Re: May we trust IntegerQ ?
- To: mathgroup at smc.vnet.net
- Subject: [mg107515] Re: May we trust IntegerQ ?
- From: David Bailey <dave at removedbailey.co.uk>
- Date: Mon, 15 Feb 2010 05:48:26 -0500 (EST)
- References: <hl8t3l$26h$1@smc.vnet.net>
Artur 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
> Some comments ?
>
> Best wishes
> Artur
>
>
>
>
IntegerQ tests to see if an expression is a Mathematica integer, but it
doesn't simplify the expression first. Thus ChebyshevT[7/2, 7] evaluates
to Cos[(7*ArcCos[7])/2], which can be resolved to 5040 by using
FullSimplify.
IntegerQ[Cos[(7*ArcCos[7])/2]] correctly returns False.
IntegerQ[FullSimplify[Cos[(7*ArcCos[7])/2]]] returns True.
I guess you could wrap FullSimplify round your ChebyshevT call, but it
is important to bear in mind that FullSimplify attempts to find a
simplification where one exists, but cannot be guaranteed to do so.
A numerical test may be safer.
David Bailey
http://www.dbaileyconsultancy.co.uk