# Dirac Delta/Mean Value Theorem Problem

• EdMel
In summary, the conversation discusses the function \delta_{\epsilon}(x) and its properties, as well as using the Mean Value Theorem to prove an integral equation involving the function. The attempted solution uses the MVT, but may not be using the most appropriate version.
EdMel

## Homework Statement

Consider the function $\delta_{\epsilon}(x)$ defined by
$$\delta_{\epsilon}(x)=\begin{cases} 0\text{,} & x<-\epsilon\text{,}\\ \frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} & \epsilon\leq x\leq\epsilon\text{,}\\ 0\text{,} & \epsilon<x\text{,} \end{cases}$$
(b) Consider a function $f$, defined on $\mathbb{R}$. which is continuous. Use
the Mean Value Theorem of integral calculus to show that
$$\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)$$
where $-\epsilon\leq\theta\leq\epsilon$.

## Homework Equations

I guess the definition of $\delta_{\epsilon}(x)$ implies $\epsilon>0$, but I will assume this for my answer anyway.

I will refer to the Mean Value Theorem as MVT.

## The Attempt at a Solution

Let,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=\overset{\epsilon}{\underset{-\epsilon}{\intop}}f(x)\delta_{\epsilon}(x)dx\quad\text{, as }\delta_{\epsilon}(x)=0\quad\forall\quad\vert x\vert\geq\epsilon\text{,}$$
$=(\epsilon-(-\epsilon))f(\theta)\delta_{\epsilon}(\theta)\text{, }$ $-\epsilon\leq\theta\leq\epsilon$, by Mean Value Theorem,
$$=2\epsilon f(\theta)\frac{3}{4\epsilon^{3}}(\epsilon^{2}-\theta^{2})\\=f(\theta)\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})$$
Then I am not sure how to proceed. I thought maybe there would be another application of MVT. The best argument I can come up with continues as ...
as $\epsilon>0$ is arbitrary let $\epsilon=\sqrt{3}\vert\theta\vert$ ($-\epsilon\leq\theta\leq\epsilon$ is still satisfied), then
$$\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})=\frac{3}{2(\sqrt{3}\vert\theta\vert)^{3}}((\sqrt{3}\vert \theta\vert)^{2}-\theta^{2})=1$$
so,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)\text{,}\quad \epsilon=\sqrt{3}\vert\theta\vert\text{.}$$
The problem I have now is the $\theta$ we know exists by MVT is dependent on $\epsilon$ we choose. Is it then valid to set $\epsilon$ as a function of $\theta$?

Last edited:
EdMel said:

## Homework Statement

Consider the function $\delta_{\epsilon}(x)$ defined by
$$\delta_{\epsilon}(x)=\begin{cases} 0\text{,} & x<-\epsilon\text{,}\\ \frac{3}{4\epsilon^{3}}(\epsilon^{2}-x^{2})\text{,} & \epsilon\leq x\leq\epsilon\text{,}\\ 0\text{,} & \epsilon<x\text{,} \end{cases}$$
(b) Consider a function $f$, defined on $\mathbb{R}$. which is continuous. Use
the Mean Value Theorem of integral calculus to show that
$$\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)$$
where $-\epsilon\leq\theta\leq\epsilon$.

## Homework Equations

I guess the definition of $\delta_{\epsilon}(x)$ implies $\epsilon>0$, but I will assume this for my answer anyway.

I will refer to the Mean Value Theorem as MVT.

## The Attempt at a Solution

Let,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=\overset{\epsilon}{\underset{-\epsilon}{\intop}}f(x)\delta_{\epsilon}(x)dx\quad\text{, as }\delta_{\epsilon}(x)=0\quad\forall\quad\vert x\vert\geq\epsilon\text{,}$$
$=(\epsilon-(-\epsilon))f(\theta)\delta_{\epsilon}(\theta)\text{, }$ $-\epsilon\leq\theta\leq\epsilon$, by Mean Value Theorem,
$$=2\epsilon f(\theta)\frac{3}{4\epsilon^{3}}(\epsilon^{2}-\theta^{2})\\=f(\theta)\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})$$
Then I am not sure how to proceed. I thought maybe there would be another application of MVT. The best argument I can come up with continues as ...
as $\epsilon>0$ is arbitrary let $\epsilon=\sqrt{3}\vert\theta\vert$ ($-\epsilon\leq\theta\leq\epsilon$ is still satisfied), then
$$\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})=\frac{3}{2(\sqrt{3}\vert\theta\vert)^{3}}((\sqrt{3}\vert \theta\vert)^{2}-\theta^{2})=1$$
so,
$$I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)\text{,}\quad \epsilon=\sqrt{3}\vert\theta\vert\text{.}$$
The problem I have now is the $\theta$ we know exists by MVT is dependent on $\epsilon$ we choose. Is it then valid to set $\epsilon$ as a function of $\theta$?

I think you aren't using the version of the MVT best suited to this problem. See http://en.wikipedia.org/wiki/Mean_value_theorem#First_mean_value_theorem_for_integration

Last edited:

## What is the Dirac Delta function?

The Dirac Delta function, denoted by δ(x), is a mathematical function that is defined as 0 everywhere except at x = 0, where it has an infinite value, and has an integral of 1. It is often used in physics and engineering to represent a point of concentration or impulse.

## What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that for a continuous and differentiable function f(x) on the closed interval [a,b], there exists a point c in the interval such that the slope of the tangent line at c is equal to the average rate of change of f(x) over the interval [a,b]. This theorem is useful for finding the average rate of change of a function over a certain interval.

## How are the Dirac Delta function and the Mean Value Theorem related?

The Dirac Delta function and the Mean Value Theorem are related through the concept of a delta function derivative. The delta function derivative of a function f(x) is defined as the limit of f(x)/δ(x) as δ(x) approaches 0. This limit is equivalent to the derivative of f(x) evaluated at the point where the delta function is concentrated, which is similar to the point c in the Mean Value Theorem.

## What is the significance of the Dirac Delta function and the Mean Value Theorem in physics?

In physics, the Dirac Delta function is often used to represent a point source or impulse, such as a point charge or a point mass. The Mean Value Theorem is used to analyze the average behavior of a system over a certain interval, which can be useful in understanding the behavior of physical systems. Furthermore, the delta function derivative is used in quantum mechanics to describe the state of a system at a specific point in time.

## What are some practical applications of the Dirac Delta function and the Mean Value Theorem?

The Dirac Delta function and the Mean Value Theorem have various practical applications in fields such as engineering, physics, and economics. For example, the Dirac Delta function is used to model electrical circuits, while the Mean Value Theorem is used in optimization problems and in finding average rates of change. Additionally, both concepts are used in signal processing and data analysis to analyze and manipulate signals and data.

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