Indeterminate result from Sinc

Question

I have an impulse train given by

(1 + Csc[(π x)/(1 + R)] Sin[(π (1 + 2 R) x)/(1 + R)])/(2 + 2 R)

Because of the Csc, evaluating this expression gives an indeterminate result for all integer multiples of R + 1, as you can see from this table:

TableForm[
  Table[
   Evaluate[(1 + Csc[(π x)/(1 + R)] Sin[(π (1 + 2 R) x)/(1 + R)])/(2 + 2 R)],
   {x, 0, 10}, {R, 1, 5}]]

(Apologies if my formatting is off; total newbie...)

Fair enough: the expression is defined in the limit, but not absolutely.

So, following advice from a previous similar question (here), I replace Sin with Sinc:

((1 + Csc[(π x)/(1 + R)] Sin[(π (1 + 2 R) x)/(1 + R)])/(2 + 2 R) // FullSimplify) 
  /. {Sin[z_] :> z*Sinc[z], Csc[z_] :> 1/(z*Sinc[z])} // Simplify

This gives me

(1 + ((1 + 2 R) Sinc[(π (1 + 2 R) x)/(1 + R)])/Sinc[(π x)/(1 + R)])/(2 + 2 R)

And this is where I get confused. The substitution should remove all indeterminate results, but it doesn't. Instead, it only removes the first indeterminate result:

TableForm[
  Table[
    Evaluate[{(1 + ((1 + 2 R) Sinc[(π (1 + 2 R) x)/(1 + R)])/Sinc[(π x)/(1 + R)]) / 
               (2 + 2 R)}],
    {x, 0, 10}, {R, 1, 5}]]

This is baffling, since the function now contains nothing but Sinc.

How do I fix this? Or am I doing something wrong mathematically?

Answer 1

You've avoided Sin[0]/0 problems, but your numerator and denominator in your more complicated expression are still both zero in some cases. One way is to take a limit in those cases:

f[xx_, RR_] := Module[{expr, trial, x, R},
  expr = (1 + ((1 + 2 R) Sinc[(\[Pi] (1 + 2 R) x)/(1 + R)])/
         Sinc[(\[Pi] x)/(1 + R)])/(2 + 2 R);
  trial = Quiet[expr /. x -> xx /. R -> RR];
  If[trial === Indeterminate, Limit[expr /. R -> RR, x -> xx], trial]]

Table[f[x, R], {x, 0, 10}, {R, 1, 5}]
(*
{{1, 1, 1, 1, 1}, {0, 0, 0, 0, 0}, {1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, 
 {1, 0, 1, 0, 0}, {0, 0, 0, 1, 0}, {1, 1, 0, 0, 1}, {0, 0, 0, 0, 0}, 
 {1, 0, 1, 0, 0}, {0, 1, 0, 0, 0}, {1, 0, 0, 1, 0}}
*)

Edit to answer questions in comments:

You can take the limit for all cases if you like.

g[xx_, R_] := 
   Module[{x}, 
      Limit[(1 + ((1 + 2 R) Sinc[(\[Pi] (1 + 2 R) x)/(1 + R)])/
         Sinc[(\[Pi] x)/(1 + R)])/(2 + 2 R), x -> xx]]

This is, however, a bit slower than the previous definition.

I call the argument pattern xx to distinguish it from the free variable x. After all, Limit[...,x->x] makes no sense. GG isn't necessary in the final code, but I was experimenting with which variable Limit preferred. Note that I localized x with Module to insure that it wouldn't conflict with either a potential global value nor a local value created by Table or something. Mathematica is an expression rewriting language: x needs to stay undefined through all the rewriting to allow Limit to use it properly.