An infinite sum of the Heaviside function

[kostoglotov]

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. Homework Statement

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

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

Homework Equations

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 $n-1 \leq t < n, \ \ n = 1,2,3,...$? 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?

 

[Nono713]

The unit step function is zero if its argument is negative, and $t - n$ will eventually become negative for sufficiently large $n$.

 

[Ray Vickson]

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. Homework Statement

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

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

Homework Equations

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 $n-1 \leq t < n, \ \ n = 1,2,3,...$? 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?

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.