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Dirac Delta/Mean Value Theorem Problem

  1. Mar 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider the function [itex]\delta_{\epsilon}(x)[/itex] defined by
    [tex]\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}[/tex]
    (b) Consider a function [itex]f[/itex], defined on [itex]\mathbb{R}[/itex]. which is continuous. Use
    the Mean Value Theorem of integral calculus to show that
    [tex]\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)[/tex]
    where [itex]-\epsilon\leq\theta\leq\epsilon[/itex].

    2. Relevant equations
    I guess the definition of [itex]\delta_{\epsilon}(x)[/itex] implies [itex]\epsilon>0[/itex], but I will assume this for my answer anyway.

    I will refer to the Mean Value Theorem as MVT.


    3. The attempt at a solution
    Let,
    [tex]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{,}[/tex]
    [itex]=(\epsilon-(-\epsilon))f(\theta)\delta_{\epsilon}(\theta)\text{, }[/itex] [itex]-\epsilon\leq\theta\leq\epsilon[/itex], by Mean Value Theorem,
    [tex]=2\epsilon f(\theta)\frac{3}{4\epsilon^{3}}(\epsilon^{2}-\theta^{2})\\=f(\theta)\frac{3}{2\epsilon^{3}}(\epsilon^{2}-\theta^{2})[/tex]
    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 [itex]\epsilon>0[/itex] is arbitrary let [itex]\epsilon=\sqrt{3}\vert\theta\vert[/itex] ([itex]-\epsilon\leq\theta\leq\epsilon[/itex] is still satisfied), then
    [tex]\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[/tex]
    so,
    [tex]I=\overset{\infty}{\underset{-\infty}{\intop}}f(x)\delta_{\epsilon}(x)dx=f(\theta)\text{,}\quad \epsilon=\sqrt{3}\vert\theta\vert\text{.}[/tex]
    The problem I have now is the [itex]\theta[/itex] we know exists by MVT is dependent on [itex]\epsilon[/itex] we choose. Is it then valid to set [itex]\epsilon[/itex] as a function of [itex]\theta[/itex]?

    Thanks in advance.
     
    Last edited: Mar 8, 2014
  2. jcsd
  3. Mar 9, 2014 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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: Mar 9, 2014
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