Trying to integrate a summation of a unit step function.

In summary, the given conversation discusses the definition of a monotone function on the interval [0,1], defined by a summation of a given function and a piecewise function. The task is to find the integral of this function over the interval [0,1] and leave the answer in the form of an infinite series. The conversation also mentions the theorem stating that a monotone function on [0,1] is Riemann integrable. The attempt at a solution involves calculating the summation under the function and seeking advice for completing the problem.
  • #1
Unassuming
167
0

Homework Statement



Define I(x)= I( x - x_n ) =

{ 0 , when x < x_n
{ 1, when x >= x_n.

Let f be the monotone function on [0,1] defined by

[tex] f(x) = \sum_{n=1}^{\infty} \frac{1}{2^n} I ( x - x_n) [/tex]

where [tex] x_n = \frac {n}{n+1} , n \in \mathbb{N} [/tex].

Find [tex] \int_0^1 f(x) dx [/tex].

Leave your answer in the form of an infinite series.

Homework Equations





The Attempt at a Solution



I know the theorem, if f is monotone on [0,1], then f is Riemann integrable on [0,1]. I am also familiar with what the graph of this function looks like.

My calculation of what is under the function is,

[tex] \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{1}{n^2 +3n +2} [/tex].


I need some advice for completing the problem (assuming that sum is correct).
Thanks
 
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  • #2
Unassuming said:

Homework Statement



Define I(x)= I( x - x_n ) =

{ 0 , when x < x_n
{ 1, when x >= x_n.

Let f be the monotone function on [0,1] defined by

[tex] f(x) = \sum_{n=1}^{\infty} \frac{1}{2^n} I ( x - x_n) [/tex]

where [tex] x_n = \frac {n}{n+1} , n \in \mathbb{N} [/tex].

Find [tex] \int_0^1 f(x) dx [/tex].

Leave your answer in the form of an infinite series.

Homework Equations





The Attempt at a Solution



I know the theorem, if f is monotone on [0,1], then f is Riemann integrable on [0,1]. I am also familiar with what the graph of this function looks like.

My calculation of what is under the function is,

[tex] \sum_{n=1}^{\infty} \frac{1}{2^n} \frac{1}{n^2 +3n +2} [/tex].
If you mean this to be f(x), shouldn't it be a function of x? If you mean it to be the integral, then aren't you done?


I need some advice for completing the problem (assuming that sum is correct).
Thanks
 

Related to Trying to integrate a summation of a unit step function.

1. What is a unit step function?

A unit step function is a mathematical function that has a value of 0 for all negative input values and a value of 1 for all positive input values. It is often represented by the symbol u(t).

2. Why is it important to integrate a summation of a unit step function?

Integrating a summation of a unit step function allows us to find the total area under the curve, which can be useful in various applications such as calculating displacement or finding the total change in a system over a period of time.

3. What is the process for integrating a summation of a unit step function?

The process for integrating a summation of a unit step function involves splitting the function into separate intervals based on where the step function changes value. Then, the integral of each interval can be calculated using basic integration techniques.

4. Can a summation of a unit step function be integrated using software or must it be done by hand?

Integrating a summation of a unit step function can be done using both software and by hand. Many software programs have built-in functions for integrating, but it is also important to understand the process and be able to do it by hand for more complex functions.

5. Are there any specific conditions or restrictions for integrating a summation of a unit step function?

There are no specific conditions or restrictions for integrating a summation of a unit step function. However, it is important to make sure the function is well-defined and continuous in order to accurately integrate it.

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