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Re: Re: May we trust IntegerQ ?

• To: mathgroup at smc.vnet.net
• Subject: [mg107536] Re: [mg107504] Re: [mg107488] May we trust IntegerQ ?
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Tue, 16 Feb 2010 03:53:11 -0500 (EST)
• Reply-to: hanlonr at cox.net

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: Andrzej Kozlowski <akozlowski@gmail.com>