Re: Help needed

• To: mathgroup at smc.vnet.net
• Subject: [mg132040] Re: Help needed
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Sat, 23 Nov 2013 02:55:26 -0500 (EST)
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```Am Donnerstag, 21. November 2013 07:47:15 UTC+1 schrieb Dr. Wolfgang Hintze:
> Am Mittwoch, 20. November 2013 11:39:31 UTC+1 schrieb Artur:
>
> > Dear Mathematica Gurus!
>
> >
>
> > Who have good computer with big RAM memory and will be able to help me
>
> >
>
> > solve following 13 eq with 13 variables (my computer have 8MB and it is
>
> >
>
> > not sufficient). If You will be reached some result write to me
>
> >
>
> > privately because from unknown reason I don't receiving posts now from
>
> >
>
> > mathgroup):
>
> >
>
> >
>
> >
>
> > Solve[{e^3 == e1 e2^2, -1 == 2 c2 e1 e2 + c1 e2^2,
>
> >
>
> >    3 b e^2 == c2^2 e1 + 2 c1 c2 e2 + 2 b2 e1 e2 + b1 e2^2,
>
> >
>
> >    0 == c1 c2^2 + 2 b2 c2 e1 + 2 b2 c1 e2 + 2 b1 c2 e2 + 2 a2 e1 e2 +
>
> >
>
> >      a1 e2^2,
>
> >
>
> >    3 b^2 e + 3 e^2 f ==
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> >
>
> >     2 b2 c1 c2 + b1 c2^2 + b2^2 e1 + 2 a2 c2 e1 + 2 b1 b2 e2 +
>
> >
>
> >      2 a2 c1 e2 + 2 a1 c2 e2 + e2^2 f1 + 2 e1 e2 f2,
>
> >
>
> >    0 == b2^2 c1 + 2 b1 b2 c2 + 2 a2 c1 c2 + a1 c2^2 + 2 a2 b2 e1 +
>
> >
>
> >      2 a2 b1 e2 + 2 a1 b2 e2 + 2 c2 e2 f1 + 2 c2 e1 f2 + 2 c1 e2 f2,
>
> >
>
> >    b^3 + 6 b e f ==
>
> >
>
> >     b1 b2^2 + 2 a2 b2 c1 + 2 a2 b1 c2 + 2 a1 b2 c2 + a2^2 e1 +
>
> >
>
> >      2 a1 a2 e2 + c2^2 f1 + 2 b2 e2 f1 + 2 c1 c2 f2 + 2 b2 e1 f2 +
>
> >
>
> >      2 b1 e2 f2,
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> >
>
> >    0 == 2 a2 b1 b2 + a1 b2^2 + a2^2 c1 + 2 a1 a2 c2 + 2 b2 c2 f1 +
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> >
>
> >      2 a2 e2 f1 + 2 b2 c1 f2 + 2 b1 c2 f2 + 2 a2 e1 f2 + 2 a1 e2 f2,
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> >
>
> >    3 b^2 f + 3 e f^2 ==
>
> >
>
> >     a2^2 b1 + 2 a1 a2 b2 + b2^2 f1 + 2 a2 c2 f1 + 2 b1 b2 f2 +
>
> >
>
> >      2 a2 c1 f2 + 2 a1 c2 f2 + 2 e2 f1 f2 + e1 f2^2,
>
> >
>
> >    0 == a1 a2^2 + 2 a2 b2 f1 + 2 a2 b1 f2 + 2 a1 b2 f2 + 2 c2 f1 f2 +
>
> >
>
> >      c1 f2^2, 3 b f^2 == a2^2 f1 + 2 a1 a2 f2 + 2 b2 f1 f2 + b1 f2^2,
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> >
>
> >    0 == 2 a2 f1 f2 + a1 f2^2, f^3 == f1 f2^2}, {f, f1, f2, e, e1, e2,
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> >
>
> >    a1, a2, b, b1, b2, c1, c2}]
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> >
>
> >
>
> >
>
> > (* best wishes Artur *)
>
>
>
> Arthur,
>
>
>
> I suggest that you try to find a numeric solution. Otherwiese Mathematica is building up a huge tower of symbolic stuff quickly filling the available memory.
>
>
>
> Try this
>
>
>
> eq = [{e^3 == e1 e2^2, -1 == 2 c2 e1 e2 + c1 e2^2,
>
>    3 b e^2 == c2^2 e1 + 2 c1 c2 e2 + 2 b2 e1 e2 + b1 e2^2,
>
>    0 == c1 c2^2 + 2 b2 c2 e1 + 2 b2 c1 e2 + 2 b1 c2 e2 + 2 a2 e1 e2 +
>
>      a1 e2^2,
>
>    3 b^2 e + 3 e^2 f ==
>
>     2 b2 c1 c2 + b1 c2^2 + b2^2 e1 + 2 a2 c2 e1 + 2 b1 b2 e2 +
>
>      2 a2 c1 e2 + 2 a1 c2 e2 + e2^2 f1 + 2 e1 e2 f2,
>
>    0 == b2^2 c1 + 2 b1 b2 c2 + 2 a2 c1 c2 + a1 c2^2 + 2 a2 b2 e1 +
>
>      2 a2 b1 e2 + 2 a1 b2 e2 + 2 c2 e2 f1 + 2 c2 e1 f2 + 2 c1 e2 f2,
>
>    b^3 + 6 b e f ==
>
>     b1 b2^2 + 2 a2 b2 c1 + 2 a2 b1 c2 + 2 a1 b2 c2 + a2^2 e1 +
>
>      2 a1 a2 e2 + c2^2 f1 + 2 b2 e2 f1 + 2 c1 c2 f2 + 2 b2 e1 f2 +
>
>      2 b1 e2 f2,
>
>    0 == 2 a2 b1 b2 + a1 b2^2 + a2^2 c1 + 2 a1 a2 c2 + 2 b2 c2 f1 +
>
>      2 a2 e2 f1 + 2 b2 c1 f2 + 2 b1 c2 f2 + 2 a2 e1 f2 + 2 a1 e2 f2,
>
>    3 b^2 f + 3 e f^2 ==
>
>     a2^2 b1 + 2 a1 a2 b2 + b2^2 f1 + 2 a2 c2 f1 + 2 b1 b2 f2 +
>
>      2 a2 c1 f2 + 2 a1 c2 f2 + 2 e2 f1 f2 + e1 f2^2,
>
>    0 == a1 a2^2 + 2 a2 b2 f1 + 2 a2 b1 f2 + 2 a1 b2 f2 + 2 c2 f1 f2 +
>
>      c1 f2^2, 3 b f^2 == a2^2 f1 + 2 a1 a2 f2 + 2 b2 f1 f2 + b1 f2^2,
>
>    0 == 2 a2 f1 f2 + a1 f2^2, f^3 == f1 f2^2}
>
>
>
> NSolve[eq//n, {f, f1, f2, e, e1, e2,
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>    a1, a2, b, b1, b2, c1, c2}]
>
>
>
> On my PC, although it is still running, the requested memory remains at about 1.5 GB.
>
>
>
> Another idea: define a reasonable subset of equations an try to solve these first.
>
>
>
> Best regards,
>
> Wolfgang

