Re: Simplify results
- To: mathgroup at smc.vnet.net
- Subject: [mg118300] Re: Simplify results
- From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
- Date: Thu, 21 Apr 2011 03:09:25 -0400 (EDT)
Hi, Berthold,
Nothing is really bad. The initial result:
expr = (8 a^2 b^3 (c^2 + 1)^4 - 6 a^3 b^2 (c^2 + 1)^3 +
14 a^4 b (c^2 + 1)^2)/(4 a^3 b^2 (c^2 + 1)^2 - 10 a^4 b^3 (c^2 + 1)^3)
expr1 = Simplify[expr]
(14 a^4 b (1 + c^2)^2 - 6 a^3 b^2 (1 + c^2)^3 + 8 a^2 b^3 (1 + c^2)^4)/(
4 a^3 b^2 (1 + c^2)^2 - 10 a^4 b^3 (1 + c^2)^3)
(-7 a^2 + 3 a b (1 + c^2) - 4 b^2 (1 + c^2)^2)/(a b (-2 + 5 a b (1 + c^2)))
and your "hand-made" result:
expr2 = (7 a^2 - 3 a b (1 + c^2) +
4 b^2 (1 + c^2)^2)/(a b (2 - 5 a b (1 + c^2)))
(7 a^2 - 3 a b (1 + c^2) + 4 b^2 (1 + c^2)^2)/(a b (2 - 5 a b (1 + c^2)))
differ from one another by multiplication by -1 both in the Numerator and in the Denominator. They are in fact equal to one another
Numerator[expr1]
Numerator[expr2]
-7 a^2 + 3 a b (1 + c^2) - 4 b^2 (1 + c^2)^2
7 a^2 - 3 a b (1 + c^2) + 4 b^2 (1 + c^2)^2
Denominator[expr1]
Denominator[expr2]
a b (-2 + 5 a b (1 + c^2))
a b (2 - 5 a b (1 + c^2))
It is probably obvious now.
Let us also make a general check:
expr1 = expr2 // Simplify
True
and numerically:
expr1 /. {a -> 1, b -> 2, c -> 3}
expr2 /. {a -> 1, b -> 2, c -> 3}
-(221/28)
-(221/28)
Have fun. Alexei
Hi,
This might be a silly question, so please bear with me, but I have been
scratching my head about it for some time now.
Is there a particular reason why Mathematica (8.01) simplifies the following
fraction reversing the signs in the result:
IN:
Simplify[(8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b
(c^2+1)^2)/(4a^3b^2(c^2+1)^2-10a^4b^3(c^2+1)^3)]
OUT:
(-7 a^2+3 a b (1+c^2)-4 b^2 (1+c^2)^2)/(a b (-2+5 a b (1+c^2)))
Reducing the fraction by hand gives me:
(7 a^2-3 a b (1+c^2)+4 b^2 (1+c^2)^2)/(a b (2-5 a b (1+c^2)))
Thanks
Berthold
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Berthold Hamburger - Cellist/Spain
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