Re: Full simplify problem
- To: mathgroup at smc.vnet.net
- Subject: [mg122295] Re: Full simplify problem
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Mon, 24 Oct 2011 05:12:49 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <D3C63570-91CA-489E-B8B7-661844B401BC@mimuw.edu.pl> <C9A65308-E4DD-4DD6-967D-FFDF15F6E7DC@mimuw.edu.pl> <1319314368.91650.YahooMailNeo@web43134.mail.sp1.yahoo.com>
This is an interesting example!
Here is the explanation and a puzzle:
In[28]:= Assuming[False, Simplify[1 == 2]]
During evaluation of In[28]:= $Assumptions::fas: Warning: One or more assumptions evaluated to False. >>
Out[28]= True
Do you see the relation ;-)
Andrzej
On 22 Oct 2011, at 22:12, dimitris anagnostou wrote:
> With your help I understood why Assuming[x ==y + z, FullSimplify[E^x - E^(y + z)]]
> does not work here.
> What is normal for me, is not normal for such advanced function like FullSimplify!
>
> What about the following approach?
>
> In[83]:= Assuming[x === y + z, FullSimplify[E^x - E^(y + z)]]
>
> During evaluation of In[83]:= $Assumptions::fas:Warning: One or more assumptions evaluated to False. >>
>
> Out[83]= 0
>
> Apart from the warning message does the requested job.
>
> I do consider however the solution with ComplexityFunction much more clever and correct of course!
>
> Dimitris
>
>
> From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
> To: dimitris anagnostou <dimmechan at yahoo.com>
> Sent: Saturday, October 22, 2011 10:45 PM
> Subject: Fwd: Full simplify problem
>
> Hello Dimitris
>
> I also sent a response to that message. It is quite similar to yours.
>
> Best regards
>
> Andrzej
>
> Begin forwarded message:
>
>> From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
>> Date: 22 October 2011 17:02:05 CEST
>> To: A. Lapraitis <ffcitatos at gmail.com>
>> Cc: mathgroup at smc.vnet.net
>> Subject: Re: Full simplify problem
>>
>> 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!
>>>
>>
>
>
>