23}, {3.8768688571383356`*^23, \
\(-3.876868857138334`*^23\)}, {\(-6.570867584057313`*^23\),
      6.570867584057313`*^23}, {9.804286814731637`*^23, \
\(-9.804286814731636`*^23\)}, {\(-1.2929086666244018`*^24\),
      1.292908666624402`*^24}, {1.51129470844226`*^24, \
\(-1.51129470844226`*^24\)}, {\(-1.5688840125551723`*^24\),
      1.5688840125551725`*^24}, {1.447609719431064`*^24, \
\(-1.447609719431064`*^24\)}, {\(-1.18679687228797`*^24\),
      1.18679687228797`*^24}, {8.630908489745314`*^23, \
\(-8.630908489745317`*^23\)}, {\(-5.551191511538765`*^23\),
      5.551191511538765`*^23}, {3.1434645609556606`*^23, \
\(-3.143464560955661`*^23\)}, {\(-1.5575043190979624`*^23\),
      1.5575043190979635`*^23}, {6.696880503554032`*^22, \
\(-6.696880503554029`*^22\)}, {\(-2.4719651614064567`*^22\),
      2.47196516140646`*^22}, {7.721985589213778`*^21, \
\(-7.721985589213745`*^21\)}, {\(-2.0024019788866655`*^21\),
      2.002401978886698`*^21}, {4.195660897089886`*^20, \
\(-4.19566089708956`*^20\)}, {\(-6.827018853514098`*^19\),
      6.827018853517361`*^19}, {8.095823791432835`*^18, \
\(-8.095823791400206`*^18\)}, {\(-6.225166642017306`*^17\),
      6.225166642343588`*^17}, {2.3299464415234008`*^16, \
\(-2.3299464382605824`*^16\)}}\)

I don't know if this has any bearing on your original situation, but I find
it rather surprising that, for example, a polynomial of degree 1 (the first
term of f[x]) has the same absolute value that a polynomial of degree 31,
and so on, for a particular value of x (x-> 2 + 2 Cos [2 Pi / 7]). On the
other hand,

In[3]:=
f[x] == Take[a1, -31].Take[a2, -31] + Take[a1, 1].Take[a2, 1]

Out[3]=
True

which would mean that whenever x =  2 + 2 Cos[2.0*Pi/7], f[x] must be zero.
I hope I am not simply adding to the confusion.

Tomas Garza
Mexico City




----- Original Message -----
From: "Tomas Garza" <tgarza01 at prodigy.net.mx>
To: mathgroup at smc.vnet.net
Subject: [mg26804] Re: [mg26778] Who can help me?


