Re: Full simplify problem
- To: mathgroup at smc.vnet.net
- Subject: [mg122289] Re: Full simplify problem
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 23 Oct 2011 06:26:01 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201110221009.GAA29840@smc.vnet.net>
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!
>
- References:
- Full simplify problem
- From: "A. Lapraitis" <ffcitatos@gmail.com>
- Full simplify problem