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Re: Re: Time needed for calculation
I use a 1.33 Ghz Powerbook (17 inch), and there are no real issues
running a calculation overnight. Both fans may come on, but there is
no problem. The limitations are the usual hardware ones -- RAM and HD
space. As long as the problem is solvable in 32 bit space, the
Powerbook can get the job done, just not as fast as some other
computers.
Personally, I'm looking forward to a quad core intel-mac. ;-)
george
On Nov 24, 2005, at 6:33 AM, Daniel Lichtblau wrote:
> JikaiRF at AOL.COM wrote:
>> Dear Sirs,
>> I have some questions about how long a computer can calculate,
>> without
>> damages on hard wear.
>> More specifically speaking, I intend to depict trajectories on (k,
>> c) axis,
>> which are derived from programming on Mathematica.
>> My program is as follows:
>>
>> << Graphics`ImplicitPlot`
>> L = 48;
>>
>> a = 0.1;
>>
>> A = 1;
>>
>> r = 0.1;
>>
>> t = 0.046296296296296294`;
>>
>> ImplicitPlot[A r L (1 - a)((1 -t) c k^a == (a+(1-a)(1- t)) L^2 (1-
>> a)^2 (1- t
>> )^3 k^(3a-1)
>> +L(1-a) c (1- t) (-A(a+(1-a)(1- t))(2- t)-(1-
>> a)(1- t)^2
>> +A a(1- t)) k^(2a-1)
>> +A c^2 (A(a+(1-a)(1- t))+(1-a)(1- t) -A a(1-
>> t)) k^(a-1)
>> +A^2 r c^2, {k, 0.8, 1.5}]
>>
>> This program seems to work well. However, I cannot obtain the
>> curve which
>> should be derived from the above program, though my computer
>> continues to
>> calculate more than fifteen minutes.
>>
>> I would like to know how long a computer can calculate without
>> damages on
>> hardware, and additionally how long it takes to obtain the
>> required curve based
>> on my computer.
>> Especially, my computer is PowerBook G4 with 1.5 GHz processor.
>>
>> Sincerely,
>> Fujio Takata.
>>
>
> I have no idea what sort of computation load a PowerBook can handle.
> Also it is not at all clear what you did to find the curve in
> question.
> The approach below will get you a polynomial reasonably fast.
>
> expr = (-A)*r*L*(1 - a)*(1 - t)*c*k^a + (a + (1 - a)*(1 - t))*
> L^2*(1 - a)^2*(1 - t)^3*k^(3*a - 1) +
> L*(1 - a)*c*(1 - t)*((-A)*(a + (1 - a)*(1 - t))*(2 - t) -
> (1 - a)*(1 - t)^2 + A*a*(1 - t))*k^(2*a - 1) +
> A*c^2*(A*(a + (1 - a)*(1 - t)) + (1 - a)*(1 - t) -
> A*a*(1 - t))*k^(a - 1) + A^2*r*c^2
>
> In[25]:= rexpr = Rationalize[expr, 0]
>
> Out[25]=
> c^2/10 + (6375024143593*c^2)/(3703618114621*k^(9/10)) -
> (388844671178*c)/(3636250307*k^(4/5)) +
> 307126999725/(197967319*k^(7/10)) -
> (165668129149671*c*k^(1/10))/40210710958658
>
> Now convert to a Laurent polynomial (only integer powers) by removing
> the denominator power in k.
>
> In[37]:= rexpr2 = PowerExpand[rexpr /. k -> l^10]
>
> Out[37]=
> c^2/10 + (6375024143593*c^2)/(3703618114621*l^9) -
> (388844671178*c)/(3636250307*l^8) +
> 307126999725/(197967319*l^7) - (165668129149671*c*l)/
> 40210710958658
>
> We can recover an implicit polynomial by eliminating l from the system
> {rexpr2,l^10-k}.
>
> imppoly =
> First[GroebnerBasis[{rexpr2, l^10 - k}, {c, k}, l,
> MonomialOrder -> EliminationOrder]]
>
> The resulting polynomial will be quite resistant to direct use of
> ImplicitPlot because the coefficients are too large to find zeros via
> e.g. subdivision with machine arithmetic. I get what appears to be a
> reasonable result from rescaling numerically as below.
>
> In[75]:= imppoly2 = N[Expand[imppoly/10^450]]
>
> Out[75]=
> -0.44710407299089117*c^20 + 3.6929000651811335*^15*c^10*k -
> 5.435249465194847*c^19*k - 1.5816021229154092*^29*k^2 +
> 5.590311570192779*^16*c^9*k^2 - 23.71722762253775*c^18*k^2 +
> 2.8406692481577024*^17*c^8*k^3 - 43.059424056179864*c^17*k^3 +
> 5.701837604927303*^17*c^7*k^4 - 28.130472729642648*c^16*k^4 +
> 3.901760263753945*^17*c^6*k^5 - 2.2597182065813595*c^15*k^5 +
> 4.17821551222496*^16*c^5*k^6 - 3.1477095526033976*c^14*k^6 +
> 13.931088647720639*c^13*k^7 - 61.13817833192989*c^12*k^8 +
> 266.15182908720067*c^11*k^9 + 1.9581785245917768*^-13*c^20*
> k^9 - 2759.462163976083*c^10*k^10
>
> That is, the plot one might obtain as below gives something visually
> similar to what I saw with the implicit plot of expr. But it has a new
> graphical component (though not a distinct algebraic component,
> because
> the polynomial is irreducible). I would guess this corresponds to
> "parasite" solutions when one goes back to the radical formulation of
> the implicit equation.
>
> ImplicitPlot[imppoly2 == 0, {k, 0.1, 100}, {c, 5, 80}]
>
>
> Daniel Lichtblau
> Wolfram Research
>
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