> First, I couldn't get a complex number, as you claim you do by simplifying
> and then taking N[].
> I copied your polynomial straight from your message and pasted it into my
> notebook. I call it f[x]. Then the same for   x-> 2 + 2 Cos [2 [Pi] / 7].
My
> computer didn't like the square brackets in [Pi] above, so I assumed that
> was a misprint and got rid of them. Then
>
> In[1]:=
> N[Simplify[f[x]] /. x -> 2 + 2 Cos[2*Pi/7]]
> Out[1]=
> 3.297766455913909`*^8
>
> which is real and one order of magnitude larger than
>
> In[2]:=
> N[f[x] /. x -> 2 + 2 Cos[2*Pi/7]]
> Out[2]=
> 3.2628184`*^7.
>
> One might think that this has to do with approximation problems. But I
give
> you still crazier results. Watch this:
>
> In[3]:=
> f[x] /. x -> 2 + 2 Cos[2.0000000000000000*Pi/7]
> Out[4]=
> 3.2628184`*^7
>
> Notice that the factor 2 has 16 zeros after the decimal point. Now type
one
> more zero, so as to have 17 instead of 16:
>
> In[5]:=
> f[x] /. x -> 2 + 2 Cos[2.00000000000000000*Pi/7]
> Out[5]=
> 0. * 10^8
>
> The effect of inserting one more zero was to change from machine precision
> numbers to arbitrary precision numbers, and apparently this created havoc
> somewhere. So there is more to this than meets the eye. In fact, it is not
a
> problem of approximation. I took your polynomial and examined its
behavior,
> truncating one by one the terms with successively smaller exponents, and
> evaluating each with and without Simplify. The results are startling:
> everything behaves nicely until you get to the last term, i.e., to the
> original polynomial, and it is only then that the difference pops out. You
> can check this:
>
> In[6]:=
> a1 = Table[Coefficient[f[x], x^j], {j, 1, 32}];
> In[7]:=
> a2 = Table[x^j, {j, 1, 32}];
> In[8]:=
> f[x] == a1.a2
> Out[8]=
> True
>
> Now take the dot products to obtain the truncated polynomials:
>
> In[9]:=
> Table[Take[a1, -j].Take[a2, -j] /. x -> 2 + 2 Cos[2.0*Pi/7], {j, 1, 32}]
> Out[9]=
> \!\({2.3299464415234008`*^16, \(-6.225166642017306`*^17\),
>     8.095823791432835`*^18, \(-6.827018853514098`*^19\),
>     4.195660897089886`*^20, \(-2.0024019788866655`*^21\),
>     7.721985589213778`*^21, \(-2.4719651614064567`*^22\),
>     6.696880503554032`*^22, \(-1.5575043190979624`*^23\),
>     3.1434645609556606`*^23, \(-5.551191511538765`*^23\),
>     8.630908489745314`*^23, \(-1.18679687228797`*^24\),
>     1.447609719431064`*^24, \(-1.5688840125551723`*^24\),
>     1.51129470844226`*^24, \(-1.2929086666244018`*^24\),
>     9.804286814731637`*^23, \(-6.570867584057313`*^23\),
>     3.8768688571383356`*^23, \(-2.0035257611223482`*^23\),
>     9.01004052138053`*^22, \(-3.495373870315845`*^22\),
>     1.1556808440293227`*^22, \(-3.200121970208588`*^21\),
>     7.230361392464404`*^20, \(-1.2804361627651423`*^20\),
>     1.664350897746699`*^19, \(-1.407456330327663`*^18\),
>     5.782285160715823`*^16, 3.2628184`*^7}\)
> In[10]:=
> Table[Simplify[Take[a1, -j].Take[a2, -j]] /. x -> 2 + 2 Cos[2.0*Pi/7], {j,
> 1,
>     32}]
> Out[10]=
> \!\({2.3299464415234008`*^16, \(-6.225166642017307`*^17\),
>     8.095823791432834`*^18, \(-6.827018853514098`*^19\),
>     4.1956608970898866`*^20, \(-2.0024019788866653`*^21\),
>     7.721985589213777`*^21, \(-2.471965161406457`*^22\),
>     6.6968805035540325`*^22, \(-1.557504319097963`*^23\),
>     3.1434645609556626`*^23, \(-5.551191511538763`*^23\),
>     8.630908489745317`*^23, \(-1.1867968722879702`*^24\),
>     1.447609719431064`*^24, \(-1.5688840125551725`*^24\),
>     1.5112947084422592`*^24, \(-1.2929086666244018`*^24\),
>     9.804286814731632`*^23, \(-6.570867584057316`*^23\),
>     3.876868857138329`*^23, \(-2.0035257611223496`*^23\),
>     9.010040521380392`*^22, \(-3.495373870315801`*^22\),
>     1.155680844029254`*^22, \(-3.20012197020816`*^21\),
>     7.230361392454027`*^20, \(-1.2804361627534713`*^20\),
>     1.6643508976590756`*^19, \(-1.4074563292969969`*^18\),
>     5.782285067577335`*^16, 3.297766455913909`*^8}\)
>
> The only term which is different on both lists is the last one! This means
> that approximation is not the issue, since all the truncated polynomials
> handle very large numbers all the time.
> Now, if you really want to have a complete picture of madness, try
> FullSimplify instead of Simplify:
>
> In[11]:=
> Simplify[f[x]] /. x -> 2 + 2 Cos[2.0*Pi/7]
> Out[11]=
> 3.297766455913909`*^8
> In[12]:=
> FullSimplify[f[x]] /. x -> 2 + 2 Cos[2.0*Pi/7]
> Out[12]=
> -2.6508595821182593`*^6
>
> The situation becomes worse every time, and my conclusion is that this
> problem deserves close examination by some of the great gurus in this
group.
>
> Tomas Garza
> Mexico City
>
>
>
>
>
>
> ----- Original Message -----
> From: "Jacqueline Zizi" <jazi at club-internet.fr>
To: mathgroup at smc.vnet.net
> To: <mathgroup at smc.vnet.net>
> Sent: Monday, January 22, 2001 2:10 AM
> Subject: [mg26804] [mg26778] Who can help me?
>
>
> > 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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