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Re: Pi Formula


Dear Jean-Marc,
Yes, I want do this expression some more complicated as Pi and less 
complicated as recent form to understand where Pi is hidden because 
after run e.g. 
a={};k=Expand[expr];Do[AppendTo[a,FullSimplify[k[[n]]]],{n,1,Length[k]}];a
we don't see yet this evidently.

Best wishes
Artur



Jean-Marc Gulliet pisze:
> Artur wrote:
>
>> Who know which another function as Simplify or FullSimplify to use to 
>> following formula
>> (Simplify do nothing but FullSimplify simplify too much).
>> 1/8 (-2 I Sqrt[-7 - I] Log[1/5 ((1 - 2 I) + 2 Sqrt[-7 - I])] -
>>    Log[(3 - 4 I)^Sqrt[7 + I]
>>       5^((3 - I) (51 - 10 Sqrt[2])^(
>>       1/4)) ((1 - 2 I) - 2 Sqrt[-7 - I])^((-1 - I) (51 - 10 Sqrt[2])^(
>>       1/4)) (-5 I + (4 - 2 I) Sqrt[-7 - I])^(-2 Sqrt[
>>       7 + I]) ((1 + 2 I) - 2 Sqrt[-7 + I])^(2 Sqrt[-7 + I])] -
>>    I Sqrt[7 - I] Log[(-1 + Sqrt[-1 - I])^2] -
>>    I Sqrt[7 - I] Log[(1 + Sqrt[-1 - I])^2] -
>>    Sqrt[7 + I] Log[(-1 + Sqrt[-1 + I])^2] -
>>    Sqrt[7 + I] Log[(1 + Sqrt[-1 + I])^2] + (1 - 2 I) Sqrt[1 - I]
>>      Log[(1 + I) - Sqrt[1 - I]] - (2 - I) Sqrt[1 + I]
>>      Log[(1 + I) - Sqrt[1 - I]] - (1 - 2 I) Sqrt[1 - I]
>>      Log[(1 + I) + Sqrt[1 - I]] + (2 - I) Sqrt[1 + I]
>>      Log[(1 + I) + Sqrt[1 - I]] +
>>    I Sqrt[7 - I] Log[(-60 - 4 I) + 8 Sqrt[-1 - I] - 24 Sqrt[7 - I]] +
>>    I Sqrt[7 - I] Log[(66 - 14 I) + 8 Sqrt[-1 - I] + 24 Sqrt[7 - I]] +
>>    Sqrt[7 + I] Log[(-60 + 4 I) + 8 Sqrt[-1 + I] - 24 Sqrt[7 + I]] +
>>    Sqrt[7 + I] Log[(66 + 14 I) + 8 Sqrt[-1 + I] + 24 Sqrt[7 + I]] +
>>    I Log[((1 + 2 I) - 2 Sqrt[-7 + I])^(
>>       2 Sqrt[7 - I]) ((-60 + 4 I) - 16 Sqrt[14 - 2 I])^-Sqrt[-7 -
>>         I] ((-60 - 4 I) - 16 Sqrt[14 + 2 I])^-Sqrt[
>>        7 - I] ((-(153/100) + (71 I)/100) + 2/25 Sqrt[287 - 359 I])^
>>       Sqrt[-7 -
>>        I] ((-(153/2500) - (71 I)/2500) + 2/625 Sqrt[287 + 359 I])^
>>       Sqrt[7 - I]])
>
> Arthur,
>
> Calling the above expression "expr", we have
>
>     In[5]:= FullSimplify[expr]
>
>     Out[5]= Pi
>
> If I have understood your correctly, you wish to have something more 
> complicated than Pi, yet simpler than the original expr, which begs 
> the  question: What do you expect? Perhaps *PowerExpand* is what you 
> are looking for (though the leaf count is not that much different)?
>
>     In[7]:= pw = PowerExpand[expr];
>
>     In[8]:= LeafCount /@ {expr, pw}
>
>     Out[8]= {590, 582}
>
> Or you may want to tweak/build your own *ComplexityFunction*. See
>
> http://reference.wolfram.com/mathematica/ref/ComplexityFunction.html
>
>
> Regards,
> -- Jean-Marc
>
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