Re: querries
- To: mathgroup at smc.vnet.net
- Subject: [mg119619] Re: querries
- From: FranD <seacrofter001 at yahoo.com>
- Date: Tue, 14 Jun 2011 06:14:11 -0400 (EDT)
The ten evaluations below might help clarify.
Francis D.
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Simplify[(a c)^b - (a^b)*(c^b), {{a, c, b} \[Element] Complexes}]
-a^b c^b + (a c)^b
Simplify[(a c)^b - (a^b)*(c^b), {{a, c, b} \[Element] Reals}]
-a^b c^b + (a c)^b
Simplify[(a c)^b - (a^b)*(c^b), {{a, c} \[Element] Reals, b \[Element] Integers}]
0
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Simplify[(a c)^b - (a^b)*(c^b), {{a, c, b} \[Element] Reals, a > 0, c > 0}]
0
Simplify[(a c)^b - (a^b)*(c^b), {{a, c, b} \[Element] Reals, a > 0, c < 0}]
0
Simplify[(a c)^b - (a^b)*(c^b), {{a, c, b} \[Element] Reals, a > 0, c > 0}]
0
Simplify[(a c)^b - (a^b)*(c^b), {{a, c, b} \[Element] Reals, a < 0, c < 0}]
-a^b c^b + (a c)^b
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Simplify[(a*c)^b - (a^b)*(c^b) /. {a -> -1, c -> -1}]
1 - (-1)^(2 b)
Simplify[(a*c)^b - (a^b)*(c^b) /. {b -> 1/2, a -> -1, c -> -1}]
2
Simplify[(a*c)^b - (a^b)*(c^b) /. {b -> 1/4, a -> -1, c -> -1}]
1 - I
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> 1) Why does Mathematica not give the value 0 to
> the expression
> (mu*lambda)^(b) - (mu^b)*(lambda^b)? It seems to
> treat the expression (mu*lambda) as a new variable. I
> used the greek letters from the palette in the actual
> expression.
>
> 2) Many times I cannot get Mathematica to
> simplify expressions involving the exponential
> functions, e.g., it won't combine the arguments of
> the Exp functions. Is there a way to force this?
>
> Thanks,
> Tom Savits
> Statistics Department
> University of Pittsburgh