Re: Help needed

• To: mathgroup at smc.vnet.net
• Subject: [mg132039] Re: Help needed
• From: Roland Franzius <roland.franzius at uos.de>
• Date: Sat, 23 Nov 2013 02:55:06 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• Delivered-to: l-mathgroup@wolfram.com
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• References: <l6i3h3\$80f\$1@smc.vnet.net>

```Am 20.11.2013 11:39, 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 ==
>      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}, {f, f1, f2, e, e1, e2,
>     a1, a2, b, b1, b2, c1, c2}]

Coupled nonlinear equations have potentially the the algebraic structure
of a rational function with powers up to product of highest powers. So
there is no chance solving for a single variable in general.

The growing complexity can be seen by eliminating linear variables:

sol = Solve[
Join[ eq[[1 ;; 2]], eq[[-4 ;; -1]]], {e1, c1, f1, a1, b1, b2}][[1]]

{e1 -> e^3/e2^2, c1 -> (-2 c2 e^3 - e2)/e2^3, f1 -> f^3/f2^2,
a1 -> -((2 a2 f^3)/f2^3),
b1 -> (5 a2^3 e2^3 f^3 + 3 a2 b e2^3 f^2 f2^2 - 2 c2 e2^3 f^3 f2^2 +
2 c2 e^3 f2^5 + e2 f2^5)/(3 a2 e2^3 f2^4),
b2 -> (4 a2^3 e2^3 f^3 + 6 a2 b e2^3 f^2 f2^2 + 2 c2 e2^3 f^3 f2^2 -
2 c2 e^3 f2^5 - e2 f2^5)/(6 a2 e2^3 f^3 f2)}

eq /. sol // Simplify

{True, True, (1/(
a2 e2 f f2))(-f2^2 (-e2^3 f^3 + e^3 f2^3) (-2 c2 e2^3 f^3 +
2 c2 e^3 f2^3 + e2 f2^3) +
a2^3 (5 e2^6 f^6 + 4 e^3 e2^3 f^3 f2^3) +
3 a2 e2^2 f^2 f2^2 (-c2 (3 c2 e^3 + 2 e2) f f2^2 +
b e2 (e2 f - e f2)^2 (e2 f + 2 e f2))) == 0, (1/(
a2 e2 f f2))(3 c2 e^3 e2 f2^8 + e2^2 f2^8 +
2 a2^3 e2^3 f^3 (5 c2 e2^3 f^3 - 2 c2 e^3 f2^3 - 2 e2 f2^3) +
a2^2 (-6 e2^7 f^6 f2 + 6 e^3 e2^4 f^3 f2^4) +
2 c2^2 (-2 e2^6 f^6 f2^2 + e^3 e2^3 f^3 f2^5 + e^6 f2^8) -
3 a2 e2^2 f^2 f2^2 (c2^2 (2 c2 e^3 + e2) f f2^2 +
2 b e2 (-c2 e2^3 f^3 + c2 e^3 f2^3 + e2 f2^3))) == 0, (1/(
a2 e2 f f2))(f2^4 (-4 e2^3 f^3 + e^3 f2^3) (-2 c2 e2^3 f^3 +
2 c2 e^3 f2^3 + e2 f2^3)^2 +
16 a2^6 (5 e2^9 f^9 + e^3 e2^6 f^6 f2^3) +
36 a2^2 e2^5 f^4 f2^3 (b^2 e2 f2 (e2 f - e f2)^2 (2 e2 f + e f2) +
e2^2 f^2 (e2 f - e f2)^2 (e2 f + 2 e f2) +
b c2 f (c2 e2^3 f^3 - 4 c2 e^3 f2^3 - 2 e2 f2^3)) +
12 a2^4 e2^5 f^5 f2 (-4 c2 e2 f f2^3 +
2 b e2 f2 (7 e2^3 f^3 + 2 e^3 f2^3) +
c2^2 (5 e2^3 f^4 - 8 e^3 f f2^3)) -
12 a2 e2^2 f^2 f2^3 (2 c2 e2^3 f^3 - 2 c2 e^3 f2^3 -
e2 f2^3) (c2 e2 f f2^3 + b e2 f2 (e2^3 f^3 - e^3 f2^3) +
