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Re: Terrible algebraic system
Posted:
Feb 22, 2012 1:33 PM
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deltaquattro@gmail.com wrote:
> Hi,
>
> couldn't solve this with Wolfram Alpha, Matlab or Maxima:
>
> (a-L)^2/a*2/(a+b)-p=0
> (b-U)^2/b*2/(a+b)-p=0
>
> with a>L>0, b>U>0. Any tips? Should I resort to numerical methods?
>
FriCAS gives the following:
(1) -> solve([(a-L)^2/a*2/(a+b)-p=0, (b-U)^2/b*2/(a+b)-p=0], [a, b])
(1)
[
[
a =
3 2 2 2
(2b + (- 2U + 2L)b + L b)p
+
3 2 2 3 2 3
(- 6b + (16U - 4L)b + (- 12U + 8L U)b + 4U - 4L U )p + 4b
+
2 2 3
- 16U b + 20U b - 8U
/
2 2 2
U p - 2U p
,
4 3 2 2 2 2
(2b + (- 2U + 2L)b + (U + L )b )p
+
4 3 2 2 3 2 4
(- 6b + (16U - 4L)b + (- 16U + 8L U)b + (8U - 4L U )b - 2U )p
+
4 3 2 2 3 4
4b - 16U b + 24U b - 16U b + 4U
=
0
]
]
This means that one has to first solve equation in one unknown
(that is b) of degree 4, and then plug in the result to get a.
The equation treating b as main variable is:
2 4 2 3
(2p - 6p + 4)b + ((- 2U + 2L)p + (16U - 4L)p - 16U)b +
2 2 2 2 2 2 3 2 3 4
((U + L )p + (- 16U + 8L U)p + 24U )b + ((8U - 4L U )p - 16U )b - 2U p
4
+ 4U
FriCAS can solve equation of degree 4 in terms of radicals, but
the answer is large and probably less useful then the above.
--
Waldek Hebisch
hebisch@math.uni.wroc.pl
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