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Re: "Assuming"
*To*: mathgroup at smc.vnet.net
*Subject*: [mg85599] Re: "Assuming"
*From*: dh <dh at metrohm.ch>
*Date*: Thu, 14 Feb 2008 06:30:20 -0500 (EST)
*References*: <fp0m11$5u7$1@smc.vnet.net>
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
markusbinder wrote:
> Hi,
>
> I would appreciate some comments about the Assuming-function. Several examples using Mathematica 6 as follows:
>
> In[1]:= Assuming[a > 0, Simplify[a == -1]]
> Out[1]= False
> In[2]:= Assuming[a \[Element] Integers, Simplify[a == -1]]
> Out[2]= a == -1
> In[3]:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[a^2< 1.1]]
> Out[3]= True
> In[4]:= Assuming[a \[Element] Reals && a > 0 && a < 2, Simplify[a^2 < 1.1]]
> Out[4]= a < 1.04881
>
> So far, everything seems pretty reasonable, but
>
> In[5]:= Assuming[a \[Element] Integers && a > 0 && a < 2, Simplify[(a^2 - 1)/(a - 1)]]
> Out[5]= 1 + a
>
> whereas
>
> In[6]:= (a^2 - 1)/(a - 1) /. a -> 1
> Out[6]= Indeterminate
>
> I consider this as blunder (or bug?); I comprehend working with domains can be pretty tricky and of course I don't expect
>
> In[7]:= Simplify[(a^2 - 1)/(a - 1)]
>
> to yield
>
> Out[7]= "Beware, a != 1 neccessary!"
>
> instead of
>
> Out[7]= 1 + a.
>
> Am I tricked by some personal misconception of how Mathematica deals with Assuming plus fractions as shown above, or is this a mere flaw?
> Thanks in advance for your corrections & suggestions.
>
> Best regards
> Markus Binder
>
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