Re: May we trust IntegerQ ?

• To: mathgroup at smc.vnet.net
• Subject: [mg107499] Re: [mg107488] May we trust IntegerQ ?
• From: Artur <grafix at csl.pl>
• Date: Mon, 15 Feb 2010 05:45:33 -0500 (EST)
• References: <20100214094853.SERF5.402884.imail@eastrmwml37>

```Bob,
aa = {}; Do[ If[PossibleZeroQ[ Round[N[ChebyshevT[x/2, x], 100]] -
N[ChebyshevT[x/2, x], 100]], ,
AppendTo[aa, x]], {x, 0, 1500}]; aa
which work always.
Best wishes
Artur

Bob Hanlon pisze:
> As stated in the documentation for IntegerQ:
>
> "IntegerQ[expr] returns False unless expr is manifestly an integer (i.e., has head Integer)."
>
> aa = {}; Do[If[IntegerQ[ChebyshevT[x/2, x]], , AppendTo[aa, x]], {x, 0, 20}];
>  aa
>
> {3,5,7,9,11,13,15,17,19}
>
> Using FunctionExpand enables Mathematica to recognize the values for 7 and 17 as integers.
>
> aa = {}; Do[
>  If[IntegerQ[FunctionExpand[ChebyshevT[x/2, x]]], , AppendTo[aa, x]], {x, 0,
>   20}]; aa
>
> {3,5,9,11,13,15,19}
>
> Oddly, FullSimplify works for 7 but not 17.  I had thought that FullSimplify would always try FunctionExpand.
>
> aa = {}; Do[
>  If[IntegerQ[FullSimplify[ChebyshevT[x/2, x]]], , AppendTo[aa, x]], {x, 0,
>   20}]; aa
>
> {3,5,9,11,13,15,17,19}
>
>
> Bob Hanlon
>
> ---- Artur <grafix at csl.pl> wrote:
>
> =============
> 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