c2^2 (e2^3 f^4 + 2 e^3 f f2^3)) -
4 a2^3 e2^3 f^3 f2^2 (19 e2^4 f^3 f2^3 + 2 e^3 e2 f2^6 +
2 c2 (17 e2^6 f^6 + 8 e^3 e2^3 f^3 f2^3 + 2 e^6 f2^6))) == 0, (
1/(36 a2^2 e2^9 f^6 f2^5))(-24 a2^5 e2^7 f^6 f2 (e2^3 f^3 +
2 e^3 f2^3) -
24 a2^4 b e2^6 f^5 f2^2 (7 c2 e2^3 f^3 - 4 c2 e^3 f2^3 -
2 e2 f2^3) +
16 a2^6 e2^6 f^6 (-5 c2 e2^3 f^3 + 2 c2 e^3 f2^3 + e2 f2^3) +
f2^4 (-2 c2 e2^3 f^3 + 2 c2 e^3 f2^3 +
e2 f2^3)^2 (4 c2 e2^3 f^3 + 2 c2 e^3 f2^3 + e2 f2^3) +
12 a2^2 e2^4 f^3 f2^3 (2 c2 (e2^3 f^3 - e^3 f2^3) (e2^3 f^3 -
3 b^2 e2^2 f f2 - e^3 f2^3) +
e2 f2^3 (2 e2^3 f^3 + 3 b^2 e2^2 f f2 + e^3 f2^3)) +
12 a2 b e2^3 f^2 f2^4 (-e2^2 f2^6 +
2 c2^2 (e2^6 f^6 + e^3 e2^3 f^3 f2^3 - 2 e^6 f2^6) +
c2 (e2^4 f^3 f2^3 - 4 e^3 e2 f2^6)) +
4 a2^3 e2^3 f^3 f2^2 (2 c2^2 (8 e2^6 f^6 + 23 e^3 e2^3 f^3 f2^3 -
4 e^6 f2^6) + c2 (23 e2^4 f^3 f2^3 - 8 e^3 e2 f2^6) -
2 (-9 b e2^7 f^5 f2 + 9 b e^3 e2^4 f^2 f2^4 + e2^2 f2^6))) ==
0, (1/(a2 e2 f f2))(80 a2^9 e2^9 f^9 + 288 a2^7 b e2^9 f^8 f2^2 +
36 a2^5 e2^7 f^6 f2^3 (2 e2^3 f^3 + 9 b^2 e2^2 f f2 +
7 e^3 f2^3) +
24 a2^6 e2^6 f^6 f2^2 (5 c2 e2^3 f^3 - 14 c2 e^3 f2^3 -
7 e2 f2^3) -
36 a2^2 e2^4 f^3 f2^5 (e2^3 f^3 - e^3 f2^3) (2 c2 e2^3 f^3 -
2 c2 e^3 f2^3 - e2 f2^3) -
9 a2 b e2^3 f^2 f2^6 (-2 c2 e2^3 f^3 + 2 c2 e^3 f2^3 +
e2 f2^3)^2 +
f2^6 (-2 c2 e2^3 f^3 + 2 c2 e^3 f2^3 + e2 f2^3)^3 -
36 a2^4 b e2^6 f^5 f2^4 (4 c2 e2^3 f^3 + 14 c2 e^3 f2^3 +
7 e2 f2^3) -
3 a2^3 e2^3 f^3 f2^4 (c2^2 (64 e2^6 f^6 + 88 e^3 e2^3 f^3 f2^3 -
44 e^6 f2^6) + 44 c2 (e2^4 f^3 f2^3 - e^3 e2 f2^6) -
e2^2 f2 (11 f2^5 +
72 b e2^2 f^2 (e2 f - e f2)^2 (2 e2 f + e f2)))) == 0, (1/(
a2 e2 f f2))(8 a2^6 e2^6 f^6 + 12 a2^4 b e2^6 f^5 f2^2 -
12 a2^3 e2^3 (2 c2 e^3 + e2) f^3 f2^5 +
6 a2 b e2^3 f^2 f2^4 (2 c2 e2^3 f^3 - 2 c2 e^3 f2^3 - e2 f2^3) -
12 a2^2 (e2^7 f^6 f2^3 - e^3 e2^4 f^3 f2^6) +
f2^4 (e2^2 f2^6 +
4 c2^2 (-2 e2^6 f^6 + e^3 e2^3 f^3 f2^3 + e^6 f2^6) +
2 c2 (e2^4 f^3 f2^3 + 2 e^3 e2 f2^6))) == 0, (1/(
a2 e2 f f2))(20 a2^6 e2^6 f^6 + 36 a2^4 b e2^6 f^5 f2^2 +
12 a2^2 e2^4 f^3 f2^3 (e2 f - e f2)^2 (2 e2 f + e f2) -
f2^4 (-2 c2 e2^3 f^3 + 2 c2 e^3 f2^3 + e2 f2^3)^2 -
8 a2^3 e2^3 f^3 f2^2 (5 c2 e2^3 f^3 + 4 c2 e^3 f2^3 +
2 e2 f2^3)) == 0, True, True, True, True}

I don't see much further improvement. Numerical methods apply if the
solution is unique and not a variety.

--

Roland Franzius

```

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