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Re: May we trust IntegerQ ?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg107500] Re: [mg107488] May we trust IntegerQ ?
  • From: danl at wolfram.com
  • Date: Mon, 15 Feb 2010 05:45:43 -0500 (EST)
  • References: <201002141316.IAA02260@smc.vnet.net>

> 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




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