Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: Re: Re: May we trust IntegerQ

• To: mathgroup at smc.vnet.net
• Subject: [mg107610] Re: [mg107582] Re: [mg107504] Re: [mg107488] May we trust IntegerQ
• From: Artur <grafix at csl.pl>
• Date: Fri, 19 Feb 2010 03:35:52 -0500 (EST)
• References: <201002181016.FAA28523@smc.vnet.net>
• Reply-to: grafix at csl.pl

Dear Bob,
Yes, I was see that Andrzej Kozlowski procedure work perfectly but 
output is True/False type. I need procedure where output will be exact 
integer number (and inside procedure will be not numerical functions 
like Round[N[]]).
Your procedure realize this aim but only up to indexes 0..7 and not for 
upper 8 and more.
Best wishes
Artur


Bob Hanlon pisze:
> 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
>>>     
>>>       
>
>
>
>   

• References:
  • Re: Re: May we trust IntegerQ ?
    • From: Bob Hanlon <hanlonr@cox.net>