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Re: Full simplify problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg122300] Re: Full simplify problem
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 24 Oct 2011 05:13:43 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201110221009.GAA29840@smc.vnet.net> <201110231026.GAA10607@smc.vnet.net> <op.v3tck4nmtgfoz2@bobbys-imac.local>
In a sense, yes. But this certianly does not answer the "why" part of the OP's question and I think that was what he really wanted to know.
Andrzej
On 23 Oct 2011, at 18:40, DrMajorBob wrote:
> Far simpler (and equivalent under the covers, I think) would be:
>
> E^x - E^(y + z) /. x -> y + z
>
> 0
>
> Bobby
>
> On Sun, 23 Oct 2011 05:26:01 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
>
>> Because FullSimplify would have to increase the default complexity to
>> obtain the cancellation. But this works:
>>
>> Assuming[x == y + z,
>> FullSimplify[E^x - E^(y + z),
>> ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]
>>
>> 0
>>
>> Note that the reason why this works is that:
>>
>> Assuming[x == y + z,
>> FullSimplify[x, ComplexityFunction -> (Count[#, x, {1, Infinity}] &)]]
>>
>> y+z
>>
>>
>> Andrzej Kozlowski
>>
>>
>> On 22 Oct 2011, at 12:09, A. Lapraitis wrote:
>>
>>> Hello,
>>>
>>> Could anyone explain why the following does not give zero?
>>>
>>> In[72]:= Assuming[
>>> x == y + z,
>>> FullSimplify[
>>> E^x - E^(y + z)
>>> ]
>>> ]
>>>
>>> Out[72]= E^x - E^(y + z)
>>>
>>> Thanks!
>>>
>>
>>
>
>
> --
> DrMajorBob at yahoo.com
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