The expression

h[n_] = -((I^-n (-1 + I^n)^2)/(n^2 π^2)) 

is real for all integers n. Although indeterminate at n = 0, Mathematica computes the limit to be 1/4 as n -> 0.

I'd like to use the expression in DiscreteConvolve, but even when I first declare h[0] = 1/4, a computation like

y[0] = DiscreteConvolve[h, DiscreteDelta[n], n, m] 

yields only h and its explicit infinite summation representation results in "Indeterminate".

How do I make operations with h usable?

  • 1
    $\begingroup$ I do not understand your code. You write h[n_] then use it as just h with no arguments. ? DiscreteConvolve[h,DiscreteDelta[n],n,m] try it with h[n], also do not understand why you use = and not := in the definition. $\endgroup$ – Nasser May 31 '14 at 17:15

I think you are trying to do the following:

h[n_] /; n >= 1 := -((I^-n (-1 + I^n)^2)/(n^2 Pi^2))

h[0] = 1/4;

y[m_] := DiscreteConvolve[h[n], DiscreteDelta[n], n, m]


(* ==> 1/4 *)

Here corrected your definition of y so it uses m as the function argument, but the important part is to understand why even with this correction your code didn't work:

The definition of h[n_] would be superseded by the specific definition h[0] if you directly asked for h[0]. However, when passed to DiscreteConvolve as an argument in the form h[n], this specific case is not used and instead h[n] is expanded to the generic expression (which is indeterminate at n=0). The remaining calculation then never knows that the expression came from h in the first place. This happens because DiscreteConvolve doesn't hold its arguments unevaluated by default.

So you have to make sure this initial expansion of h[n] doesn't occur. I do this above by adding a condition ;n>=1 to the left-hand side of the line defining h[n_]. This makes it clear that this definition is not to be used for all n. Then the generic term h[n] remains un-expanded when it is passed to DiscreteConvolve, giving it a chance to see that there is a different definition for h[0].


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.