# Dirac Delta using periodic functions

• SisypheanZealot
In summary: So, we have finally solved our original problem.In summary, the problem is that in order to find the function f(y+2\pi m) from the equation \int dx f(x)\delta(x-y) we need to solve for c_n, and then multiply by f(y+2\pi m).
SisypheanZealot
Homework Statement
Exercise1.3: Using the periodic form (1.14) for the Dirac delta, show
$$\int^{\infty}_{-\infty}dx\:f(x)\delta(x-y)=\sum^{\infty}_{n=-\infty}f(y+2nL)$$
Relevant Equations
$$\delta(x-y)=\frac{1}{2L}\sum^{\infty}_{n=-\infty}\lbrace sin(\frac{n\pi x}{L})sin(\frac{n\pi y}{L})+cos(\frac{n\pi x}{L})cos(\frac{n\pi y}{L})\rbrace$$
I know it is something simple that I am missing, but for the life of me I am stuck. So, I used the identity ##sin(a)sin(b)+cos(a)cos(b)=cos(a-b)## which gives me $$\int^{\infty}_{-\infty}dx\:f(x)\delta(x-y)=\int^{\infty}_{-\infty}dx\:f(x)\frac{1}{2L}\sum^{\infty}_{n=-\infty}\lbrace sin(\frac{n\pi x}{L})sin(\frac{n\pi y}{L})+cos(\frac{n\pi x}{L})cos(\frac{n\pi y}{L})\rbrace$$ $$=\int^{\infty}_{-\infty}dx\:f(x)\frac{1}{2L}\sum^{\infty}_{n=-\infty}cos(\frac{n\pi}{L}(x-y))$$ Then I performed the transformation ##u=\frac{\pi}{L}(x-y)## giving $$=\frac{1}{2L}\sum^{\infty}_{n=-\infty}\frac{L}{\pi}\int^{\infty}_{-\infty}du\:f(u\frac{L}{\pi}+y)cos(u\cdot n)$$ $$=\int^{\infty}_{-\infty}du\:f(u\frac{L}{\pi}+y)\frac{1}{2\pi}\sum^{\infty}_{n=-\infty}cos(u\cdot n)$$ $$\int^{\infty}_{-\infty}du\:f(u\frac{L}{\pi}+y)\lbrace\frac{1}{2\pi}+\frac{1}{\pi}\sum^{\infty}_{n=1}cos(u\cdot n)\rbrace$$ Now we have ##u=2\pi## as the solutions for ##cos(u\cdot n)=1## which to me looks like it gives ##f(2L+y)##. What am I not seeing?

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New thought. What if I work backwards on this. so start with $$\sum^{\infty}_{n=-\infty}f(2nL+y)$$ Then set this equal to $$\int dx f(x)\sum^{\infty}_{n=-\infty}\delta(x-(2nL+y))$$ Now look at the delta term as a periodic term, and look at the Fourier transform definition of the Dirac Delta $$\delta(x-y)=\int\frac{dz}{2\pi}e^{iz(x-y)}$$ The Fourier will be invariant under rotation of multiples of ##2\pi##. So then, I can rewrite the integral as $$\int\frac{dz}{2\pi}e^{iz(x-y)}e^{-i2n\pi}$$ This will then give the new definition of the delta function $$\sum^{\infty}_{n=-\infty}\delta(x-(2n\pi-y))=\sum^{\infty}_{n=-\infty}\int\frac{dz}{2\pi}e^{iz(x-y)}e^{-i2n\pi}$$. But, then I need to worry about the modes of the system. I don't know if this moved me forward, or just in circles.

So, for the interested. I think I finally got this thing figured out. The first thing that we are going to do is transition to the complex Fourier series definitions of ##f(x)## and ##\delta(x-y)## and set ##L=\pi##. i.e. $$f(x)=\sum_{n=-\infty}^{\infty}c_{n}e^{inx}$$ and $$\delta(x-y)=\int\frac{dz}{2\pi}e^{iz(x-y)}$$ So that the formula takes the form $$\int dx f(x)\delta(x-y)=\int dx \sum_{n=-\infty}^{\infty}c_{n}e^{inx}\int\frac{dz}{2\pi}e^{iz(x-y)}$$ then doing some algebra we get $$\int dx \sum_{n=-\infty}^{\infty}c_{n}\int\frac{dz}{2\pi}e^{i[z(x-y)+nx]}$$ performing the integration we then get $$\sum_{n=-\infty}^{\infty}c_{n}\int dx\frac{e^{i[z(x-y)+nx]}}{2\pi i(x-y)}$$ Now we apply Cauchy's formula ##\oint\frac{g(w)}{w-w_0}dz=2\pi ig(w_0)## to this. Doing so we notice that the translations are ##w=x,w_0=y##, and that there is a pole at ##x=y##. So, the problem becomes $$\sum_{n=-\infty}^{\infty}c_{n}\int dx\frac{e^{i[z(x-y)+nx]}}{2\pi i(x-y)}=\sum_{n=-\infty}^{\infty}c_{n}e^{i[z(y-y)+ny]}$$ Which then gives us the common result of the original problem $$\int dx f(x)\delta(x-y)=\sum_{n=-\infty}^{\infty}c_{n}e^{iny}=f(y)$$ Now, to progress from here we need to introduce yet another definition for the Dirac delta ##\delta(n)=\sum^{\infty}_{m=-\infty}e^{i2\pi nm}## we now multiple this by the new Fourier series giving $$\sum_{n=-\infty}^{\infty}c_{n}e^{iny}\sum^{\infty}_{m=-\infty}e^{i2\pi nm}=\sum^{\infty}_{m=-\infty}\sum_{n=-\infty}^{\infty}c_{n}e^{in(y+2\pi m)}=\sum^{\infty}_{m=-\infty}f(y+2\pi m)$$

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Abhishek11235

## 1. What is the Dirac Delta function?

The Dirac Delta function, denoted by δ(x), is a mathematical function that is defined as zero for all values of x except at x=0, where it is infinite. It is often referred to as a "function" because it is used to represent a point in a function, even though it is not a true function in the traditional sense.

## 2. How is the Dirac Delta function related to periodic functions?

The Dirac Delta function can be expressed as a periodic function, specifically as the sum of an infinite series of periodic functions. This is because the Dirac Delta function has a repeating pattern of infinite height at x=0, which is similar to the behavior of periodic functions.

## 3. What is the significance of using periodic functions to represent the Dirac Delta function?

Using periodic functions to represent the Dirac Delta function allows for a more intuitive understanding of its properties and behavior. It also allows for easier manipulation and analysis of the Dirac Delta function in mathematical equations.

## 4. How is the Dirac Delta function used in physics and engineering?

The Dirac Delta function is commonly used in physics and engineering to represent point-like objects or events, such as point charges in electromagnetism or point masses in mechanics. It is also used in signal processing and control systems to model impulses or sudden changes in a system.

## 5. Can the Dirac Delta function be used in practical applications?

Yes, the Dirac Delta function has many practical applications in various fields, including signal processing, control systems, and quantum mechanics. It is a powerful mathematical tool that allows for the precise representation and analysis of point-like phenomena in various systems.

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