# An infinite sum of the Heaviside function

• kostoglotov
In summary, the unit stair-case function can be represented as the infinite sum of unit step functions, with a finite number of nonzero terms for each finite value of t. This contradicts the idea of having an infinite sum, but it can be explained by the fact that the unit step function drops down to zero for sufficiently large values of n.
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 $t \geq 0$

## 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?

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

kostoglotov
kostoglotov said:
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 $t \geq 0$

## 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.

Last edited:

## What is the Heaviside function?

The Heaviside function, also known as the unit step function, is a mathematical function that is defined as 0 for negative values and 1 for positive values.

## What is an infinite sum?

An infinite sum is a series of numbers that continues infinitely, without a defined endpoint. In mathematics, it is denoted by the symbol ∑ and is used to find the sum of an infinite number of terms.

## How is the Heaviside function used in an infinite sum?

The Heaviside function is often used in an infinite sum to represent a discontinuous function. It is commonly used in Fourier series and Laplace transforms to simplify the calculations.

## What is the value of an infinite sum of the Heaviside function?

The value of an infinite sum of the Heaviside function depends on the specific series being evaluated. In some cases, the sum may converge to a finite value, while in others it may diverge to infinity.

## What are some real-world applications of an infinite sum of the Heaviside function?

An infinite sum of the Heaviside function has various applications in physics, engineering, and economics. For example, it can be used to model the behavior of electrical circuits, signal processing, and pricing options in finance.

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