No result obtained.

Now the idea of usind a subset

1) Use all 13 equations but set some Parameter -> 0

r = {b -> 0, a1 -> 0, b1 -> 0};
eqx = eq /. r
sol = Solve[eqx];
Join[r, #] & /@ sol;
{a1, a2, b, b1, b2, c1, c2, f, f1, f2, e, e1, e2} /. %;
Union[%]

The non trivial part of the solutions is of the type
{e -> e1^(1/3) e2^(2/3), c1 -> -(1/e2^2)}
with two free Parameters (e1 and e2)

2) Take a subset of equations

2a) use 7 equations, leaving 6 Parameters free

sol = Solve[
Join[Take[eq, {7, 13}], {f != 0, a1 != 0, a2 != 0, c2 != 0}]] //
Simplify;
{Length[sol], Length[sol[[1]]]};

2b) use this intermediate result with one further equation to fix one additional Parameter

eqx = eq /. sol[[-1]];
sol1 = Solve[
Join[Take[eqx, {1}], {f != 0, a1 != 0, a2 != 0, c2 != 0}]] //
Simplify // Short

Gives typically
c2 -> Root[-2187 a2^14 b2^2 f1^8 -
36450 a2^12 b2^3 f1^8 f2 + <<60>> + 118098 a2^3 f1^8 f2^8 #1^5 &,
1]

That is, c2 is the root of an equation of grade 5.

Summarizing, we have seen that there are several exact Solutions, but it will be very diffucult to get all Solutions.

Can you say some words on what these equations mean?

Best regards,
Wolfgang

```

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