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An infinite sum of the Heaviside function

  1. Mar 25, 2016 #1
    I'm not sure where to put this question. It is by itself pretty basic, but it's a preamble to a Laplace Transform exercise, and I'll probably want to ask some follow up questions once the current query is resolved.

    1. The problem statement, all variables and given/known data


    Unit stair-case function: [itex]f(t) = n, \ if \ \ n-1 \leq t < n, \ \ n = 1,2,3,...[/itex]

    Show that [itex]f(t) = \sum_{n=0}^{\infty} u(t-n) \ [/itex] for all [itex]t \geq 0[/itex]

    2. Relevant equations


    3. The attempt at a solution

    I can see how as we move through the values of n, each unit step function will just be adding one, which just builds the stair-case, but after any given t is reached, won't that sum just keeping adding an infinite number of 1's to result?

    Does the sum to infinity actually stop at some finite point because [itex]n-1 \leq t < n, \ \ n = 1,2,3,...[/itex]? Wouldn't that contradict the idea of having an infinite sum?

    Or does the unit step function used in the sum drop down to zero after a certain n is reached? How would that work?
     
  2. jcsd
  3. Mar 25, 2016 #2
    The unit step function is zero if its argument is negative, and [itex]t - n[/itex] will eventually become negative for sufficiently large [itex]n[/itex].
     
  4. Mar 26, 2016 #3

    Ray Vickson

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    For each finite ##t## there will be only finitely many nonzero terms in the sum. For ##n > t## all the ##u(t-n)## are zero.

    You cannot write the result for ALL ##t## as a single finite sum, since different ##t## need different numbers of terms.
     
    Last edited: Mar 26, 2016
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