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Re: querries

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  Re: querries FranD <seacrofter001@yahoo.com> - 2011-06-14 10:14 +0000

#3091 — Re: querries

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

=================================

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


================================

> 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

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