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Re: "Assuming"
*To*: mathgroup at smc.vnet.net
*Subject*: [mg85607] Re: "Assuming"
*From*: "Mariano Suárez-Alvarez" <mariano.suarezalvarez at gmail.com>
*Date*: Thu, 14 Feb 2008 18:51:31 -0500 (EST)
*References*: <fp0m11$5u7$1@smc.vnet.net> <fp192b$gju$1@smc.vnet.net>
On Feb 14, 9:35 am, dh <d... at metrohm.ch> wrote:
> HI Markus,
>
> I think you are fooling yourself.
>
> Consider Simplify[(a^2 - 1)/(a - 1)]
>
> As a^2 - 1)== (a+1)(a-1) this will give (a+1). This is true in general,
>
> for a==1 understood as limit. Therefore, this has nothing to do with
>
> assumptions.
>
> On the other hand, (a^2 - 1)/(a - 1) /. a -> 1 can not be calculated as
>
> it gives 0/0. Note that "/. and ->" mean replacement, not limit. If
>
> you want a limit, you have to say so: Limit[(a^2 - 1)/(a - 1), a -> 1]
>
> hope this helps, Daniel
I've always guessed that Simplify[e] is always an
expression with the same value as e, for valuations
of the variables appearing in e for which e is
defined. Hence Simplify[(1 - a^2)/(1 - a)] evaluates
to 1 + a, which has the same value as
(1 - a^2)/(1 - a) for all values of a for which the
last expression is defined.
I do not recall having seen an explicit specification
of what Simplify does.
A more extreme case of the OP's situation is simply
Assuming[a == 1, Simplify[(1 - a^2)/(1 - a)]]
which evaluates to 2.
According to the docs, Simplify tries to use
Refine to acheve simplifications, but clearly
it does not take into account the information
that Refine produces: for example,
Assuming[a == 1, Refine[(1 - a^2)/(1 - a)]]
tells me
Power::infy: Infinite expression 1/0 encountered.
Infinity::indet: Indeterminate expression 0 ComplexInfinity
encountered.
Out[1]= Indeterminate
-- m
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