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Re: Re: verification

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74768] Re: [mg74724] Re: verification
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 4 Apr 2007 04:08:35 -0400 (EDT)
  • References: <eunpso$772$1@smc.vnet.net> <200704021057.GAA08445@smc.vnet.net>

On 2 Apr 2007, at 19:57, David W.Cantrell wrote:

> "dimitris" <dimmechan at yahoo.com> wrote:
>> Hello.
>>
>> foo = {ArcTan[8/(1 - Sqrt[-15 - 4*I])] + ArcTan[8/(1 + Sqrt[-15 -
>> 4*I])] + ArcTan[8/(1 - Sqrt[-15 + 4*I])] +
>>      ArcTan[8/(1 + Sqrt[-15 + 4*I])], ArcTan[3] + ArcTan[5] +
>> ArcTan[41/3] + ArcTan[21], 2*Pi - ArcTan[1/4] - ArcTan[5/12]};
>>
>> The elements of foo list are equal
>>
>> Chop[N[foo, 30]]
>> {5.64341552435296080601310475496,5.64341552435296080601310475496,5.\
>> 64341552435296080601310475496}
>>
>> Block[{Message}, FullSimplify[foo[[2]] == foo[[3]]]]
>> Block[{Message}, FullSimplify[foo[[1]] == foo[[3]]]]
>> Block[{Message}, FullSimplify[foo[[1]] == foo[[2]]]]
>>
>> True
>
> OK, so you easily showed that foo[[2]] and foo[[3]] are the same.  
> Here's
> the easiest way I know to show that foo[[1]] and foo[[3]] are the  
> same:
>
> In[4]:= FullSimplify[TrigToExp[foo[[1]]]] == FullSimplify[foo[[3]]]
>
> Out[4]= True
>
> What really disturbs me is that I can also "show" that foo[[1]] and
> foo[[3]] are NOT the same:
>
> In[6]:= FullSimplify[TrigToExp[foo[[1]]] == foo[[3]]]
>
> messages regarding Internal precision limit snipped
>
> Out[6]= False
>
> It seems that Mathematica is asserting that two _equal_ expressions
> are _not equal_. What am I missing?! (Of course, if Mathematica had  
> merely
> left the logical expression unevaluated, I wouldn't have been  
> disturbed...)

I get the same result which certainly suggests some curious problems  
in FullSimplify. However, in general and for several good reasons it  
is better in such situations to use:

FullSimplify[TrigToExp[foo[[1]]] -foo[[3]]]

0

On my 1 gigaretz PowerBook the above took a very long time (I forgot  
to use Timing but now I do want to run it again).


Andrzej Kozlowski




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