Re: May we trust IntegerQ ?
- To: mathgroup at smc.vnet.net
- Subject: [mg107504] Re: [mg107488] May we trust IntegerQ ?
- From: Artur <grafix at csl.pl>
- Date: Mon, 15 Feb 2010 05:46:27 -0500 (EST)
- References: <201002141316.IAA02260@smc.vnet.net> <49834.204.180.236.11.1266171168.squirrel@webmail.wolfram.com>
- Reply-to: grafix at csl.pl
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:
- May we trust IntegerQ ?
- From: Artur <grafix@csl.pl>
- May we trust IntegerQ ?