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Re: Re: May we trust IntegerQ ?

• To: mathgroup at smc.vnet.net
• Subject: [mg107582] Re: [mg107504] Re: [mg107488] May we trust IntegerQ ?
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 18 Feb 2010 05:16:21 -0500 (EST)
• Reply-to: hanlonr at cox.net

As recommended by Andrzej Kozlowski in this thread, use Element to test

And @@ Table[
  FullSimplify[
   Element[
    (-1/2 (x - Sqrt[-1 + x^2])^(2 x) + 1/2 (x + Sqrt[-1 + x^2])^(2 x))^2,
    Integers]],
  {x, 0, 20}]

True


Bob Hanlon

---- Artur <grafix at csl.pl> wrote: 

=============
Dear Bob,
Your procedure do full simplify only up to index x=7 (and not from 8 and up)

Table[FullSimplify[(-1/2 (x - Sqrt[-1 + x^2])^(2 x) +
     1/2 (x + Sqrt[-1 + x^2])^(2 x))^2], {x, 0, 10}]

Do you know what to do more ?
Best wishes
Artur

Bob Hanlon pisze:
> Use FullSimplify in this case
>
> Table[IntegerQ[
>   FullSimplify[
>    1/2 (x - Sqrt[-1 + x^2])^(2 x) + 1/2 (x + Sqrt[-1 + x^2])^(2 x)]], {x, 0, 
>   10}]
>
> {True,True,True,True,True,True,True,True,True,True,True}
>
>
> Bob Hanlon
>
> ---- Artur <grafix at csl.pl> wrote: 
>
> =============
> Dear Daniel,
> What to do in following case:
> Table[IntegerQ[FunctionExpand[1/2 (x - Sqrt[-1 + x^2])^(2 x) + 1/2 (x + 
> Sqrt[-1 + x^2])^(2 x)]], {x, 0, 10}]
> Best wishes
> Artur
>
>
> danl at wolfram.com pisze:
>   
>>> Procedure: find such x that ChebyshevT[x/2, x] isn't integer
>>> aa = {}; Do[ If[IntegerQ[ChebyshevT[x/2, x]], , AppendTo[aa, x]], {x, 0,
>>> 20}]; aa
>>> and answer Mathematica is set:
>>> {3, 5, 7, 9, 11, 13, 15, 17, 19}
>>> where occered e.g. number 7
>>> N[ChebyshevT[7/2, 7],100]
>>> 5042.00000000000000000000000000000000000000000000000000000000000000000\
>>> 0000000000000000000000000000000
>>> evidently is integer 5042
>>> Some comments ?
>>>
>>> Best wishes
>>> Artur
>>>     
>>>       
>> Trust IntegerQ? Mais oui.
>>
>> Documentation Center entry for IntegerQ, first item under More Information:
>>
>> "IntegerQ[expr] returns False unless expr is manifestly an integer (i.e.,
>> has head Integer)."
>>
>> Example:
>> In[16]:= IntegerQ[Sin[3]^2 + Cos[3]^2]
>> Out[16]= False
>>
>> For your example, it might be better to use
>> FunctionExpand[ChebyshevT[x/2,x]]. Then try to figure out which cases
>> involving half-integer powers (Puiseux polynomials) will still evaluate to
>> integers.
>>
>> Daniel Lichtblau
>> Wolfram Research
>>     

• Follow-Ups:
  • Re: Re: Re: May we trust IntegerQ
    • From: Artur <grafix@csl.pl>