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Re: Full simplify problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg122348] Re: Full simplify problem
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Wed, 26 Oct 2011 17:37:41 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <D3C63570-91CA-489E-B8B7-661844B401BC@mimuw.edu.pl> <201110251018.GAA05951@smc.vnet.net>
On 10/25/2011 05:18 AM, A. Lapraitis wrote:
> First of all, thanks to all the people answering!
>
> The problem is more complicated, indeed, this is just the minimal working
> example. However, I still do not understand Mathematica's behaviour.
>
> 1. I thought that the whole point of Count[#, x, Infinity]& was to
> distinguish between Exp[x] and Exp[y+z] in favour of Exp[x]. However, the
> same is true for the LeafCount function (it gives 3 and 5 respectively). Why
> is not a good complexity function in this case?
>
> 2. I was trying a different workaround and have noticed a dependence of the
> result on the names of the variables:
>
> In[13]:= Clear["Global`*"]
> In[14]:= Assuming[x == d&& d == y + z, FullSimplify[E^x - E^(y + z)]]
> Out[14]= E^x - E^(y + z)
> In[15]:= Assuming[a == d&& d == b + c, FullSimplify[E^a - E^(b + c)]]
> Out[15]= 0
>
> Can anyone explain this?
http://forums.wolfram.com/mathgroup/archive/2005/Dec/msg00549.html
http://forums.wolfram.com/mathgroup/archive/2005/Jan/msg00237.html
Daniel Lichtblau
Wolfram Research
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