       Re: Re: How to declare Integers?

• To: mathgroup at smc.vnet.net
• Subject: [mg13433] Re: [mg13167] Re: How to declare Integers?
• From: "Fred Simons" <wsgbfs at win.tue.nl>
• Date: Thu, 23 Jul 1998 03:33:34 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```With some interest I read the messages on type declaration in
Mathematica. I might have missed some, so if someone else already
posted a similar solution I apologize.

I think that Sean Ross in his last message hits the root of the
desire for type declarations:
>
> Perhaps the term variable "typing" is ill-advised since it sounds
> the same as what is done in C++ or Fortran.  No, what we are after
> is something like this:
>
>
> Declare[symbol,Integer];
>
> Sin[symbol Pi x]
>
> and have it return zero for all x even with no explicit value
> assigned to symbol.
>
> I want to be able to tell mathematica that a certain symbol is Real,
> Complex, Imaginary, greater than 2, Integer etc. and have every
> single function in the language react appropriately taking that
> declaration as an assumption.  (...)

I have no idea how realistic this last statement is. Can anyone give a
complete list of results he wants to have when a variable has to be
declared of integer type? Usually it is only a much smaller problem,
like Sean Ross's problem with the trigonometric functions, that we are
interested in. I have often met that problem when computing Fourier
coefficients with Mathematica in a traditional pen and paper course on
Fourier series. It can be solved in a straightforward way.

Of course I will declare a variable n to be an integer by the  statement
IntegerQ[n] ^= True. Unfortunately, IntegerQ does not know (why not?)
that the sum, the product and positive powers of integers are integers
so I extend IntegerQ in the following way:

Unprotect[IntegerQ];
IntegerQ[ a_?IntegerQ +b_?IntegerQ] = True; IntegerQ[ a_?IntegerQ *
b_?IntegerQ] = True; IntegerQ[a_?IntegerQ ^ b_Integer?Positive] = True;
Protect[IntegerQ];

Then the following definitions are entered:

Unprotect[ Sin, Cos, Exp];
Sin[ n_?IntegerQ Pi] = 0;
Cos[ n_?IntegerQ Pi] := (-1)^n;
Exp[ n_?IntegerQ Pi Complex[a_, b_]] := Exp[n a Pi] (-1)^(b n); Protect[
Sin, Cos, Exp ];

Having done this, we adapt the function (-1)^n. The first rule states
that for integer a, n and natural p we have (-1)^( a n^p) = (-1)^ (a
n); the second rule that for integer a and n we have (-1)^(a n) = 1
when a is even and (-1)^n when a is odd. The last rule reduces for
example (-1)^(n+3) to -(-1)^n when n is an integer.

Unprotect[Power];
(-1)^((a_Integer:1) b_?IntegerQ^c_Integer?Positive +d_.) :=
(-1)^(a b + d);
(-1)^(a_Integer n_?IntegerQ + b_.) := (-1)^(Mod[a,2]n + b);
(-1)^(a_Integer + b_) := (-1)^a (-1)^b;  Protect[Power];

For me, this works very satisfactory. You could try the following
examples:

IntegerQ[n] ^= True
(-1)^(4 n^3 + 3n+7 )
Cos[ (m + 3n + 7)Pi ] // ExpandAll // TrigExpand Integrate[x  Exp[ I n x
], {x, 0, Pi}]

Fred Simons
Eindhoven University of Technology

```

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